Chapter 17: Electrostatics - PDF

– + – Electrostatics 1 + – + – – E3 60 + Electric Charge. electrical and magnetic effects. ID (1) Definition : Charge is the property associated with matter due to which it produces and experiences (2) Origin of electric charge : It is known that every atom is electrically neutral, containing as man...

– + – Electrostatics 1 + – + – – E3 60 + Electric Charge. electrical and magnetic effects. ID (1) Definition : Charge is the property associated with matter due to which it produces and experiences (2) Origin of electric charge : It is known that every atom is electrically neutral, containing as many U electrons as the number of protons in the nucleus. Charged particles can be created by disturbing neutrality of an atom. Loss of electrons gives positive D YG charge (as then np > ne) and gain of electrons gives negative charge (as then ne > np) to a particle. When an object is negatively charged it gains electrons and therefore its mass increases negligibly. Similarly, on charging a body with positive electricity its mass decreases. Change in mass of object is equal to n  me. Where, n is the + ST Electron < Proton Electron = Proton Mass M Mass M U number of electrons transferred and m e is the mass of electron  9.1  10 31 Kg. – Neutral Positively charged M < M (3) Type : There exists two types of charges in nature (i) Positive charge Electron > Proton Negatively charged M > M (ii) Negative charge 2 Electrostatics Charges with the same electrical sign repel each other, and charges with opposite electrical sign attract each + + – – – E3 + 60 other. dQ  dt dQ  idt , hence S.I. unit of charge is – Ampere  sec = coulomb (C), smaller S.I. units are mC, C, nC ID (4) Unit and dimensional formula : Rate of flow of electric charge is called electric current i.e., i  Benjamin Franklin was the first to assign positive and negative sign of charge. D YG Note :  U (1mC  10 3 C, 1 C  10 6 C, 1nC  10 9 C). C.G.S. unit of charge is – Stat coulomb or e.s.u. Electromagnetic unit 1 of charge is – ab coulomb 1C  3  10 9 stat coulomb  ab coulomb. Dimensional formula [Q]  AT  10  The existence of two type of charges was discovered by Dufog.  Franklin ( i.e. , e.s.u. of charge) is the smallest unit of charge while faraday is largest (1 Faraday = 96500 C ). U  The e.s.u. of charge is also called stat coulomb or Franklin ( Fr) and is related to e.m.u. of charge emu of charge  3  10 10 esu of charge ST through the relation (5) Point charge : A finite size body may behave like a point charge if it produces an inverse square electric field. For example an isolated charged sphere behave like a point charge at very large distance as well as very small distance close to it’s surface. (6) Properties of charge (i) Charge is transferable : If a charged body is put in contact with an uncharged body, uncharged body becomes charged due to transfer of electrons from one body to the other. Electrostatics 3 (ii) Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. 60 (iii) Charge is conserved : Charge can neither be created nor be destroyed. e.g. In radioactive decay the uranium nucleus (charge   92 e ) is converted into a thorium nucleus (charge  90 e ) and emits an  -particle (charge  2e ) 238 90 Th 234  2 He 4. Thus the total charge is  92 e both before and after the decay. E3 92 U (iv) Invariance of charge : The numerical value of an elementary charge is independent of velocity. It is proved ID by the fact that an atom is neutral. The difference in masses on an electron and a proton suggests that electrons move much faster in an atom than protons. If the charges were dependent on velocity, the neutrality of atoms would be violated. U (v) Charge produces electric field and magnetic field : A charged particle at rest produces only electric field D YG in the space surrounding it. However, if the charged particle is in unaccelerated motion it produces both electric and magnetic fields. And if the motion of charged particle is accelerated it not only produces electric and magnetic fields but also radiates energy in the space surrounding the charge in the form of electromagnetic waves.  + v 0 +  E  v = constant +   E , B and Radiates energy   E and B but no Radiation ST U  v  constant (vi) Charge resides on the surface of conductor : Charge resides on the outer surface of a conductor because like charges repel and try to get as far away as possible from one another and stay at the farthest distance from each other which is outer surface of the conductor. This is why a solid and hollow conducting sphere of same outer radius will hold maximum equal charge and a soap bubble expands on charging. (vii) Charge leaks from sharp points : In case of conducting body no doubt charge resides on its outer surface, if surface is uniform the charge distributes uniformly on the surface and for irregular surface the + + + + + + + + + + + + + + + + + + + 4 Electrostatics distribution of charge, i.e., charge density is not uniform. It is maximum where the radius of curvature is ID E3 60 minimum and vice versa. i.e., σ  1 /R . This is why charge leaks from sharp points. (viii) Quantization of charge : When a physical quantity can have only discrete values rather than any value, the quantity is said to be quantised. The smallest charge that can exist in nature is the charge of an electron. If U the charge of an electron (  1.6  10 19 C ) is taken as elementary unit i.e. quanta of charge the charge on any body will be some integral multiple of e i.e., D YG Q   ne with n  1, 2, 3.... Charge on a body can never be  Note :  2 e ,  17.2e or  10 5 e etc. 3 Recently it has been discovered that elementary particles such as proton or neutron are composed of quarks having charge  1 / 3  e and  2 / 3  e. However, as quarks do not exist in free U state, the quanta of charge is still e.  Quantization of charge implies that there is a maximum permissible magnitude of charge. ST Comparison of Charge and Mass. We are familiar with role of mass in gravitation, and we have just studied some features of electric charge. We can compare the two as shown below Charge Mass (1) Electric charge can be positive, negative or zero. (1) Mass of a body is a positive quantity. (2) Charge carried by a body does not depend upon (2) Mass of a body increases with its velocity as velocity of the body. Electrostatics 5 m m0 where c is velocity of light in 1  v2 / c2 vacuum, m is the mass of the body moving with 60 velocity v and m 0 is rest mass of the body. (3) Charge is quantized. (3) The quantization of mass is yet to be established. (4) Electric charge is always conserved. (4) Mass is not conserved as it can be changed into (5) Force between charges can be attractive or repulsive, (5) The gravitational force between two masses is always attractive. ID according as charges are unlike or like charges. E3 energy and vice-versa. Methods of Charging. U A body can be charged by following methods : (1) By friction : In friction when two bodies are rubbed together, electrons are transferred from one body to D YG the other. As a result of this one body becomes positively charged while the other negatively charged, e.g., when a glass rod is rubbed with silk, the rod becomes positively charged while the silk negatively. However, ebonite on rubbing with wool becomes negatively charged making the wool positively charged. Clouds also become charged by friction. In charging by friction in accordance with conservation of charge, both positive and negative charges in equal amounts appear simultaneously due to transfer of electrons from one body to the U other. (2) By electrostatic induction : If a charged body is brought near an uncharged body, the charged body will ST attract opposite charge and repel similar charge present in the uncharged body. As a result of this one side of neutral body (closer to charged body) becomes oppositely charged while the other is similarly charged. This process is called electrostatic induction. Q +Q + + + + + ++ + + + + – – – – – – – + + + + + +Q + + + + +  + + + + + + – – – – – +Q – – + + + + +  + + + + + + – – – – – – –  – – – – – – ––– – – – – – – – 6 Electrostatics Inducting body neither gains nor loses charge. 60 Note :   Induced charge can be lesser or equal to inducing charge (but never greater) and its maximum E3 1  value is given by Q'  Q 1   where Q is the inducing charge and K is the dielectric constant K  of the material of the uncharged body. Dielectric constant of different media are shown below K ID Medium 1 Water 80 Mica 6 Glass 5–10 Metal  D YG U Vacuum / air  Dielectric constant of an insulator can not be   For metals in electrostatics K   and so Q'   Q; i.e. in metals induced charge is equal and U opposite to inducing charge. ST (3) Charging by conduction : Take two conductors, one charged and other uncharged. Bring the conductors in contact with each other. The charge (whether  ve or  ve ) under its own repulsion will spread over both the conductors. Thus the conductors will be charged with the same sign. This is called as charging by conduction (through contact). + + + + + + + + + + + + + + Uncharged Charged  + + + + + + + + + + + + + + + + + + + + + Bodies in contact  + + + + + + + + + + + + + + + + + + + + + + + + + + + + Both are positively charged Note :  60 Electrostatics 7 A truck carrying explosives has a metal chain touching the ground, to conduct away the charge E3 produced by friction. Electroscope. It is a simple apparatus with which the presence of electric charge on a body is detected (see figure). When ID metal knob is touched with a charged body, some charge is transferred to the gold leaves, which then diverges due to repulsion. The separation gives a rough idea of the amount of charge on the body. If a charged body brought near a charged electroscope the leaves will further diverge. If the charge on body is similar to that on U electroscope and will usually converge if opposite. If the induction effect is strong enough leaves after converging D YG may again diverge. (1) Uncharged electroscope + + + + + + + + + + + + + + + + U + – – – + + D (A) + + + – + – – – + + + + + + – C (B) – – – – – – – – – –– – – + – – + – – + – – – D (C) Charging by conduction ST Charging by conduction + (2) Charged electroscope + + + + (A) + + – – – + + + + + D + – (B) – – – – – – – – – – –– – – – – – – – – – – D – + + + + – (C) + + + – –– – – + – – – – + + + + C + E3 Concepts 60 8 Electrostatics  After earthing a positively charged conductor electrons flow from earth to conductor and if a negatively charged conductor is earthed then electrons flows from conductor to earth. + + + + + + e – – – + – – – – – – e– D YG U + + ID + +Qinsulated conductor and connected if through a fine conducting  When a charged spherical conductor placed inside a hollow U wire the charge will be completely transferred from the inner conductor to the outer conductor.  Lightening-rods arrestors are made up of conductors with one of their ends earthed while the other sharp, and protects a ST building from lightening either by neutralising or conducting the charge of the cloud to the ground.  With rise in temperature dielectric constant of liquid decreases.  Induction takes place only in bodies (either conducting or non-conducting) and not in particles.  If X-rays are incident on a charged electroscope, due to ionisation of air by X-rays the electroscope will get discharged and hence its leaves will collapse. However, if the electroscope is evacuated. X-rays will cause photoelectric effect with gold and so the leaves will further diverge if it is positively charged (or uncharged) and will converge if it is negatively charged.  If only one charge is available than by repeating the induction process, it can be used to obtain a charge many times greater than it’s equilibrium. (High voltage generator) + + + + + + + + + + Examples based on properties of charge Electrostatics 9 A soap bubble is given negative charge. Its radius will (a) Increase Solution: (a) [DCE 2000; RPMT 1997; CPMT 1997; MNR 1988] (b) Decrease (c) Remain unchanged (d) Fluctuate 60 Example: 1 Due to repulsive force. Example: 2 Which of the following charge is not possible Solution: (c) (b) 1.6  10 19 C 1.6  10 20 C, because this is (c) 1.6  10 20 C (d) None of these E3 (a) 1.6  10 18 C 1 of electronic charge and hence not an integral multiple. 10 Example: 3 Five balls numbered 1 to 5 balls suspended using separate threads. Pair (1,2), (2,4) and (4,1) show electrostatic [NCERT 1980] (a) Positively charged (d) Made of metal ID attraction, while pair (2,3) and (4,5) show repulsion. Therefore ball 1 must be (b) Negatively charged (c) Neutral Solution: (c) Since 1 does not enter the list of repulsion, it is just possible that it may not be having any charge. Moreover, U since ball no. 1 is being attracted by 2 and 4 both. So 2 and 4 must be similarly charged, but it is also given that 2 and 4 also attract each other. So 2 and 4 are certainly oppositely charged. D YG Since 1 is attracting 2, either 1 or 2 must be neutral but since 2 is already in the list of balls repelling each other, it necessarily has some charge, similarly 4 must have some charge. It means that though 1 is attracting 2 and 4 it does not have any charge. Example: 4 If the radius of a solid and hollow copper spheres are same which one can hold greater charge (a) Solid sphere [BHU 1999; KCET 1994; IIT-JEE 1974] (b) Hollow sphere (d) None of these U (c) Both will hold equal charge Solution: (c) Charge resides on the surface of conductor, since both the sphere having similar surface area so they will ST hold equal charge. Example: 5 Number of electrons in one coulomb of charge will be (a) 5.46  10 29 (b) 6.25  10 18 [RPET 2001; MP PMT/PET 1998] (c) 1.6  10 19 1 Q  n  6.25  10 18 19 e 1.6  10 Solution: (b) By using Q  ne  n  Example: 6 The current produced in wire when 10 7 electron/sec are flowing in it (a) 1.6  10–26 amp Solution: (d) i (d) 9  1011 (b) 1.6  1012 amp Q ne   10 7  1.6  10 19  1.6  10 12 amp t t (c) 1.6  1026 amp [CPMT 1994] (d) 1.6  10–12 amp 10 Electrostatics Example: 7 A table-tennis ball which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metal plates. One plate is earthed. When the other plate is connected to a high (a) Is attracted to the high voltage plate and stays there (b) Hangs without moving (c) Swings backward and forward hitting each plate in turn E3 (d) None of these 60 voltage generator, the ball Solution: (c) The table tennis ball when slightly displaced say towards the positive plate gets attracted towards the positive plate due to induced negative charge on its near surface. ID The ball touches the positive plate and itself gets positively charged by the process of conduction from the plate connected to high voltage generator. On getting positively charged it is repelled by the positive + + + + + + + + + – – – – – U plate and therefore the ball touches the other plate (earthed), which has + + + + + negative charge due to induction. On touching this plate, the positive charge of the ball gets neutralized D YG and in turn the ball shares negative charge of the earthed plate and is again repelled from this plate also, and this process is repeated again and again. Here it should be understood that since the positive plate is connected to high voltage generator, its potential and hence its charge will always remain same, as soon as this plate gives some of its charge to ball, excess charge flows from generator to the plate, and an equal negative charge is always induced on U the other plate. Tricky example: 1 ST In 1 gm of a solid, there are 5  1021 atoms. If one electron is removed from everyone of 0.01% atoms of the solid, the charge gained by the solid is (given that electronic charge is 1.6  10–19 C) (a) + 0.08 C Solution: (a) (b) + 0.8 C (c) – 0.08 C (d) – 0.8 C To calculate charge, we will apply formula Q = ne for this, we must have number of electrons. Here, number of electrons n . 01 % of 5  1021 i.e. n  5  10 21 .01  5  10 21  10 4 = 5  1017 100 So Q = 5  1017  1.6  10–19 = 8  10–2 = 0.08 C Since electrons have been removed, charge will be positive i.e. Q = + 0.08 C Electrostatics 11 Coulomb’s Law. If two stationary and point charges Q 1 and Q 2 are kept at a distance r, then it is found that force of Q1 60 attraction Q2 Q1 Q 2 r i.e., F  2 kQ1 Q 2 r2 ; (k = Proportionality constant) ID or repulsion between them is F  E3 r (1) Dependence of k : Constant k depends upon system of units and medium between the two charges. (b) In S.I. for air k  Note :  Q1 Q 2 r2 Dyne N m 2 1 Q1 Q 2 , F Newton (1 Newton = 105 Dyne). 2 40 r 2 C D YG (a) In C.G.S. for air k  1, F  U (i) Effect of units 1 4 0  9  10 9  0  Absolute permittivity of air or free space = 8.85  10 12 C2  Farad   . It’s 2 m  N m  Dimension is [ML3 T 4 A 2 ] U   0 Relates with absolute magnetic permeability (  0 ) and velocity of light (c) according to the 1 0 0 ST following relation c  (ii) Effect of medium (a) When a dielectric medium is completely filled in between charges rearrangement of the charges inside the dielectric medium takes place and the force between the same two charges decreases by a factor of K known as dielectric constant or specific inductive capacity (SIC) of the medium, K is also called relative permittivity r of the medium (relative means with respect to free space). Hence in the presence of medium Fm  Fair Q Q 1 . 12 2 K 4 0 K r Q1 K r Q2 12 Electrostatics Here  0 K   0  r   (permittivity of medium) (b) If a dielectric medium (dielectric constant K, thickness t) is partially filled between the charges then effective air separation between the charges becomes (r  t  t K ) r Q1 Q 2 1 Q2 60 Hence force F  K Q1 4 0 (r  t  t K ) 2 q1 q 2 r 12  K. E3 (2) Vector form of coulomb’s law : Vector form of Coulomb’s law is F 12  K. r 3 q1 q 2 r2 rˆ12 , where r̂12 is the unit vector from first charge to second charge along the line joining the two charges. (i) Force Force of Nature and formula gravitation (ii) Range Attractive F = Gm1m2/r2, obey’s Long range (between planets Newton’s third law of motion, and between electron and D YG between two masses U S.No. ID (3) A comparative study of fundamental forces of nature Relative strength 1 it’s a conservative force proton) Electromagnetic force Attractive as well as repulsive, Long (upto few kelometers) 10 37 (for obey’s Newton’s third law of stationary and moving charges) motion, it’s a conservative force Nuclear force (between Exact expression is not known Short (of the order of nuclear 1039 nucleons) till date. However in some cases size 10–15 m) (strongest) Short (upto 10–15m) 1024 ST U (iii) (iv) Weak force empirical formula U0 e r / r0 can be utilized for nuclear potential energy U0 and r0 are constant. (for Formula not known processes like  decay) Note :  Coulombs law is not valid for moving charges because moving charges produces magnetic field also.  Coulombs law is valid at a distance greater than 10 15 m. Electrostatics 13  A charge Q1 exert some force on a second charge Q 2. If third charge Q 3 is brought near, the force of Q 1 exerted on Q 2 remains unchanged. protons is 10–36/1 (iii) One proton and one electron 10–39/1.  Decreasing order to fundamental forces FNuclear  FElectromag 60  Ratio of gravitational force and electrostatic force between (i) Two electrons is 10–43/1. (ii) Two  FWeak  FGravitatio nal netic E3 (4) Principle of superposition : According to the principle of super Q position, total force acting on a given charge due to number of charges is r1 the vector sum of the individual forces acting on that charge due to all the Q1 ID charges. Consider number of charge Q 1 , Q 2 , Q 3 …are applying force on a Net force on Q will be Q2 Q3 Qn Qn – 1 D YG      Fnet  F1  F2 ..........  Fn 1  Fn r3 U charge Q r2 Concepts  Two point charges separated by a distance r in vacuum and a force F acting between them. After filling a dielectric medium having dielectric constant K completely between the charges, force between them decreases. To maintain the force as before ST U separation between them changes to r K. This distance known as effective air separation. +Q –Q Examples based on Coulomb’s law Example: 8 Two point charges 3 C and 8 C repel each other with a force of 40 N. If a charge of 5 C is added to each of them, then the force between them will become (a) 10 N Solution: (a) Initially (b) 10 N F k (c) 20 N 2  3  10 12 3  8  10 12 F '   k and Finally r2 r2 (d) 20 N so F' 1   F '  10 N F 4 14 Electrostatics Example: 9 Two small balls having equal positive charge Q (coulomb) on each are suspended by two insulated string of equal length L meter, from a hook fixed to a stand. The whole set up is taken in satellite into space where there is no gravity (state of weight less ness). Then the angle between the string and tension in the string is 4 0 (d) 180 o , Solution: (a) Q2 4 0 (2 L)2 1 (c) 180  ,. 1. Q2 L2. Q L L E3 (b) 90  , 1 +Q +Q 2 4 0 2 L2 1 QL. 4 0 4 L2 ID (a) 180 o , 60 [IIT-JEE 1986] In case to weight less ness following situation arises +Q 1 Q2 U So angle   180 and force F   4 0. L 2 L 2 +Q 180o L D YG Example: 10 Two point charges 1 C & 5 C are separated by a certain distance. What will be ratio of forces acting on these two (a) 1 : 5 Solution: (c) [CPMT 1979] (b) 5 : 1 (c) 1 : 1 (d) 0 Both the charges will experience same force so ratio is 1:1 Example: 11 Two charges of 40 C and 20 C are placed at a certain distance apart. They are touched and kept at the same distance. The ratio of the initial to the final force between them is (b) 4 : 1 U (a) 8 : 1 (c) 1 : 8 Since only magnitude of charges are changes that’s why ST Solution: (a) [MP PMT 2001] F  q1q 2 (d) 1 : 1  F1 q q 40  20 8  1 2   F2 q'1 q'2 10  10 1 Example: 12 A total charge Q is broken in two parts Q1 and Q 2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when (a) Q2  Solution: (d) Q Q , Q1  Q  R R (b) Q 2  Force between charges Q1 and Q 2 For F to be maximum, Hence Q1  Q 2  Q 2 dF 0 dQ1 3Q 2Q Q Q Q Q , Q1  Q  (c) Q 2  , Q1  (d) Q1  , Q 2  4 4 4 2 3 2 Q1 Q 2 Q1 Q  Q1  Fk k R2 R2 i.e., d dQ1  2   Q1Q  Q1 k  R2    0 or Q  2Q   1  0, Q1  Q 2 Electrostatics 15 Example: 13 The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become Solution: (b) (b) 11.25 N (c) 22.50 N Initial separation between the charges = 0.06 m Final separation between the charges = 0.04m 2  5  0.04  4       F2  11.25 N F2  0.06  9  2 E3 r F 1 Since F  2  1   2 F2  r1 r (d) 45.00 N 60 (a) 7.20 N [SCRA 1994] Example: 14 Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the F 16 Solution: (a) (b) 9F 16 (c) F +Q –Q A B D YG r Initially F  k Q2 r2 15 F 16 (d) +Q/4 U (a) ID charges becomes – Q/4 B A r 2 Q k.  F 4 Finally F '   2   16 r Example: 15 Three equal charges each +Q, placed at the corners of on equilateral triangle of side a what will be the force  1 on any charge  k  4 0  kQ 2 a2 (b) U (a)     2kQ 2 a2 [RPET 2000] (c) 2 kQ 2 a2 3 kQ 2 a2 (d) Solution: (d) Suppose net force is to be calculated on the charge which is kept at A. Two charges kept at B and C are ST applying force on that particular charge, with direction as shown in the figure. Since Fb  Fc  F  k FB 60o A Q2 a FC +Q 60o 2 So, Fnet  FB2  FC2  2 FB FC cos 60 Fnet  3 F  3 kQ 2 a2 +Q +Q B C Example: 16 Equal charges Q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be [MP PMT 1994] 16 Electrostatics (a) 3Q 2 4  0 a 2 1  1 2  Q2  1  Q2   (c)  (d)  2     4 a 2 2 2  4 0 a 2  0   4Q2 (b) 4  0 a 2 Solution: (c) After following the guidelines mentioned above FC 2kQ 2 Fnet  a 2  a2 kQ 2 2a 2 kQ 2 and FD   (a 2 )2 kQ 2  1 Q2 2     2  4 0 a 2 a2  1  2 2      2   FAC +Q +Q A FA B D C E3 Since FA  FC  kQ 2 FD 60 Fnet  FAC  FD  FA2  FC2  FD +Q Example: 17 Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance, will experience maximum coulomb force when d (b) x  2 d 2 d ID (a) x  (c) x  2 3 2 2 Solution: (c) Suppose third charge is similar to Q and it is q . and cos   4 0  2 d 2  x    4   Fnet  2  1 Qq. 4 0  2 d  x     4   U for Fnet to be maximum 2 dFnet dx  q d2 x  4 2 x  2 d2  x    4   1/2  3 / 2 5 / 2   2  2  d 2 2  x 2  d  0   3 x  x     4  4     i.e. x 2Qqx  d 2  4 0  x 2   4     d  2Qqx  0 i.e.  dx   2 d2  4 0  x  4   ST or   x D YG Where F  Qq F F U So net force on it Fnet = 2F cos 1 d (d) x  3/2 2   x 2 d /4 Q 2 x 2 d /4 Q B d d 2 2 C    0 3/2        x  d 2 2 Example: 18 ABC is a right angle triangle in which AB = 3 cm, BC = 4 cm and ABC  . The three charges +15, +12 and – 2 20 e.s.u. are placed respectively on A, B and C. The force acting on B is (a) 125 dynes (b) 35 dynes Solution: (c) Net force on B Fnet  FA2  FC2 FA  15  12 3 2  20 dyne (c) 25 dynes (d) Zero A +15 esu 3 cm B FC C +12 esu 4 cm FA Fnet  2 – 20 esu 2 FA  FC Electrostatics 17 FC  12  20 4 2  15 dyne Fnet  25 dyne 60 Example: 19 Five point charges each of value + Q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value – q placed at the centre of the hexagon Q2 L2 (b) k (c) Zero Q2 4 L2 E3 (a) k [IIT-JEE 1992] (d) Information is insufficient Solution: (a) Four charges cancels the effect of each other, so the net force on the charge placed at centre due to remaining +Q fifth charge is –Q Q2 L2 ID Fk +Q +Q +Q +Q U L D YG Example: 20 Two small, identical spheres having +Q and – Q charge are kept at a certain distance. F force acts between the two. If in the middle of two spheres, another similar sphere having +Q charge is kept, then it experience a force in magnitude and direction as [MP PET 1996] (b) 8F towards +Q charge (a) Zero having no direction (c) 8F towards – Q charge Solution: (c) Initially, force between A and C Fk 4F towards +Q charge (d) Q2 r2 U When a similar sphere B having charge +Q is kept at the mid point ST of line joining A and C, then Net force on B is Fnet  FA  FC  k Q2  kQ 2 r 2 2 r 2 2 kQ 2  8 2  8 F. (Direction is shown r +Q +Q A B r/2 r –Q FA FC r/2 C in figure) Tricky example: 2 Two equal spheres are identically charged with q units of electricity separately. When they are placed at a distance 3R from centre-to-centre where R is the radius of either sphere the force of repulsion between them is + + + + + + + + + 18 Electrostatics Solution: (a) 1. q2 4 0 R 2 (b) 1. q2 (c) 4 0 9 R 2 1 Generally students give the answer q2 4 0 (3 R)2 1. q2 4 0 4 R 2 (d) None of these but it is not true. Since the charges are not 60 (a) uniformly distributed, they cannot be treated as point charges and so we cannot apply coulombs ID E3 law which is a law for point charges. The actual distribution is shown in the figure above. Electrical Field. U A positive charge or a negative charge is said to create its field around itself. If a charge Q 1 exerts a force D YG on charge Q 2 placed near it, it may be stated that since Q 2 is in the field of Q 1 , it experiences some force, or it may also be said that since charge Q 1 is inside the field of Q 2 , it experience some force. Thus space around a charge in which another charged particle experiences a force is said to have electrical field in it.  (1) Electric field intensity (E) : The electric field intensity at any point is defined as the force experienced by   F +Q (q0) a unit positive charge placed at that point. E   F q0 U P ST Where q 0 0 so that presence of this charge may not affect the source charge Q and its electric field is   F not changed, therefore expression for electric field intensity can be better written as E  Lim q 0 0 q 0 (2) Unit and Dimensional formula : It’s S.I. unit – Newton volt Joule and   coulomb meter coulomb  meter C.G.S. unit – Dyne/stat coulomb. Dimension : [ E ] =[ MLT 3 A 1 ]  (3) Direction of electric field : Electric field (intensity) E is a vector quantity. Electric field due to a positive charge is always away from the charge and that due to a negative charge is always towards the charge Electrostatics 19  E +Q –Q  E 60  (4) Relation between electric force and electric field : In an electric field E a charge (Q) experiences a force F  QE. If charge is positive then force is directed in the direction of field while if charge is negative force acts +Q E E –Q ID F E3 on it in the opposite direction of field (5) Super position of electric field (electric field at a point due to various charges) : The resultant electric U field at any point is equal to the vector sum of electric fields at that point due to various charges.     E  E1  E 2  E 3 ... The magnitude of the resultant of two electric fields is given by and the direction is D YG E  E12  E 22  2 E1 E 2 cos  E 2 sin  tan   E1  E 2 cos  E E2 given by   E1 (6) Electric field due to continuous distribution of charge : A system of closely spaced electric charges forms a continuous charge distribution U Continuous charge distribution Surface charge distribution Volume charge distribution In this distribution charge distributed In this distribution charge distributed In this distribution charge distributed on a line. on the surface. in the whole volume of the body. For example : charge on a wire, For charge on a ring etc. Relevant conducting sphere, charge on a charged sphere. Relevant parameter parameter is  which isQ called linear + sheet etc. Relevant Qparameter is + is  which is called volume charge Q+ charge density i.e.,  which is called +surface R +charge density i.e., ST Linear charge distribution  charge length + + R + + + + + + Circular charged ring example : Charge + + density i.e., + on + + a For + Spherical shall  example charge volume : Non conducting + + + + + + R + +++ + + + + Non conducting sphere 20 Electrostatics Q 2R   charge area   Q  Q 4 R 3 3 4R 2 60  To find the field of a continuous charge distribution, we divide the charge into infinitesimal charge E3 elements. Each infinitesimal charge element is then considered, as a point charge and electric field dE is determined due to this charge at given point. The Net field at the given point is the summation of fields of all  the elements. i.e., E  dE ID Electric Potential. (1) Definition : Potential at a point in a field is defined as the amount of work done in bringing a unit U positive test charge, from infinity to that point along any arbitrary path (infinity is point of zero potential). W q0 D YG Electric potential is a scalar quantity, it is denoted by V; V  (2) Unit and dimensional formula : S. I. unit – 1 Joule Stat  volt C.G.S. unit – Stat volt (e.s.u.); 1 volt  300 Coulomb volt Dimension – [V ]  [ML2 T 3 A 1 ] (3) Types of electric potential : According to the nature of charge potential is of two types (ii) Negative potential : Due to negative charge. U (i) Positive potential : Due to positive charge. (4) Potential of a system of point charges : Consider P is a point at which net electric potential is to be ST determined due to several charges. So net potential at P  Q 4  Q Q Q V k 1 k 2 k 3 k ... r1 r2 r3 r4 In general V  X  i1 Note :  k Qi ri P r1 r2 – Q1 – Q2 At the centre of two equal and opposite charge V = 0 but E  0 r3 + Q3 r4 – Q4 Electrostatics 21  At the centre of the line joining two equal and similar charge V  0, E  0 (5) Electric potential due to a continuous charge distribution : The potential due to a continuous charge  divided i.e., V  dV ,  60 distribution is the sum of potentials of all the infinitesimal charge elements in which the distribution may be dQ  4 πε 0r E3 (6) Graphical representation of potential : When we move from a positive charge towards an equal negative charge along the line joining the two then initially potential decreases in magnitude and at centre become zero, but this potential is throughout positive because when we are nearer to positive charge, overall ID potential must be positive. When we move from centre towards the negative charge then though potential remain always negative but increases in magnitude fig. (A). As one move from one charge to other when both U charges are like, the potential first decreases, at centre become minimum and then increases Fig. (B). –q +q D YG Y V O x +q +q Y v X (B) x ST U (A) X O (7) Potential difference : In an electric field potential difference between two points A and B is defined as equal to the amount of work done (by external agent) in moving a unit positive charge from point A to point B. i.e., VB  VA  W in general W  Q. V ; V  Potential difference through which charge Q moves. q0 Electric Field and Potential Due to Various Charge Distribution. (1) Point charge : Electric field and potential at point P due to a point charge Q is 22 Electrostatics Ek  Q Q or E  k 2 rˆ 2 r r  1  k  4 0   Q  , V  k r  Q P Note :  60 r Electric field intensity and electric potential due to a point charge q, at a distance t1 + t2 where K2 are : E 1 Q ; 4 πε 0 (t 1 K 1  t 2 K 2 ) 2 1 Q 4 πε 0 (t 1 K 1  t 2 K 2 ) ID (2) Line charge V  E3 t1 is thickness of medium of dielectric constant K1 and t2 is thickness of medium of dielectric constant (i) Straight conductor : Electric field and potential due to a charged straight conducting wire of length l and If  = ; Ex  2k  sin  and Ey = 0 r If l  i.e.  =  =  ; 2  2 ; Ex  | E x| | E y|  ST (b) Potential : V  Ey l + + + + + r    2k  and Ey = 0 so Enet  20r r 2 k k so Enet  Ex2  Ey2  r r  r2  l2 1     for infinitely long conductor V  log e r  c log e  2 0 2 0  r 2  l 2  1  U If  = 0,   k k (cos   cos  ) (sin   sin  ) and E y  r r D YG (a) Electric field : E x  U charge density  P Ex Electrostatics 17 (ii) Charged circular ring : Suppose we have a charged circular ring of radius R and charge Q. On it’s axis electric field and potential is to be determined, at a point ‘x’ away from the centre of the ring. + + dQ + + + + + +  x + + + + P  dE sin  + dE cos  x  dE  O R 2 x R 2 E3 R + E 60 + KdQ R2  x 2 directed as shown. It’s component along the axis is dE cos  and perpendicular to the axis is kdQ x dE sin . By symmetry dE sin   0 , hence E  dE cos  . 2 2 2 (R  x ) (R  x 2 )1 2 (a) Electric field : Consider an element carrying charge dQ. It’s electric field dE   R kQx 2  x2  3 2  directed away from the centre if Q is positive 1 Q. 2 4 0 x  R2 D YG (b) Potential : V  U E  ID   Note :  At centre x = 0 so Ecentre= 0 and Vcentre  kQ R  At a point on the axis such that x >> R E   At a point on the axis if x   R 2 , E max  kQ x2 Q and V  kQ x 6 3 0 a 2 (3) Surface charge : U (i) Infinite sheet of charge : Electric field and potential at a point P as shown E  2 0 (E  r o ) + + + + + + + + + + + + + + + + ST r V  C 2 0 and P r (ii) Electric field due to two parallel plane sheet of charge : Consider two large, uniformly charged parallel. Plates A and B, having surface charge densities are  A and  B respectively. Suppose net electric field at points P, Q and R is to be calculated. At P, E P  ( E A  E B )  1 ( A   B ) 2 0 A EA EB + + + + + + + + +  + + + ++ P + + + B EB + + + + + EA + + + + + +   + + + + + R Q + EA EB 18 Electrostatics At Q, E Q  ( E A  E B )  At R, E R  ( E A  E B )    A   and  B   then E p  0, E Q  1 ( A   B ) 2 0  , E R  0. Thus in case of two infinite 0 60 Note :  If 1 ( A   B ) ; 2 0 plane sheets of charges having equal and opposite surface charge densities, the field is non-zero only in the space between + + the two sheets and is independent of the distance between them + E=/0 – i.e., field is uniform in this region. It should be noted that this result will hold good for finite plane sheet also, if they are held at a distance much smaller then the dimensions of sheets i.e., + (iii) Conducting sheet of charge : – + –  0 ID E – + E3 parallel plate capacitor. – – + + + + + + r V  C 0 + + + r P U + D YG (iv) Charged conducting sphere : If charge on a conducting sphere of radius R is Q as shown in figure then electric field and potential in different situation are – +Q + + +Q + + + R + + + + + + + + U Solid + + + + + R + + + + + + + + Hollo w (a) Out side the sphere : P is a point outside the sphere at a distance r from the centre at ST which electric field and potential is to be determined. Electric field at P E out  1 Q R 2. 2  4  0 r 0r2 and Vout  1 Q R 2 Q    A.   4  0 r  0 r     4 R 2 (b) At the surface of sphere : At surface r  R So, Es  1 Q  1 Q R. 2  and Vs .  4  0 R 0 4  0 R  0 (c) Inside the sphere : Inside the conducting charge sphere electric field is zero and potential remains constant every where and equals to the potential at the surface. Electrostatics 19 E in  0 and Vin = constant  Vs 60 Note :  Graphical variation of electric field and potential of a charged spherical conductor with distance V-r graph R O O R E3 E-r graph VS E E out  1 ID r Vout  2 Ein=0 r O O 1 r r r =R U (4) Volume charge (charged non-conducting sphere) : Charge given to a non conducting spheres spreads uniformly throughout it’s volume. D YG (i) Outside the sphere at P E out  E out  Q 1 Q 1 Q. and Vout . by using   4 4  0 r 2 4  0 r R 3 3 R 3 R 3 and V  out 3 0 r 3 0 r 2 + + + +Q + + + + + + + + + R + ++ + + + + + + + + + + + + + + + + P r (ii) At the surface of sphere : At surface r  R R 1 Q. 2  4  0 R 3 0 U Es  Vs  and 1 Q R 2.  4  0 R 3 0 ST (iii) Inside the sphere : At a distance r from the centre E in  Note :  At r 1 Qr.  4  0 R 3 3 0 centre r  0 Ein  r and Vin  1 Q [3 R 2  r 2 ] (3 R 2  r 2 )  4  0 6 0 2R 3 Vcentre  So, 3 1 Q 3 .  Vs 2 4  0 R 2 i.e., Vcentre  V surface  Vout  Graphical variation of electric field and potential with distance E-r graph V-r graph + + + + + + + R O + + + + + + + + + + + ++ + + R O + + + + + + + + VC E Ein  r E out  1 r 2 VS Vout  1 r 60 20 Electrostatics (i) Uniformly charged semicircular ring :   charge length E3 (5) Electric field and potential in some other cases +Q At centre : + + 2 K Q E  2 R 2  0 R 2 + + + + + ID V  + + + KQ Q  R 4  0 R + R U (iii) Charged cylinder of infinite length D YG (a) Conducting r (b) Non-conducting P U conducting E in  0 and for non-conducting E in  + + + + + + + + + + + + ST V  4 0 R 2 0 + + +r P + r. (we can also write formulae in form 2 0 R 2 (ii) Hemispherical charged body : E +   , and E suface  but for 2 0 r 2 0 R R 2 etc.) 2 0 r At centre O, R + + For both type of cylindrical charge distribution E out  of  i.e., E out  + + + + + + + + + + + + + + + + + + + + + O + +  (iv) Uniformly charged disc At a distance x from centre O on it’s axis  R O x Electrostatics 21 V    x 2  R 2  x   2 x  R2  x  2 0   Total charge on disc Q = R2  If x 0, E ~–  i.e. for points situated near the disc, it behaves as an infinite 2 0 sheet of charge. ID Concepts E3 Note :   1  2 0  60 E  No point charge produces electric field at it’s own position.  Since charge given to a conductor resides on it’s surface hence electric field inside it is zero. + + + + + + E=0 + + D YG + +  + E=0 + + + + + + Two charged spheres having radii + + r1 and r2 charge densities  1 and  2 respectively, then the ratio of r2 E1   1  2 E2  2 r 2 1  Q   4 r 2  In air if intensity of electric field exceeds the value 3  10 6 N/C air ionizes. U  + + E  electric field on their surfaces will be  + + The electric field on the surface of a conductor is directly proportional to the surface charge density at that point i.e,  + + + + + + U + + + A small ball is suspended in a uniform electric field with the help of an insulated thread. If a high energy x– ray bean falls on the ball, x-rays knock out electrons from the ball so the ball is positively charged and  ST therefore the ball is deflected in the direction of electric field. E F= QE X– Ray  Electric field is always directed from higher potential to lower potential.  A positive charge if left free in electric field always moves from higher potential to lower potential while a negative charge moves from lower potential to higher potential. +Q + + + + – – + – +Q – + + + +q 22 Electrostatics  The practical zero of electric potential is taken as the potential of earth and theoretical zero is taken at infinity. An electric potential exists at a point in a region where the electric field is zero and it’s vice versa.  A point charge +Q lying inside a closed conducting shell does not exert force another point charge q placed outside the shell as shown in figure E3 60  Actually the point charge +Q is unable to exert force on the charge +q because it can not produce electric field at the position of +q. All the field lines emerging from the point charge +Q terminate inside as these lines cannot penetrate the conducting medium (properties of lines of force). ID The charge q however experiences a force not because of charge +Q but due to charge induced on the outer surface of the shell. E0 V=0 +e Examples based on electric field and electric potential U –e  1 k   4  0      D YG Example: 21 A half ring of radius R has a charge of  per unit length. The electric field at the centre is (a) Zero (b) [CPMT 2000; CBSE PMT 2000; REE 1999] k R (c) Solution: (c) dl  Rd  2k  R (d) k  R dl d    U Charge on dl  Rd . C Rd . dE dl  dE R2 We need to consider only the component dE cos  , as the component dE sin  will cancel out because of the field at C due to the symmetrical element dl, ST Field at C due to dl  k The total field at C is  2  2  0 dE cos   2 k R  2  0 cos  d   2k  R  Q   2 0 R 2    Example: 22 What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2 newton/coulomb [1 / 4  0  9  10 9 Nm 2 ] (a) 2  10 11 coulomb 1 Q Solution: (a) By using E . ; 4  0 r 2 (b) 3  10 11 coulomb Q 2  9  10 9  30  10  2  (c) 5  10 11 coulomb  2  Q  2  10 11 C (d) 9  10 11 coulomb Electrostatics 23 Example: 23 Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the location of Q is E, then at the locality of – 3Q, it is (a)  E (b) E/3 (c) 3 E (d) – E/3 Solution: (b) Let the charge Q and – 3Q be placed respectively at A and B at a distance x 3Q E x 2 60 Now we will determine the magnitude and direction to the field produced by charge – 3Q at B A, this is E as mentioned in the Example. A –3Q Q (along AB directed towards negative charge) x Q x 2  E3 Now field at location of – 3Q i.e. field at B due to charge Q will be E'  directed away from positive charge) E (along AB 3 Example: 24 Two charged spheres of radius R1 and R 2 respectively are charged and joined by a wire. The ratio of electric field of the spheres is R1 R2 (b) R2 R1 (c) R 12 Solution: (b) After connection their potential becomes equal i.e., k. R E1 Q  1   2 E2 Q 2  R1 R 22 R 12 Q1 R  1 Q2 R2 Q1 k. Q2 ;   R1 R2 2  R   2.  R1  U Ratio of electric field (d) R 22 ID (a) D YG Example: 25 The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of 0.036 N/C just above its surface is (b) 2. 6  10 5 (a) 2.7  10 5 Solution: (c) By using E  k Q R2 (c) 2. 5  10 5 (d) 2. 4  10 5 , where R = radius of sphere so 0.036 = 9  10 9  ne 0.1 2  n  2. 5  10 5 Example: 26 Eight equal charges each +Q are kept at the corners of a cube. Net electric field at the centre U  1 will be  k   4 0  kQ (b) r2 ST (a)    8 kQ (c) r2 2kQ (d) Zero r2 Solution: (d) Due to the symmetry of charge. Net Electric field at centre is zero. Note +q : q a a a a E= 0 E=0 +q a +q q q a a q Square Equilateral triangle Example: 27 q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field at the centre O of the square has the direction along. q 2q A B O (a) AB (b) CB 60 24 Electrostatics (c) AC (d) BD E3 Solution: (b) By making the direction of electric field due to all charges at centre. Net electric field has the direction along CB Example: 28 Equal charges Q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is Q 4  0 a (b) 2 2Q 4  0 a 2 (c) 3Q 4  0 a 2 ID (a) E EB2  EC2  2 EB EC cos 60 1 Q. 4  0 a 2 D YG E B  EC  U Solution: (c) As shown in figure Net electric field at A So, E  3Q 4  0 a 2 (d) E FC Q 2 0 a 2 FB A 60 a o +Q a +Q a B C Example: 29 Four charges are placed on corners of a square as shown in figure having side of 5 cm. If Q is one micro coulomb, then electric field intensity at centre will be (a) 1.02  10 7 N / C upwards Q – 2Q –Q + 2Q (b) 2.04  10 7 N / C downwards U (c) 2.04  10 7 N / C upwards (d) 1.02  10 7 N / C downwards ST Solution: (a) | E C|  | E A| so resultant of EC & E A is ECA  EC  E A directed toward Q Q A Also | E B |  | E D| so resultant of EB and ED i.e. ECA Enet EC EBD  EB  ED directed toward – 2Q charge hence Net electric field at centre is E ECA  2 EB  EBD  2 ED EA.… (i) D By proper calculations | E A |  9  10 9  O – 2Q B EBD – 10 6  5    10  2    2  2  0. 72  10 7 N /C Q C +2Q Electrostatics 25 2  10 6  5    10  2    2  | E D |  9  10 9  2 10 6  5    10  2    2  2  1. 44  10 7 N /C ; | E C |  9  10 9   0. 72  10 7 N /C ;  1. 44  10 7 N /C Hence from equation – (i) E3 7  5    10  2    2  2 So, |E CA|  |E C|  |E A|  0.72  10 7 N/C | E BD | | E B |  | E D |  0. 72  10 7 N /C. and 2  10 6 60 | E B |  9  10 9  E  1. 02  10 N /C upwards (a) 12  10 9 Q N/C, 1.8  104 V (c) 6  10 9 Q N/C, 9  103 V (b) Zero, 1.2  104V (d) 4  10 9 Q N/C , 6  103 U V ID Example: 30 Infinite charges are lying at x = 1, 2, 4, 8…meter on X-axis and the value of each charge is Q. The value of intensity of electric field and potential at point x = 0 due to these charges will be respectively Solution: (a) By the superposition, Net electric field at origin D YG 1 1 1 1  E  kQ  2  2  2  2 ...  2 4 8 1  1 1 1   E  kQ 1    ...   4 16 64   x=0 x=1 x=2 x=4 x=8 1 1 1   ...  is an infinite geometrical progression it’s sum can be obtained by using 4 16 64 a the formula S   ; Where a = First term, r = Common ratio. 1r 1 U Here a  1 and r  1 1 1 1 1 4 so, 1    .....   . 4 4 16 64 1 1/ 4 3 ST Hence E  9  10 9  Q  4  12  10 9 Q N /C 3 Electric potential at origin V  9  9  10  10 6  1 1  10 6 1  10 6 1  10 6 1  10 6    .......   4  0  1 2 4 8     1  1 1 1   3 4   1.8  10 volt 1  2  4  8 ............  9  10    1  1   2  26 Electrostatics Note :  In the arrangement shown in figure +Q and – Q are alternatively and equally spaced from each other, the net potential at the origin O is Q log e 2 1. x 4  0 +Q –Q +Q –Q x 2x 3x 4x O 60 [IIT 1998] E3 V Example: 31 Potential at a point x-distance from the centre inside the conducting sphere of radius R and charged with charge Q is [MP PMT 2001] (a) Q R (b) Q x Q (c) ID Solution: (a) Potential inside the conductor is constant. x2 (d) xQ Example: 32 The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9  10 5 V is (b) 8  10 6 V (a) 80 V (d) 9  10 5 V ne 50  1.6  10 19  9  10 9   8  10 6 V 15 r 9  10 U Solution: (b) V  9  10 9  (c) 9 V D YG Example: 33 Eight charges having the valves as shown are arranged symmetrically on a circle of radius 0.4m in air. Potential at centre O will be +5 C – 5 C – 7 C +15 C +7 C O – 5 C +7 C U (a) 63  10 4 volt +11 C (c) 63  10 6 volt (b) 63  10 10 volt (d) Zero Solution: (a) Due to the principle of superposition potential at O ST V  1 28  10 6 28  10 6   9  10 9   63  10 4 volt 4  0 0.4 0.4 Example: 34 As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is A a B b +q b –q C Electrostatics 27 (a) 1 2q. 4  0 a2  b 2 (b) 1 q. 4  0 a2  b 2 (c) 1 (q ). 4  0 a2  b 2 (d) Zero Solution: (d) Potential at A = Potential due to (+q) charge + Potential due to (– q) charge (q ) 1 q 1.  2 2  4  0 4 0 a b 0 a2  b 2 60  E3 Example: 35 A conducting sphere of radius R is given a charge Q. consider three points B at the surface, A at centre and C at a distance R/2 from the centre. The electric potential at these points are such that [DCE 1994] (a) VA = VB = VC (b) VA = VB  VC (c) VA  VB  VC (d) VA  VB = VC Solution: (a) Potential inside a conductor is always constant and equal to the potential at the surface. 10  10 9 coulomb are lying on the corners of a square of side 8 cm. The 3 electric potential at the point of intersection of the diagonals will be Example: 36 Equal charges of (b) 900 2 V (c) 150 2 V (d) 1500 2 V ID (a) 900 V Solution: (d) Potential at the centre O Q 1 10 V 4. given Q   10 9 C  a  8 cm  8  10 2 m 4  0 a / 2 3 D YG 2 U 10  10  9 9 3 V  5  9  10   1500 2 volt 8  10  2 Q Q a 2 O Q a Q Tricky example: 3 A point charge Q is placed outside a hollow spherical conductor of radius R, at a distance (r > R) from its centre C. The field at C due to the induced charges on the conductor is  1   K   4  0   (b) K (a) Zero Q r  R 2 Q Q directed towards Q (d) directed K 2 r2 r away from Q Solution: (c) A according to the figure shown below. The total field at C must be zero. The field at C – + Q + – due to the point charge is E  K 2 towards left. The field at C due to the induced + – r C +Q + KQ charges must be 2 towards right i.e. Q. + directed towards – R r + – ST U (c) K + – r Tricky example: 4 A point charge q is placed at a distance of r from the centre of an uncharged 28 Electrostatics conducting sphere of radius R (< r). The potential at any point on the sphere is (a) Zero (b) 1 q. 4  0 r (c) 1 qR. 4  0 r 2 (d) 1 qr 2. 4  0 R q 1.. 4 0 r E3 net induced charge will be zero)  V  60 Solution: (c) Since, potential V is same for all points of the sphere. Therefore, we can calculate its value at the centre of the sphere. 1 q.  V ' ; where V = potential at centre due to induced charge = 0 (because  V  4 0 r Potential Due to Concentric Spheres. To find potential at a point due to concentric sphere following guideline are to be considered ID Guideline 1: Identity the point (P) at which potential is to be determined. Guideline 2: Start from inner most sphere, you should know where point (P) lies w.r.t. concerning sphere/shell (i.e. outside, at surface or inside) U Guideline 3: Then find the potential at the point (P) due to inner most sphere and then due to next and so on. D YG Guideline 4: Using the principle of superposition find net potential at required shell/sphere. Standard cases Case (i) : If two concentric conducting shells of radii r1 and r2(r2 > r1) carrying uniformly distributed charges Q1 and Q2 respectively. What will be the potential of each shell To find the solution following guidelines are to be taken. Here after following the above guideline potential at the surface of inner shell is Q Q 1 1. 1 . 2 4  0 r1 4  0 r2 U V1  and potential at the surface of outer shell Q Q 1 1. 1 . 2 4  0 r2 4 0 r2 ST V2  Q2 Q1 r2 r1 Electrostatics 29 Case (ii) : The figure shows three conducting concentric shell of radii a, b and c (a < b < c) having charges Qa, Qb and Qc respectively what will be the potential of each shell After following the guidelines discussed above 1  Qa Qb Q c    b c  4  0  a Qa c 60 Potential at A; V A  Qc Qb b Potential at C; VC  1  Q a Qb Q c    c c  4 0  c a A B E3 1  Q a Qb Q c  Potential at B; VB    b c  4  0  b Case (iii) : The figure shows two concentric spheres having radii r1 and r2 respectively (r2 > r1). If charge on inner sphere is +Q and outer sphere is earthed then determine. (a) The charge on the outer sphere (b) Potential of the inner sphere Q r2 1 1 Q Q'. . 0 4  0 r2 4 0 r2 ID (i) Potential at the surface of outer sphere V 2   Q'  Q Q 1 Q 1 (Q ).   4  0 r1 4  0 r2 4  0 U (ii) Potential of the inner sphere V1  +Q r1 1 1    r r 2  1 Case (iv) : In the case III if outer sphere is given a charge +Q and inner sphere is earthed then D YG (a) What will be the charge on the inner sphere (b) What will be the potential of the outer sphere (i) In this case potential at the surface of inner sphere is zero, so if Q' is the charge induced on inner sphere then V1  1  Q' Q      0 i.e., 4 0  r1 r2  Q'   r1 Q r2 r2 (Charge on inner sphere is less than that of the outer sphere.) +Q r1 (ii) Potential at the surface of outer sphere 1 Q' 1 Q. . 4  0 r2 4 0 r2 U V2  1 4  0 r2 ST V2    r1 Q  Q  Q   r 4  2 0 r2    r1  1   r 2   Examples based on concentric spheres Example: 37 A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 volts. The potential at the centre of the sphere is (a) Zero (b) 10 V (c) Same as at a point 5 cm away from the surface (d) cm away from the surface Same as at a point 25 Solution: (b) Inside the conductors potential remains same and it is equal to the potential of surface, so here potential at the centre of sphere will be 10 V 30 Electrostatics Example: 38 A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is [MP PMT 1991] (b) 1 e.s.u. 4 (c) 30 e.s.u. (d) 36 e.s.u. 60 (a) 54 e.s.u. Solution: (d) Let charge on inner sphere be +Q. charge induced on the inner surface of outer sphere will be –Q. So potential at the surface of inner sphere (in CGS) Q Q  4 6 E3 3 +Q 4cm  Q  36 e.s.u. 6cm (a) Q(R 2  r 2 ) 4  0 (R  r) (b) ID Example: 39 A charge Q is distributed over two concentric hollow spheres of radii r and (R  r) such that the surface densities are equal. The potential at the common centre is Q R r (c) Zero (d) Q(R  r) 4  0 (R 2  r 2 ) U Solution: (d) If q1 and q 2 are the charges on spheres of radius r and R respectively, in accordance with conservation of charge Q  q1  q 2 q2 ….(i) D YG and according to the given problem  1   2 q1 i.e., 4 r 2  q2 4 R 2 q1 r2  2 q2 R  q1  So equation (i) and (ii) gives Potential at common centre V  q1 Qr 2 2 …. r(ii) R 2 (R  r ) and q 2  QR 2 (R 2  r 2 ) QR  Q(R  r) 1  q1 q 2  1  Qr 1   2  . 2  2   2 2  4  0  r R  4  0  (R  r ) (R  r )  4  0 (R  r 2 ) U Example: 40 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the two surfaces is ST (a) V (b) 2V (c) 4V (d) –2V Solution: (a) If a and b are radii of spheres and spherical shell respectively, potential at their surfaces will be Vsphere  1 Q. and 4  0 a Vshell  1 Q. 4  0 b Q and so according to the given problem. V  Vsphere  Vshell  Q 1 1   4  0  a b  …. (i) + + + + + Sphere + + + a + + + + + b Now when the shell is given a charge –3Q the potential at its surface and also inside will 1  3Q  change by V0   4  0  b  Electrostatics 31 So that Vsphere  Vshell  now Q 4  0 Vsphere  1  Q 3Q   4  0  a b  Vshell  and 1  Q 3Q   4  0  b b  hence 1 1  a  b   V   (a)   (a  b  c), 0 0 60 Example: 41 Three concentric metallic spheres A, B and C have radii a, b and c (a  b  c) and surface charge densities on them are  ,   and  respectively. The valves of VA and VB will be  a2    b  c  b    C B a2 (b) (a  b  c), c (d)  0 E3 0  (a  b  c), 0     a    b  c  c    2  a2 b 2      c  and (a  b  c)  c  c 0    a ID (c) c A b Solution: (a) Suppose charges on A, B and C are q a , q b and q c qc 4 a 2  q a    4 a 2 ,  B    4 c 2 qb 4 b 2  qb    4 b 2  q c    4 c 2 D YG and  C    qa U Respectively, so  A    Potential at the surface of A 1  qa qb qc  1      b c 4  0  a 4  0 VA  (VA )surface  (VB )in  (VC )in  VA     4 a 2 ( )  4 b 2   4 c 2      a b c    a  b  c] 0 U Potential at the surface of B ST VB  (VA )out  (VB )surface  (VC )in  1  qa qb qc  1     4 0  b b c  40   4a2   4b 2   4c2    a 2        b  c b c  b   0  b  Electric Lines of Force. (1) Definition : The electric field in a region is represented by continuous lines (also called lines of force). Field line is an imaginary line along which a positive test charge will move if left free. Electric lines of force due to an isolated positive charge, isolated negative charge and due to a pair of charge are shown below + – + – +  N + 60 32 Electrostatics (2) Properties of electric lines of force E3 (i) Electric field lines come out of positive charge and go into the negative charge. (ii) Tangent to the field line at any point gives the direction of the field at that point. EA EB ID B A U (iii) Field lines never cross each other. (iv) Field lines are always normal to conducting surface. D YG + + + + + + (A) (B) – – – – – – (v) Field lines do not exist inside a conductor. (vi) The electric field lines never form closed loops. (While magnetic lines of forces form ST U closed loop) – + S N (vii) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In the following figure electric lines of force are originating from A and terminating at B hence QA is positive while QB is negative, also number of electric lines at force linked with QA are more than those linked with QB hence | Q A | | Q B | A B QA QB Electrostatics 33 (viii) Number of lines of force per unit area normal to the area at a point represents 60 magnitude of intensity (concept of electric flux i.e.,   EA ) (ix) If the lines of forces are equidistant and parallel straight lines the field is uniform and if either lines of force are not equidistant or straight line or both the field will be non uniform, E3 also the density of field lines is proportional to the strength of the electric field. For example see the following figures. X Y (B ) EX > E D YG U (A ) EX = E Y ID X (3) Electrostatic shielding : Electrostatic shielding/screening is the phenomenon of protecting a certain region of space from external electric field.  E Sensitive instruments and appliances are affected seriously with strong external electrostatic fields. Their working suffers and + + + + – E=0 q =0 Hollow space they may start misbehaving under the effect of unwanted fields.  E The electrostatic shielding can be achieved by protecting and enclosing the sensitive instruments inside a  E –– hollow –  E – –– Conductor + + + – +  E zero. U conductor because inside hollow conductors, electric fields is ST (i) It is for this reason that it is safer to sit in a car or a bus during lightening rather than to stand under a tree or on the open ground. (ii) A high voltage generator is usually enclosed in such a cage which is earthen. This would prevent the electrostatic field of the generator from spreading out of the cage. (iii) An earthed conductor also acts as a screen against the electric field. When conductor is not earthed field of the charged body C due to C C electrostatic induction continues beyond AB. If AB is earthed, induced positive charge neutralizes and the field in the region beyond AB disappears. – – A + + – – + – – – B + + A – – – – B 34 Electrostatics 60 Equipotential Surface or Lines. If every point of a surface is at same potential, then it is said to be an equipotential surface or for a given charge distribution, locus of all points having same potential is called E3 “equipotential surface” regarding equipotential surface following points should keep in mind : (1) The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density larger the field strength. (2) The direction of electric field is perpendicular to the equipotential surfaces or lines. ID (3) The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres. V1 V2 V3 V4 V5 U V = V2 D YG V = V1 Equipotentia l surface V1 > V2 > V3 > V4 > V5 (4) For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field lines. (5) A metallic surface of any shape is an equipotential surface e.g. When a charge is given ST U to a metallic surface, it distributes itself in a manner such that its every point comes at same potential even if the object is of irregular shape and has sharp points on it. + + + + + + + O +  V= + + const. + + + + + + + Metallic charged sphere + + + + + + + + V= const. + + + + + + + Charged metallic body of irregular shape If it is not so, that is say if the sharp points are at higher potential then due to potential difference between these points connected through metallic portion, charge will flow from Electrostatics 35 points of higher potential to points of lower potential until the potential of all points become same. (7) Equipotential surface for pair of charges + + + E3 – 60 (6) Equipotential surfaces can never cross each other Pair of two equal and opposite charges ID Pair of two equal and similar charges Concepts Unit field i.e. 1N/C is defined arbitrarily as corresponding to unit density of lines of force.  Number of lines originating from a unit charge is  It is a common misconception that the path traced by a positive test charge is a field line but actually the path traced by a unit positive test charge represents a field full line only when it moves along a straight line.  Both the equipotential surfaces and the lines of force can be used to depict electric field in a certain region of space. The advantage of using equipotential surfaces over the lines of force is that they give a visual picture of both the magnitude and direction of the electric field. U  U D YG 1 0 Examples based on electric lines of force ST Example: 42 Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in (a) (b) (c) (d)    Solution (c) Option (a) shows lines of force starting from one positive charge and terminating at another. Option (b) has one line of force making closed loop. Option (d) shows all lines making closed loops. All these are not correct. Hence option (c) is correct 36 Electrostatics  +q +q +q  60  E3 Example: 43 A metallic sphere is placed in a uniform electric field. The lines of force follow the path (s) shown in the figure as 1 1 2 2 3 3 4 (b) 2 ID (a) 1 4 (c) 3 (d) 4 U Solution: (d) The field is zero inside a conductor and hence lines of force cannot exist inside it. Also, due to induced charges on its surface the field is distorted close to its surface and a line of force must deviate near the surface outside the sphere. D YG Example: 44 The figure shows some of the electric field lines corresponding to an electric field. The figure suggests A (a) E A  E B  E C B C (c) E A  E C  E B (d) E A  E C  E B U Solution: (c) (b) E A  E B  E C [MP PMT 1999] ST Example: 45 The lines of force of the electric field due to two charges q and Q are sketched in the figure. State if (a) Q is positive and Q  q (b) Q is negative and Q  q Q q (c) q is positive and Q  q (d) q is negative and Q  q Solution: (c) q is +ve because lines of force emerge from it and Q  q because more lines emerge from q and less lines terminate at Q. Example: 46 The figure shows the lines of constant potential in a region in which an electric field is present. The magnitude of electric field is maximum at Electrostatics 37 20 V A  B  50 V (b) B (c) C (d) Equal at A, B and C E3 (a) A 60 C 30 V 40 V Solution: (b) Since lines of force are denser at B hence electric field is maximum at B Example: 47 Some equipotential surface are shown in the figure. The magnitude and direction of the electric field is y  40 V    20 30 x cm U 10 30 V ID 20 V  = 30o D YG (a) 100 V/m making angle 120o with the x-axis (b) 100 V/m making angle 60o with the xaxis (c) 200 V/m making angle 120o with the x-axis (d) None of the above Solution: (c) By using dV  E dr cos  suppose we consider line 1 and line 2 then (30 – 20) = E cos 60o (20 – 10) × 10–2  E y 20 V So E  200 volt / m making in angle 120o with x-axis 40 V 120 30o U ST 30 V 10 1 dr o x 20 40 30 2 Tricky example: 5 Which of the following maps cannot represent an electric field (a) (b) (c) (d) a d b c 38 Electrostatics Solution: (a) If we consider a rectangular closed path, two parallel sides of it considering with lines of force 60 as shown, then we find that work done along the closed path abcd is abE1 – cdE2  0. Hence the field cannot represent a conservative field. But electric field is a conservative field. Hence a field represented by these lines cannot be an electric field. (c) ID (b) (d) U (a) E3 A charge Q is fixed at a distance d in front of an infinite metal plate. The lines of force are represented by D YG Solution: (a) Metal plate acts as an equipotential surface, therefore the field lines should act normal to the surface of the metal plate. Relation Between Electric Field and Potential. In an electric field rate of change of potential with distance is known as potential gradient. It is a vector quantity and it’s direction is opposite to that of electric field. Potential gradient relates with electric dV ; This relation dr +Q A volt gives another unit of electric field is. In the above meter relation negative sign indicates that in the direction of electric field potential decreases.  In space around a charge distribution we can also write E  E x ˆi  E y ˆj  E z kˆ field according to the following relation E   dr  E ST U B where E x   dV dV dV , Ey   and E z   dy dx dz dV , potential difference between any two points in an dr determined by knowing the boundary conditions With the help of formula E   electric dV    r2 r1 field E. dr   can  r2 r1 be E. dr cos . Electrostatics 39 For example: Suppose A, B and C are three points in an uniform electric field as shown in figure. (i) Potential difference between point A and B is  B A E  dr d 60 VB  V A   A Since displacement is in the direction of electric field, hence  = B C o So, VB  VA    B A E  dr cos 0    B A E  dr   Ed E3 0  In general we can say that in an uniform electric field E   + + – E – + – + + V1 = V – d ID Another example  E V2 = 0 V d U + – – V V or | E |  d d D YG (ii) Potential difference between points A and C is : VC  V A    C A E dr cos    E( AC ) cos    E( AB ) = – Ed Above relation proves that potential difference between A and B is equal to the potential difference between A and C i.e. points B and C are at same potential. Concept Negative of the slope of the V-r graph denotes intensity of electric field i.e. tan   U  + V  E r – ST E = V/d Example based on E = – dV/dr Example: 48 The electric field, at a distance of 20 cm from the centre of a dielectric sphere of radius 10 cm is 100 V/m. The ‘E’ at 3 cm distance from the centre of sphere is (a) 100 V/m (b) 125 V/m (c) 120 V/m Solution: (c) For dielectric sphere i.e. for non-conducting sphere E out  E out  100 KQ (20  10  2 )2  KQ = 100  (0.2)2 so E in  k.q r 2 (d) Zero and E in  100  (0. 2) 2  (3  10 2 ) 2 (10  10 2 )3 kqr R3 = 120 V/m 40 Electrostatics Example: 49 In x-y co-ordinate system if potential at a point P(x, y) is given by V  axy ; where a is a constant, if r is the distance of point P from origin then electric field at P is proportional to [RPMT 2000] dV dr Ex   Electric field at point P E  dV   ay , dx Ey   (d) r2 dV  ax dy E x2  E y2  a x 2  y 2  ar i.e., E  r E3 Solution: (a) By using E   (c) r—2 60 (b) r–1 (a) r Example: 50 The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volt. The electric field at the point (1m, 0, 2m) in volt/metre is (a) 8 along negative X-axis Solution: (a) By using E   8 along positive X-axis (d) 16 along positive Z- ID (c) 16 along negative X-axis axis (b) dV d  E   (4 x 2 )  8 x. Hence at point (1m, 0, 2m). E = – 8 volt/m i.e. 8 along dx dx U – ve x-axis. D YG Example: 51 The electric potential V is given as a function of distance x (metre) by V = (5x2 + 10x – 9) volt. Value of electric field at x = 1m is (a) – 20 V/m Solution: (a) By using E   at (b) 6 V/m dV ; dx E (c) 11 V/m (d) – 23 V/m d (5 x 2  10 x  9)  (10 x  10 ) , dx E  20 V /m x = 1m Example: 52 A uniform electric field having a magnitude E0 and direction along the positive X-axis exists. If the electric potential V, is zero at X = 0, then, its value at X = +x will be U (a) V(x)= +xE0 (V  V1 ) V  2 ; r (r2  r1 ) ST Solution: (b) By using E   (b) V(x)= – xE0 E0   (c) V(x)= x2E 0 (d) V(x)= – x2E0 {V (x )  0}  V(x) = – xE0 x 0 Example: 53 If the potential function is given by V = 4x + 3y, then the magnitude of electric field intensity at the point (2, 1) will be (a) 11 (b) 5 E x2  E y2 ; E x   Solution: (b) By using i.e., E  and Ey    E  (c) 7 dV d  (4 x  3 y )  4 dx dx dV d   (4 x  3 y)  3 dy dy (4 ) 2  ( 3 ) 2  5 N /C (d) 1 Electrostatics 41 60 Tricky example: 7 (a) 2.5 volt/m (b) – 2.5 volt/m 2 volt /m 5 (d)  3 2 1 0 2 volt /m 5 1 2 3 4 5 6 Distance R in metres ID (c) 5 4 E3 Potential in volts The variation of potential with distance R from a fixed point is as shown below. The electric field at R  5 m is (0  5) V  2.5 64 m Note :  At R = 1m , E   (5  0 ) V  2. 5 (2  0 ) m D YG E Potential in volts U Solution: (a) Intensity at 5 m is same as at any point between B and C because the slope of BC is same throughout (i.e. electric field between B and C is uniform). Therefore electric dV field at R = 5m is equal to the slope of line BC hence by 5E  A; B dr and at R  3 m potential is constant so E = 0. 4 3 2 1 C O 1 2 3 4 5 6 Distance R in metres Work Done in Displacing a Charge. U (1) Definition : If a charge Q displaced from one point to another point in electric field then work done in this process is W  Q  V where V = Potential difference between the two position of charge Q. ( V  E. r  Er cos  where  is the angle between direction of electric ST field and direction of motion of charge). (2) Work done in terms of rectangular component of E and r : If charge Q is given a displacement r  (r1ˆi  r2 ˆj  r3 kˆ ) in an electric field E  ( E1ˆi  E 2 ˆj  E 3 kˆ ). The work done is W  Q( E. r )  Q(E1 r1  E2 r2  E3 r3 ). Conservation of Electric Field. As electric field is conservation, work done and hence potential difference between two point is path independent and depends only on the position of points between. Which the charge is moved. I A II III B WI = WII = WIII 60 42 Electrostatics Concept  No work is done in moving a charge on an equipotential surface. II A B III E3 I WI = WII = W Examples based on work done 1 Qq 4  0 l Solution: (d) Since V A  VC  (b) kQ l (c) 1 Qql 4  0 (d) Zero A –q l D YG so W  q (VC  V A )  0 1 Qq 4  0 l 2 U (a) ID Example: 54 A charge (– q) and another charge (+Q) are kept at two points A and B respectively. Keeping the charge (+Q) fixed at B, the charge (– q) at A is moved to another point C such that ABC forms an equilateral triangle of side l. The network done in moving the charge (– q) is [MP PET 2001] B +Q l C l Example: 55 The work done in bringing a 20 coulomb charge from point A to point B for distance 0.2 m is 2 Joule. The potential difference between the two points will be (in volt) (a) 0.2 (b) 8 (c) 0.1 (d) 0.4 Solution: (c) W  Q.V  2 = 20  V  V = 0.1 volt ST U Example: 56 A charge +q is revolving around a stationary +Q in a circle of radius r. If the force between charges is F then the work done of this motion will be (a) F × r [CPMT 1975, 90, 91, 97; NCERT 1980, 83; EAMCET 1994; MP PET 1993, 95; MNR 1998; AIIMS 1997; DCE 1995; RPET 1998] (b) F  2r (c) F 2r (d) 0 Solution: (d) Since +q charge is moving on an equipotential surface so work done is zero. +Q +q Example: 57 Four equal charge Q are placed at the four corners of a body of side ‘a’ each. Work done in removing a charge – Q from its centre to infinity is Electrostatics 43 (a) 0 (b) 2 Q2 4  0 a (c) 2 Q2  0 a Solution: (c) We know that work done in moving a charge is W = QV Q 60 1 Q 4 2Q 2Q.   4  0 a / 2 4  0 a  0 a W So, B O –Q a 2Q 2  0 a D a C a Q Q E3 Also V0  4  Q a A W  Q (V0  V )  V  0  W = Q × V0 Here Q2 2 0 a (d) (b) at 9 VB  9  10  Potential VC  9  10 9  B due 100  10 40  10 at C 6 2 100  10 6 9   10 6 volt 4 9 J 25 charge (d)  (c) C +100 U Solution: (d) Potential 81 J 20 to C due to +100 9   10 6 volt 5 charge is + 100 C is 50  10  2 Hence work done in moving charge +5 C from B to C W  5  10 6 (VC  VB ) D YG A 9 J 4 50 cm 40 cm (a) 9 J ID Example: 58 Two point charge 100 C and 5 C are placed at point A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 C from B to C, where BC = 30  1 cm, angle ABC  and [MP PMT 1997]  9  10 9 Nm 2 /C 2 2 4  0 /2 B + 50 C 30 cm C 9 9 9  W  5  10 6   10 6   10 6    J 5 4 4   Example: 59 There is an electric field E in x-direction. If the work done in moving a charge 0.2 C through a distance of 2 metres along a line making an angle 60 o with the x-axis is 4J, what is the value of E [CBSE 1995] U (a) 4 N/C (b) 8 N/C (c) 3 N /C (d) 20 N/C Solution: (d) By using W  q  V and V  Er cos  A 0.2 C ST So, W  qE r cos  2m W  4 j  0. 2  E  2  cos 60 60o  E = 20 N/C x O Example: 60 An electric charge of 20 C is situated at the origin of X-Y co-ordinate system. The potential difference between the points. (5a, 0) and (– 3a, 4a) will be (a) a Solution: (c) V A   (b) 2a kQ kQ and VB  5a 5a V A  VB  0 (c) Zero (d) B (–3a, 4a) 5a A (5a, 0) Q 5a a 2 44 Electrostatics Example: 61 Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is q(Q1  Q 2 )( 2  1) 4 0 R 2 Solution: (b) Potential at the centre of first ring VA  60 (b) q(Q 1  Q 2 ) 2 4 0 R (c) E3 (a) Zero Q1 Q2  4  0 R 4  R 2  R 2 0 Q1 Q2 Q1  4  0 R 4 R 2  R 2 0 ID Potential at the centre of second ring VB  R 2 1 A B 4  0 R 2 R U q( 2  1)(Q1  Q 2 ) 4  0 R 2 D YG Tricky example: 8 R Q2 ( 2  1)(Q1  Q 2 ) Potential difference between the two centres V A  VB   Work done W  Q  q  1  ( 2  1) Q2  (d)  4  0 R 2 A point charge q moves from point A to point D along the path ABCD in a uniform electric field. If the co-ordinates of the points A, B, C and D are (a, b, 0), (2a, 0, 0), (a, – b, 0) and (0, 0, 0) then the work done by the electric field in this process will be Y  E (a) – qEa A (b) Zero (c) 2E (a + b)q D qEa 2b U (d) C ST Solution: (a) As electric field is a conservative field Y Hence the work done does not depend on path   E A (a,b,0) a2 + b2 b a D a2 + b2 W ABCD  W AOD  W AO  WOD = Fb cos 90o + Fa cos 180o = 0 + qEa (– 1)= – qEa Equilibrium of Charge. X B  O b   B a2 + b2 C X Electrostatics 45 (1) Definition : A charge is said to be in equilibrium, if net force acting on it is zero. A system of charges is said to be in equilibrium if each charge is separately in equilibrium. (2) Type of equilibrium : Equilibrium can be divided in following type: case of stable equilibrium 60 (i) Stable equilibrium : After displacing a charged particle from it's equilibrium position, if it returns back then it is said to be in stable equilibrium. If U is the potential energy then in d 2U is positive i.e., U is minimum. dx 2 E3 (ii) Unstable equilibrium : After displacing a charged particle from it's equilibrium position, if it never returns back then it is said to be in unstable equilibrium and in unstable d 2U equilibrium is negative i.e., U is maximum. dx 2 (iii) Neutral equilibrium : After displacing a charged particle from it's equilibrium ID position if it neither comes back, nor moves away but remains in the position in which it was d 2U is zero i.e., U is dx 2 kept it is said to be in neutral equilibrium and in neutral equilibrium U constant (3) Guidelines to check the equilibrium (i) Identify the charge for which equilibrium is to be analysed. D YG (ii) Check, how many forces acting on that particular charge. (iii) There should be atleast two forces acts oppositely on that charge. (iv) If magnitude of these forces are equal then charge is said to be in equilibrium then identify the nature of equilibrium. (v) If all the charges of system are in equilibrium then system is said to be in equilibrium (4) Different cases of equilibrium of charge U Case – 1 :Suppose three similar charge Case – 2 : Two similar charge Q1 and Q 2 Q 1 , q and Q 2 are placed along a straight are placed along a straight line at a A line as shown below F O 2 ST Q1 Charge q | F1 |  | F2 | i.e., B F1 q x1 dissimilar Acharge q is placed in between B F1 O F2 them asQshown below Q2 1 x2 q x x1 x2 x will x  Q1   1  Q2  x2  distance x from each other and a third Q2 be in equilibrium if Charge q will be in | F1 |  | F2 | 2 2 ; This is the condition of i.e., x  Q1   1 . Q2  x2  equilibrium if 46 Electrostatics e.g. if two charges +4C and +16 C are separated by a distance of 30 cm from each E3 60 equilibrium of charge q. After following Note :  Same short trick can be used here the guidelines we can say that charge q is to find the position of charge q as we in stable equilibrium and this system is not discussed in Case–1 i.e., in equilibrium x x and x 2  x1  x Note :  x 1  1  Q 1  Q 2 /Q1 1 /Q 2 1  Q 2 /Q1  It is very important to know that x magnitude of charge q can be determined if and x2  1  Q 1 /Q 2 one of the extreme charge (either Q1 or Q 2 ) is in equilibrium i.e. if Q2 is in 2 U ID x  equilibrium then | q |  Q 1  2  and if Q1 is other then for equilibrium a third charge  x  should be placed between them at a 2 x  in equilibrium then | q |  Q 2  1  (It should 30 distance or x1   10 cm  x  1  16 / 4 be remember that sign of q is opposite to that x 2  20 cm of Q 1 (or Q 2 ) ) Two dissimilar charge Q1 and Q 2 are placed along a straight line at a D YG Case – 3 : distance x from each other, a third charge q should be placed out side the line joining Q1 and Q 2 for it to experience zero net force. Q1 x – Q2 q d (Let |Q2| < |Q1|) Short Trick : U For it's equilibrium. Charge q lies on the side of chare which is smallest in magnitude x and d  Q 1 /Q 2  1 ST (5) Equilibrium of suspended charge in an electric field (i) Freely suspended charged particle : To suspend a charged a particle freely in air under the influence of electric field it’s downward weight should be balanced by upward electric force for example if a positive charge is suspended freely in an electric field as shown then  E or F = QE F = QE +Q +Q or – – – – – – – – – F = QE d V +Q mg + + + + + + + + + Q  ne  mg  E mg mg mgd  E V Electrostatics 47 Note :  In mg Q 60 In equilibrium QE  mg  E  the above case if direction of electric field is suddenly reversed in any figure then acceleration of charge particle at that instant will be a = 2g. E3 (ii) Charged particle suspended by a massless insulated string (like simple pendulum) : Consider a charged particle (like Bob) of mass m, having charge Q is suspended in an electric field as shown under the influence of electric field. It turned through an angle (say ) and comes in equilibrium. T sin   QE ….(i) T cos   mg ….(ii) ID So, in the position of equilibrium (O position) QE  U By squaring and adding equation (i) and (ii) T  2  l  E T cos   T T sin  QE O  mg  2 mg O D YG QE QE Dividing equation (i) by (ii) tan    θ  tan 1 mg mg (iii) Equilibrium of suspended point charge system : Suppose two small balls having charge +Q on each are suspended by two strings of equal length l. Then for equilibrium position as shown in figure. ….(i) T cos   mg ….(ii) U T sin   Fe T 2  Fe   mg  2 l T cos  T +Q x  T sin  ST 2   l and tan   +Q Fe mg Fe 1 Q2 x ; here Fe  and  l sin  mg 4  0 x 2 2 (iv) Equilibrium of suspended point charge system in a liquid : In the previous discussion if point charge system is taken into a liquid of density  such that  remain same then Fe' Q2   tan   (mg  Vg) 4  0 K (mg  Vg)x 2   In equilibrium Fe'  T ' sin  and (mg  Vg)  T ' cos  l l T +Q x T sin   T cos  +Q (mg – Vg) F e 48 Electrostatics When this system was in air tan   If  is the density of material of ball then K  QE Q 1 ρ   1  σ    Examples based on equilibrium of charge ID mg 60 1 1 m 1  K  m k (m  V ) m  V  V  1    m   E3  So equating these two gives us Fe Q2  mg 4  0 mgx 2 Example: 62 A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium. If q is equal to (a)  Q 2 U [CPMT 1999; MP PET 1999, MP PMT 1999; CBSE 1995; Bihar MEE 1995; IIT 1987] (b)  Q 4 2 D YG  x /2 Solution: (b) By using Tricky formula q  Q    x  (c)  Q Q since q should be negative so q  . 4 4 ST U  q Q 4 (d)  Q 2 46 Electrostatics Example: 63 Two point charges +4q and +q are placed at a distance L apart. A third charge Q is so placed that all the three charges are in equilibrium. Then location and magnitude of third charge will be [IIT-JEE 1975] (b) At a distance L 4q from +4q charge,  3 9 (c) At a distance 2L 4q from +4q charge,  3 9 (d) At a distance 2L 4q from +q charge,  3 9 60 L 4q from +4q charge, 3 9 E3 (a) At a distance Solution: (c) Let third charge be placed at a distance x 1 from +4q charge as shown 1  q 4q 2L L  x2  3 3 Q +q +4q ID L Now x 1  x1 x2 L 2 U 4q 4q L/3 For equilibrium of q, Q  4 q.  Q   9 9  L  D YG Example: 64 A drop of 10 6 kg water carries 10 6 C charge. What electric field should be applied to balance it’s weight (assume g = 10 m/sec2) (a) 10 V / m, Upward (b) 10 V / m, Downward (c) 0.1 V/m Downward (d) 0.1V / m, Upward Solution: (a) In equilibrium QE = mg E mg 10 6  10 = = 10 V/m; Since charge is positive so electric field will be upward. Q 10 6 Example: 65 A charged water drop of radii 0.1 m is under equilibrium in some electric field. The charge U on the drop is equivalent to electronic charge. The intensity of electric field is ST (a) 1.61 N / C (b) 25.2 N / C Solution: (c) In equilibrium QE = mg ; (c) 262 N / C (d) 1610 N / C 4 3   r  .g 6 3 3 mg 3   4  (3.14 ) (0.1  10 )  10  10 = 262 N/C E  3 Q Q 1. 6  10 19 Example: 66 The bob of a pendulum of mass 8 g carries an electric charge of 39.2  10 10 coulomb in an electric field of 20  10 3 volt / meter and it is at rest. The angle made by the pendulum with the vertical will be (a) 27o (b) 45o (c) 87o (d) 127o Solution: (b) T sin  qE , T cos  mg  qE tan   mg  E  T cos  T  T sin  qE mg Electrostatics 47 tan   39. 2  10 10  20  10 3 8  10 6  9.8 1    45 o Example: 67 Two small spherical balls each carrying a charge Q = 10 C (10 micro-coulomb) are 60 suspended by two insulating threads of equal lengths 1 m each, from a point fixed in the ceiling. It is found that in equilibrium threads are separated by an angle 60 o between them, 1 as shown in the figure. What is the tension in the threads. (Given :  9  10 9 Nm / C 2 ) 4 0  E3 (a) 18 N (b) 1.8 N 60 o (c) 0.18 N Q Solution: (b) From the geometry of figure r = 1m ID (d) None of these 1 (10  10 6 ) 2 T   9  10 9. 2 12 o 30o 30 1m 1m T +10  C D YG  T= 1.8 N U In the condition of equilibrium T sin 30 o  Fe Q 30o Fe +10  C mg T sin 30o r T cos 30o Example: 68 Two similar balloons filled with helium gas are tied to L m long strings. A body of mass m is tied to another ends of the strings. The balloons float on air at distance r. If the amount of charge on the balloons is same then the magnitude of charge on each balloon will be 1/2  2k  tan   (b)  2  mgr  1/2 r Q Q U  mgr 2  tan   (a)   2k  ST  mgr  (c)  cot    2k  L 1/ 2  2k  tan   (d)  mgr   m 1/2 Solution: (a) In equilibrium 2 R  mg …. (i) Fe  T sin …. (ii) L   R  T cos  From equation (i) and (iii) R r Fe T sin  2T cos  mg …. (iv) Dividing equation (ii) by equation (iv) R …. (iii) T T cos  L   m mg L 48 Electrostatics k Q2 1/2 60 2  mgr 2  F 1 1 tan   tan   e  tan   r     2 mg 2 mg  2k  Time Period of Oscillation of a Charged Body. (1) Simple pendulum based : If a simple pendulum having length l and mass of bob m l g ID E3 oscillates about it's mean position than it's time period of oscillation T  2 π D YG l O QE mg mg  U O On displacing the bob from it’s equilibrium ST position 0. It will oscillate under the effective acceleration g, where mg'  Effective acceleration g'  g  QE / m  T2  2 π l g  QE/m  mg + QE T2 < T Case – 3 : In case 2 if electric field is applied in upward direction then, effective acceleration. g'  g  QE / m l  E l T3  2 π g  QE/m  T3 > T Case – 4 if T3  Hence the new time period is T1  2 l 2 QE  So new time period mg  g'  g 2  QE / m 2 g T  E mg 2  QE 2 T1  2 π l So new time period  E d  O Case – 2 : If electric field is applied in the downward direction then. U Case – 1 : If some charge say +Q is given to bob and an electric field E is applied in the direction as shown in figure then equilibrium position of charged bob (point charge) changes from O to O. l  QE/m 2 Since g' >g, hence T1 < T l g' : In the case 3, T i.e., 2 2 1 l  2 2 g l QE O g  QE / m   QE = 3 mg  E mg  1 2 i.e., effective vertical force (gravity + electric) Electrostatics 49 on the bob = mg – 3 mg = – 2 mg, hence the equilibrium position O of the bob will be above the point of suspension and bob will oscillate under on effective acceleration 2g directed upward. i.e. time period of pendulum will decrease. l Hence new time period T4  2 π , T4 < T 60 2g E3 (2) Charged circular ring : A thin stationary ring of radius R has a positive charge +Q unit. If a negative charge – q (mass m) is placed at a small distance x from the centre. Then motion of the particle will be simple harmonic motion. 1 Qx. Electric field at the location of – q charge E  3 4  0 x2  R2 2 + + + +Q + + 1 Qx + Since x a i.e. point P must lies on negative x-axis as shown at a distance x from origin a hence x  d  a   a  2 a. Actually P lies on negative x-axis so x   2 a. 2 1 D YG Example: 71 Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero is (a) charge (c) charge  r  from 9e charge (b) 1  3  from 3e charge (d) 3 1 r r 1 1 3 r 1 1 3 from 9e from 3e U Solution: (b) Suppose neutral point is obtained at a distance x 1 from charge 9e and x 2 from charge 3e ST By using x 1  1 x Q2 Q1 r = 1  3e 9e r N 9e  1  1   3  x1 3e x2 r Example: 72 Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained at (a) A distance of (b) A distance of (c) A distance of R ( 2  1) R ( 2  1) R ( 2  1) from – Q charge and lies between the charges. from – Q charge on the left side of it from 2Q charge on the right side of it 52 Electrostatics (d) A point on the line which passes perpendicularly through the centre of the line joining – Q and 2Q charge. Solution: (b) As already we discussed neutral point will be obtained on the side of charge which is smaller in magnitude i.e. it will obtained on the left side of – Q charge and at a distance. 2Q 1 Q l R ( 2  1) 60 R l Example: 73 A charge of + 4C is kept at a distance of 50 cm from a charge of – 6C. Find the two points where the potential is zero E3 (a) Internal point lies at a distance of 20 cm from 4C charge and external point lies at a distance of 100 cm from 4C charge. (b) Internal point lies at a distance of 30 cm from 4C charge and external point lies at a distance of 100 cm from 4C charge (c) Potential is zero only at 20 cm from 4C charge between the two charges x  Q2    1   Q1   point x1  X, x 50   20 cm  Q2  6 1   1  4  Q1  50  100 cm 6 1 4 and Y for external X 4C point – 6C 20cm 50c m 100cm D YG x1  internal U Solution: (a) For ID (d) Potential is zero only at 20 cm from – 6C charge between the two charges Tricky example: 9 Two equal negative charges – q are fixed at points (0, a) and (0, – a) on the y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will [IIT-JEE 1984, Bihar MEE 1995, MP PMT 1996] (a) Execute simple harmonic motion about the origin U (b) Move to the origin and remains at rest (c) Move to infinity ST (d) Execute oscillatory but not simple harmonic motion. Solution: (d) By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. A If at any time charge Q –q is at a distance x from O. F  2 F cos   2 1 qQ a x  4 0 (a 2  x 2 ) (a 2  x 2 )1 F i.e., 1 4 0. a 2 x 2qQx 2  x2  (with amplitude 2a) but not 2a a 3 2 As the restoring force F is not linear, motion will be oscillatory  O simple –q B Q C Y, Electrostatics 53 harmonic. Electric Potential Energy. 60 (1) Potential energy of a charge : Work done in bringing the given charge from infinity to E3 a point in the electric field is known as potential energy of the charge. Potential can also be W U V . written as potential energy per unit charge. i.e. Q Q W  U2  1 40. Q1Q2 r Q1 Q2 A B r U So, ID (2) Potential energy of a system of two charges : Since work done in bringing charge Q2 1 Q1 from  to point B is W  Q2 VB , where VB is potential of point B due to charge Q1 i.e. VB  4 0 r D YG This is the potential energy of charge Q2, similarly potential energy of charge Q1 will be 1 Q1Q2 U1 . 40 r Hence potential energy of Q1 = Potential energy of Q2 = potential energy of system U  k C.G.S. U  Q1Q2 (in r Q1Q2 ) r Note :  Electric potential energy is a scalar quantity so in the above formula take sign of Q1 and Q2. U (3) Potential energy of a system of n charges : In a system of n charges electric potential energy is calculated for each pair and then all energies so obtained are added algebraically. i.e.  Q1Q2 Q2Q3   .........  and in case of continuous distribution of charge. As dU  dQ.V  40  r12 r23  1 ST U  Q1 U  V dQ e.g. Electric potential energy for a system of three charges Potential energy   Q1Q2 Q2Q3 Q3 Q1      40  r12 r23 r31  r12 r31 1 While potential energy of any of the charge say Q1 is Q2  Q1Q2 Q3 Q1     40  r12 r31  1 r23 Q3  54 Electrostatics Note :  For the expression of total potential energy of a system of n charges consider n (n  1) 2 number of pair of charges. (4) Electron volt (eV) : It is the smallest practical unit of energy used in atomic and 60 nuclear physics. As electron volt is defined as “the energy acquired by a particle having one 1J  1.6  10 19 J = 1.6  quantum of charge 1e when accelerated by 1volt” i.e. 1eV  1.6  10 19 C  C 10–12 erg E3 Energy acquired by a charged particle in eV when it is accelerated by V volt is E = (charge in quanta) × (p.d. in volt) Commonly asked examples : Charge Accelerated by p.d. Gain in K.E. (i) Proton 5  104 V K = e  5  104 V = 5  104 eV = 8  10–15 J (ii) Electron 100 V K = e  100 V = 100 eV = 1.6  10–17 J ID S.No. [JIPMER 1999] [MP PMT 2000; AFMC Proton (iv) 0.5 C (v) - K = e  1 V = 1 eV = 1.6  10–19 J 1V 2000 V 106 V [CBSE 1999] K = 0.5  2000 = 1000 J [JIPMER 2002] K = (2e)  106 V = 2 MeV [MP PET/PMT 1998] D YG (iii) U 1999] particle (5) Electric potential energy of a uniformly charged sphere : Consider a uniformly charged sphere of radius R having a total charge Q. The electric potential energy of this sphere is equal to the work done in bringing the charges from infinity to assemble the sphere. U U 3Q 2 20 0 R ST (6) Electric potential energy of a uniformly charged thin spherical shell :