Electrostatics PDF

10. Electrostatics Can you recall? 1. Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend? 2. Ever heard a crackling sound while taking out your sweater in winter? 3. Have you seen the l...

10. Electrostatics Can you recall? 1. Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend? 2. Ever heard a crackling sound while taking out your sweater in winter? 3. Have you seen the lightning striking during pre-monsoon weather? 10.1 Introduction: the nucleus so as to make an atom electrically Electrostatics deals with static electric neutral. Thus, most matter around us is charges, the forces between them and the electrically neutral. effects produced in the form of electric fields and electric potentials. We have already studied some aspects of electrostatics in earlier standards. In this Chapter we will review some of them and then go on to study some aspects in details. Current electricity, which plays a major role in our day to day life, is produced by moving Fig. 10.1 a Fig. 10.1 b charges. Charges are present everywhere around us though their presence can only be felt under special circumstances. For example, when we remove our sweater in winter on a dry day, we hear some crackling sound and the sweater appears to stick to our body. This is because of the electric charges produced due to friction between our body and the sweater. Similarly, the lightening that we see in the sky is also due Fig. 10.1 c Fig. 10.1 d to the flow of large amount of electric charges Fig. 10.1 (a): Insulated conductor that develop on the clouds due to friction. Fig. 10.1 (b): +ve charge is neutralized by 10.2 Electric Charges: electron from Earth Historically, opposite electric charges Fig. 10.1 (c): Earthing is removed -ve were known to the Greeks in the 600 BC. charge still stays on the conductor due to They realized that equal and opposite charges +ve charged rod develop on amber and fur when rubbed against Fig. 10.1 (d): Rod removed -ve charge is each other. Now we know that electric charge distributed over the surface of the conductor is a basic property of elementary particles of When certain dissimilar substances, like which matter is made of. These elementary fur and amber or comb and dry hair are rubbed particles are proton, neutron, and electron. against each other, electrons get transferred Atoms are made of these particles and matter is to the other substance making them charged. made from atoms. A proton is considered to be The substance receiving electrons develops a positively charged and electron to be negatively negative charge while the other is left with an charged. Neutron is electrically neutral, i.e., it equal amount of positive charge. This can be has no charge. An atomic nucleus is made up called charging by conduction as charges are of protons and neutrons and hence is positively transfered from one body to another. Charges charged. Negatively charged electrons surround can be separated by other means as well, like 188 chemical reactions (in cells), convection (in clouds), diffusion (in living cells) etc. Gold Leaf Electroscope: If an uncharged conductor is brought near This is a classic instrument for detecting a charged body, (not in physical contact) the presences of electric charge. A metal disc nearer side of the conductor develops opposite is connected to one end of a narrow metal charge to that on the charged body and the far rod and a thin piece of gold leaf is fixed to side of the conductor develops charge similar the other end. The whole of this part of the to that on the charged body. This is called electroscope is insulted from the body of induction. This happens because the electrons the instrument. A glass front prevents air in a conductor are free and can move easily in draughts but allows to observe the effect of presence of a charged body. This can be seen charge on the leaf. from Fig. 10.1. When a charge is put on the disc at the A charged body attracts or repels electrons top it spreads down to the plate and leaf in a conductor depending on whether the charge moves away from the plate. This happens on the body is positive or negative respectively. because similar charges repel. The more the Positive and negative charges are redistributed charge on the disc, more is the separation of and are accumulated at the ends of the conductor the leaf from the plate. near and away from the changed body. From The leaf can be made to fall again by the above discussion it can be inferred that touching the disc. This is done by earthing there are only two types of charges found in the electroscope. An earth terminal prevents nature, namely, positive and negative charges. the case from accumulating any stray In induction, there is no transfer of charges charge. The electroscope can be charged in between the charged body and the conductor. two ways. So when the charged body is moved away from (a) by contact- a charged rod is brought the conductor, the charges in the conductor are in contract with the disc and charge free again. is transferred to the electroscope. This method gives the gold leaf the same Can you tell? charge as that on the conductor. This is 1. When a petrol or a diesel tanker is not a very effective method of charging emptied in a tank, it is grounded. the electroscope. 2. A thick chain hangs from a petrol or a (b) by induction- a charged rod is brought diesel tanker and it is in contact with close to the disc (not touching it) and ground when the tanker is moving. the electroscope is earthed. The rod is then removed. This method give the 10.3 Basic Properties of Electric Charge: gold leaf opposite charges. 10.3.1 Additive Nature of Charge: The following diagrams show how the Electric charge is additive, similar to mass. charges spread to the gold leaf and lift it. The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on different parts of the object. It may be pointed out that while taking the algebraic sum, the sign (positive or negative) of the electric charges must be taken into account. Thus if two bodies have equal and opposite charges, the net charge on the system of the two bodies is zero. This is similar to that in case of atoms where the nucleus is positively charged and this charge is equal to the negative charge 189 of the electrons making the atoms electrically Example 10.1: How much positive and neutral. negative charge is present in 1gm of water? It is interesting to compare the additive How many electrons are present in it? Given, property of charge with that of mass. molecular mass of water is 18.0 g. 1) The masses of the particles constituting Solution: Molecular mass of water is 18.0 gm, an object are always positive, whereas the that means the number of molecules in 18.0 gm charges distributed on different parts of the of water is 6.02×1023. abject may be positive or negative.... Number of molecules in 1gm of water 2) The total mass of an object is always = 6.02×1023/18. One molecule of water (H2O) positive whereas, the total charge on the contains two hydrogen atoms and one oxygen object may be positive, zero or negative. atom. Thus the number of electrons in H2O 10.3.2 Quantization of Charge: is sum of the number of electrons in H2 and oxygen. There are 2 electrons in H2 and 8 The minimum value of the charge on an electrons oxygen. electron as determined by the Milikan's oil drop experiment is e = 1.6×10-19 C. This is called the ∴Number of electrons in H2O = 2+8 = 10. elementary charge. Here, C stands for coulomb Total number of protons / electrons in 1.0 which is the unit of charge in SI system. Unit of 6.02  1023 gm of water   10  3.34  1023 charge is defind in article 10.4.3. Since protons 18 (+ve) and electrons (-ve) are the charged Total positive charge = 3.34×1023 × charge particles constituting matter, the charge on on a proton an object must be an integral multiple of ±e. = 3.34×1023 ×1.6×10-19 C = 5.35×104 C q = ± ne, where n is an integer. This positive charge is balanced by equal Further, charge on an object can be amount of negative charge so that the water increased or decreased in multiples of e. It molecule is electrically neutral. is because, during the charging process an integral number of electrons can be transferred Do you know ? from one body to the other body. This is known According to recent advancement in physics, as quantization of charge or discrete nature of it is now believed that protons and neutrons charge. are themselves built out of more elementary The discrete nature of electric charge is units called quarks. They are of six types, usually not observable in practice. It is because having fractional charge (-1/3)e or +(2/3)e. A the magnitude of the elementary electric charge, proton or a neutron consists of a combination e, is extremely small. Due to this, the number of three quarks. It may be clearly understood of elementary charges involved in charging an that even in the quark model, quantization of object becomes extremely large. Suppose, for charge is not affected. It is only the step size example, when a glass rod is rubbed with silk, of the charge that decreases from e to e/3. a charge of the order of one µC (10-6 C) appears Quarks are always present in bound states on the glass rod or silk. Since elementary charge and no free quarks are known to exist. e = 1.6×10-19 C, the number of elementary In modern day experiments it is possible to charges on the glass rod (or silk) is given by observe the discrete nature of charge in very 106 C sensitive divides such as single electron n=  6.25  1012 1.6  1019 C transistor Since it is a tremendously large number, 10.3.3 Conservation of Charge: the quantization of charge is not observed and We know that when a glass rod is rubbed one usually observes a continuous variation of with silk, it becomes positively charged and charge. silk becomes negatively charged. The amount 190 of positive charge on glass rod is found to be proportional to the square of the distance exactly the same as negative charge on silk. between them. This force acts along the line Thus, the systems of glass rod and silk together joining the two charges. possesses zero net charge after rubbing. Let q1 and q2 be two point charges at Result and conclusion of this experiment rest with respect to each other and separated can be generalized and we can say that "in by a distance r. The magnitude F of the force any given physical process, charge may get between them is given by, transferred from one part of the system to qq another, but the total charge in the system Fα 122 r remains constant" or, for an isolated system q1q2 total charge cannot be created nor destroyed. F=K 2 In simple words, the total charge of an isolated r --- (10.1) system is always conserved. where K is the constant of proportionality. Its 10.3.4 Forces between Charges: magnitude depends on the units in which F, q1, It was observed in carefully conducted q2 and r are expressed and also on the properties experiments with charged objects that they of the medium around the charges. experience force when brought close (not The force between the two charges will be touching) to each other. This force can be attractive if they are unlike (one positive and one attractive or repulsive. Like charges repel negative). The force will be repulsive if charges each other and unlike charges attract each are similar (both positive or both negative). other. Figure 10.2 describes this schematically. Figure 10.3 describes this schematically. This is the reason for charging by induction as described in section 10.2 and Fig. 10.1. Fig. 10.2: Attractive and repulsive force. 10.4 Coulomb’s Law: The electric interaction between two charged bodies can be expressed in terms of the forces they exert on each other. Coulomb (1736- 1806) made the first quantitative investigation of the force between electric charges. He used Fig. 10.3: Coulomb’s law. point charges at rest to study the interaction. 10.4.2   Relative Permittivity or Dielectric A point charge is a charge whose dimensions Constant: are negligibly small compared to its distance While discussing the coulomb’s law it was from another bodies. Coulomb’s law is a assumed that the charges are held stationery fundamental law governing interaction in vacuum. When the charges are kept in between charges at rest. a material medium, such as water, mica or 10.4.1 Scalar form of Coulomb’s Law: parafined paper, the medium affects the force Statement : The force of attraction or between the charges. The force between the two repulsion between two point charges at rest charges placed in a medium may be written as, is directly proportional to the product of 1  q1q 2  Fmed =  r 2  --- (10.2) the magnitude of the charges and inversely 4   191 where ε is called the absolute permittivity of force between charges by a factor of εr, its the medium. The force between the same two relative permittivity. charges placed in free space or vacuum at While using Eq. (10.5) we assume that distance r is given by, the medium is homogeneous, isotropic and 1  q1q 2  infinitely large. Fvac =   --- (10.3) 10.4.3 Definition of Unit Charge from the 4 0  r  2 Dividing Eq. (10.3) by (10.2) Coulomb’s Law: 1  q1q 2  The force between two point charges q1 and q2, separated by a distance r in free space, is Fvac 4 0  r   2 = = written by using Eq. (10.2), Fmed 1  q1q 2   0 4  r  2 F 1 qq qq  1 2 2  9  109  1 2 2 4 0 r r ε The ratio is the relative permittivity ε0 = 8.85 Í10 C N m -12 2 -1 -2 ε0 or dielectric constant of the medium and is If q1= q2= 1C and r = 1.0 m denoted by εr or K. Then F = 9.0Í109 N  F From this, we define, coulomb (C) the unit K or  r   vac --- (10.4)  0 Fmed of charge in SI units. Thus, One coulomb is the amount of charge which, when placed at a distance of one (i) εr is the ratio of absolute permittivity of a metre from another charge of the same medium to the permittivity of free space. magnitude in vacuum, experiences a force of (ii) εr is the ratio of the force between two point 9.0 × 109 N. This force is a tremendously large charges placed a certain distance apart in force realisable in practical situations. It is, free space or vacuum to the force between therefore, necessary to express the charge in the same two point charges when placed at smaller units for practical purpose. Subunits the same distance in the given medium. of coulomb are used in electrostatics. For εr is a dimensionless quantity. example, micro-coulomb (10-6 C, µC), nano- (iii) εr is also called specific inductive capacity. coulomb (10 C, nC) or pico-coulomb. (10 C, -9 -12 The force between two point charges q pC) are normally used units. 1 and q2 placed at a distance r in a medium of Do you know ? relative permittivity εr, is given by 1 q1q 2 Force between two charges of 1.0 C each, F --- (10.5) separated by a distance of 1.0 m is 9.0×109 N 4 0 r r 2 or, about 10 million metric tonne. A normal For water, εr = 80 then from Eq. (10.4) truck-load is about 10 metric tonne. So, this Fvac force is equivalent to about one million truck-   r  80 Fwater loads. A tremendously large force indeed ! Fvac Example 10.2: Charge on an electron is Fwater  80 1.6×10-19 C. How many electrons are required This means that when two point charges to accumulate a charge of one coulomb? are placed some distance apart in water, the Solution: 1.6Í10-19 C = 1 electron th force between them is reduced to  1  of the 1C  1 electrons  80  1.6  1019 force between the same two charges placed at  0.625  1019  6.25  1018 electron ns the same distance in vacuum. 6.25×1018 electrons are required to Thus, a material medium reduces the accumulate a charge of one coulomb. 192  It is now possible to measure a very Here, q1 = q2 = +1.6×10-19 C, r = 10-15m small amount of current in otto-amperes (1.6  1019 ) (1.6  1019 )  Fe  9  109 which measures flow of single electron. (1015 ) 2 10.4.4 Coulomb’s Law in Vector Form:  9  1.6  1.66  101 N As shown in Fig 10.4, q1 and q2 are two Fe  2.3  102 N --- (10.8.a) similar point charges situated at points A and B. The gravitational force between the r12 is the distance of separation between them. protons is given by F 21 denotes the force exerted on q2 by q1 mm Fg = G 1 2 2  r 1 q1q 2  F21    r 21 --- (10.6) 6.674  1011  1.67  1027  1.67  1027 4 0 r 21 2  (1015 ) 2 Fg  1.86  1034 N --- (10.8.b) Comparing 10.8. (a) and 10.8.(b) Fe 2.30  102 N  = 1.23  1036 Fg 1.86  1034 N 1 Thus, the electrostatic force is about 36 orders of magnitude stronger than the gravitational force. Fig. 10.4: Coulomb’s law in vector form r 21 is the unit vector along AB , away Comparison of gravitational and  electrostatic forces: from B. Similarly, the force F12 exerted on q1 Similarities by q2 is given by 1. Both forces obey inverse square law :  1 qq 1 F12   1 22  r12 --- (10.7) F∝ 2 4 0 r12 r 2. Both are central forces : act along the r12 line joining the two objects.  is the unit vector along BA , away from Differences A. F12 acts on q1at A and is directed along BA, 1. Gravitational force between two objects away from A. The unit vectors r12 and r 21 are is always attractive while electrostatic oppositely directed i.e., r12   r 21 hence, force between two charges can be   F 21 = - F12 either attractive or repulsive depending Thus, the two charges experience force on the nature of charges. of equal magnitude and opposite in direction. 2. Gravitational force is about 36 These orders of magnitude weaker than the  two forces  form an action- reaction pair. electrostatic force. As F 21 and F12 act along the line joining the two charges, the electrostatic force is a 10.5 Principle of Superposition: central force. The principle of superposition states that Example 10.3: Calculate and compare the when a number of charges are interacting, electrostatic and gravitational forces between the resultant force on a particular charge is two protons which are 10-15 m apart. Value of given by the vector sum of the forces exerted by G = 6.674×10-11 m3 kg-1 s-2 and mass of the individual charges. proton is 1.67×10-27 kg Consider a number of point charges q1, q2, Solution: The electrostatic force between the q3 ------- kept at points A1, A2, A3--- as shown 1 q1q 2 in Fig. 10.5. The force exerted  on the charge protons is given by Fe   4 0 r 2 q1 by q2 is F12. The value of F12 is calculated 193 by ignoring the presence   of other charges.  F AC Similarly, we find F13 , F14 etc, one at a time,  2µ C 3µ C B 4 cm F AB using the coulomb’s law. A θ F14 3 cm F15 F13 Fig. a: Position of charges. 4µC C q2 q1 F12 Solution : Given, A2 A1 AB = 4.0 cm, BC = 3.0 cm q3 q5  AC = 4 2  32  5.0 cm q4 A5  A3 Magnitude of force F AB on A due to B is, A4   1  2  106  3  106 F AB     4 0  (4  10 ) 2 2 Fig. 10.5: Principle  of superposition. Total force F 1 on charge q1 is the vector 9  109  6 sum of all such forces.  4  1012     16  10 F1  F12  F13  F14 ... = 3.37  10   = 33.7 N  1  q1q 2 r12  q1q 3 r13 ... , 4 0  2  2  This force acts at point A and is directed  r12 r13  along BA (Fig. (b)).    where r12 , r13 etc., are unit vectors directed to F  q1 from q2, q3 etc., and r12, r13, r14,etc., are the F AC distances from q1 to q2, q3 etc respectively. Let there be N point charges q1, q2,q3  A F AB etc., qN. The force F exerted by these charges on a test charge q0 can be written using the Fig. b: Forces acting at point A.  summation  notation    Σ as follows,  Magnitude of force F AC on A due to C is, F test  F1  F2  F3      F N --- (10.9)   1  2  106  4  106 N 1 N q 0q n  F      Fn =  AC r 0n --- (10.10) 4 0  (5  102 ) 2 4 0 n=1 r0 n2  n=1 9  109  8.0  1012 Where r 0n is a unit vector directed from the  25  104 nth charge to the test charge qo and r0n is the  72 separation between them, r 0 n = r0 n r 0n   10  28.8 N 25 Can you tell? This force acts at point A and is directed Three charges, q each, are placed at the along CA. (Fig. 10.6.(b)) vertices of an equilateral triangle. What will    be the resultant force on charge q placed at F = F AB + F AC the centroid of the triangle? Magnitude of resultant force is, Example 10.4: Three charges of 2µC, 3µC F =  FAC2  FAB2  2 FAC  FAB  cos   1/ 2 and 4µC are placed at points A, B and C respectively, as shown in Fig. a. Determine the  59.3 N force on A due to other charges. 194  Direction of the resultant force is 16.9° electric field of a charge is also a vector and north of west. (Fig. c) is directed along the direction of the coulomb N force, experienced by a test charge. The magnitude of electric field at a distance  F r from a point charge Q is same at all points on  the surface of a sphere of radius r as shown in F AC Fig. 10.6. Its direction is along the radius of 16.9° the sphere, pointing away from its centre if the W  A F AB charge is positive.  Fig. c: Direction of the resultant force. E 10.6 Electric Field: Space around a charge Q gets modified so Q that when a test charge is brought in this region, E 4 0 r 2 it experiences a coulomb force. This region + around a charged object in which coulomb force is experienced by another charge is called electric field. Fig. 10.6: Electric field due to a point charge Mathematically, electric field is defined as (+Q). the force experienced per unit charge. Let Q and q be two charges separated by a distance r. SI unit of electric intensity is newton per coulomb (NC-1). Practically, electric field The coulomb force between them is given  1 Qq  is expressed in volt per metre(Vm-1). This is by F   r , where, r is the unit vector discussed in article 10.6.2. 4 0 r 2 Dimensional formula of E is, along the line joining Q to q. F [LMT  2 ] Therefore, electric field due to charge Q is E =  q0 [IT] given by,   3 E = [LMT I 1 ]  F Q  E  r --- (10.11) 10.6.1 Electric Field Intensity due to a Point q 4 0 r 2 Charge in a Material Medium: The coulomb force acts across an empty space (vacuum) and does not need any Consider a point charge q placed at point O intervening medium for its transmission. in a medium of dielectric constant K as shown in Fig. 10. 7. The electric field exists around a charge irrespective of the presence of other charges. Since the coulomb force is a vector, the q 0 A precise definition of electric field is: k Electric field is the force experienced by a test charge in presence of the given charge at q the given distance from it.  F Fig. 10.7: Field in a material medium. E  lim q 0 q Consider the point P in the electric field of Test charge is a positive charge so point charge q at distance r from it. A test charge small in magnitude that it dose not affect the q0 placed at the point P will experience a force surroundings of the given charge. which is given by the Coulomb’s law, 195   1 q q0  F r 4 0 K r 2 where r is the unit vector in the direction of force i.e., along OP. Fig. 10.9 (a): By the definition of electric field intensity uniform electric  field.  F 1 q  E  r q0 4 0 K r 2  The direction of E will be along OP when q is positive and along PO when q is negative. Fig. 10.9 (b): non uniform The magnitude of electric field intensity in electric field. a medium is given by 1 q E --- (10.12 ) 10.6.2 Practical Way of Calculating Electric 4 0 K r 2 Field For air or vacuum K = 1 then A pair of charged parallel plates is 1 q arranged as shown in Fig. 10.10. The electric E= 4 0 r 2 field between them is uniform. A potential The coulomb force between two charges difference V is applied between two parallel and electric field E of a charge both follow the plates separated by a distance ‘d’. The electric inverse square law, (F∝1/r2, E∝1/r2) Fig. 10.8. field between them is directed from plate A to plate B as shown. F∝1/r2, E∝1/r2 E A B Fig 10.10: Electric field or between two parallel F plates. r A charge +q placed between the plates Fig. 10.8: Variation of Coulomb force/ experiences a force F due to the electric field. If Electric field due to a point charge. we have to move the charge against the direction of field, i.e., towards the positive plate, we have 1. Uniform electric field: A uniform electric to do some work on it. If we move the charge field is a field whose magnitude and +q from the negative plate B to the positive direction is same at all points. For example, plate A, the work done against the field is field between two parallel plates. Fig 10.9.a W = Fd; where ‘d’ is the separation between the 2. Non uniform electric field: A field whose plates. The potential difference V between the magnitude and direction is not the same at two plates is given by all points. For example, field due to a point W = Vq, but W = Fd charge. In this case, the magnitude of field is same at distance r from the point charge ∴Vq = Fd ∴ F/q = V/d = E in any direction but the direction of the ∴ Electric field can be defined as field is not same. Fig 10.9.b E = V/d --- (10.13) This is the commonly used definition of electric field. 196 Example 10.5: Gap between two electrodes of magnitudes   and are opposite in direction,   the spark-plug used in an automobile engine E A = - EC. E A + EC = 0. Thus, the field at P is is 1.25 mm. If the potential of 20 V is applied only to the charge at B and can be written as across the gap, what will be the magnitude of   2  106 Ep  EB  electric field between the electrodes? 4 0 (BP) 2  2  106  9  109 Solution: Ep  (5 / 2 ) 2 V E 2  9  103  2 d  25 20V V 36 E  16  103  1.6  104   103 1.25  103 m m 25   1.44  1015 NC-1 along BP This electric field is sufficient to ionize the gaseous mixture of fuel compressed in the To calculate BP  5 cylinder and ignite it. BP  (BA) cos (45)  2 Can you tell? Example 10.7: A simplified model of hydrogen atom consists of an electron revolving about a Why a small voltage can produce a proton at a distance of 5.3×10-11m. The charge reasonably large electric field? on a proton is +1.6×10-19 C. Calculate the Example 10.6: Three point charges are placed intensity of the electric field due to proton at at the vertices of a right isosceles triangle as this distance. shown in the Fig. a. What is the magnitude and Solution: direction of the resultant electric field at point P q E which is the mid point of its hypotenuse? 4 o r 2 C  +10 µC  EB q  1.6  1019 C EA r  5.3  1011 m P 1 5cm  9.0  109 Nm2C 1  4 o EC 1.6  1019 E  9.0  109  5.1 1011 NC 1 +2 µC +10 µC (5.3  1011 ) 2 B 5cm A The force between electron and proton in Fig (a): Position of charges.  hydrogen atom can be calculated by using the EB F  electric field. We have, E   F  qE EA q F = -1.6×10-19 C× 5.1×1011 NC-1 = -8.16×108 N. P This force is attractive.  Using the Coulomb's law, EC 1 q1q2 F Fig (b): Electric field at point P. 4 0 r 2 19 19 Solution: Electric field is the force an a unit  9.0  109 Nm2C 1  (1.6  10 C )  (1.6  10 C ) positive charge, the fields at P due to thecharges (5.33  1011 m) 2 b. E A is the  8.6  10 N 8 at A, B and C are shown in the Fig.  field at P due to charge at A and EC is the field Knowing electric field at a point is useful at P due to charge at C. Since P is the midpoint to estimate the force experienced by a charge of AC and the fields at A and C are equal in at that point. 197 10.6.3 Electric Lines of Force: (6) Electric lines of force are crowded in a Michael Faraday (1791-1867) introduced region where electric intensity is large. the concept of lines of force for visualising (7) Electric lines of force are widely separated electric and magnetic fields. An electric line of from each other in a region where electric force is an imaginary curve drawn in such a intensity is small way that the tangent at any given point on this (8) The lines of force of an uniform electric curve gives the direction of the electric field field are parallel to each other and are at that point. See Fig.10.11. If a test charge is equally spaced. placed in an electric field it would be acted upon The lines of force are purely a geometric by a force at every point in the field and will construction which help us visualise the nature move along a path. The path along which the of electric field in a region. The lines of force unit positive charge moves is called a line of have no physical existence. force. Fig. 10.13 (a): Lines of force due to positive charge. Fig. 10.11: Electric line of force. A line of force is defined as a curve such that the tangent at any point to this curve gives Fig. 10.13 (b): Lines of the direction of the electric field at that point. force due to negative The density of field lines indicates the charge. strength of electric fields at the given point in space. Figure 10.12. Fig. 10.13 (c): Lines of force due to (High field) opposite charge. Fig. 10.12: density Fig. 10.13; (d): of field lines and Lines of force strength of electric due to similar (Low field) field. charge. Characteristics of electric lines of force Fig. 10.13 (e): Lines (1) The lines of force originate from a of force terminate positively charged object and terminate on on a conductor. a negatively charged object. (2) The lines of force neither intersect nor meet Fig. 10.13 (f): Intensity each other, as it will mean that electric field of a electric field is more has two directions at a single point. at point A and less at B. (3) The lines of force leave or terminate on a More lines cross the area conductor normally. at A and less at the same (4) The lines of force do not pass through area at B. conductor i.e. electric field inside a Fig. 10.13: The lines of force due to various conductor is always zero, but they pass geometrical arrangement of electrical charges. through insulators. (5) Magnitude of the electric field intensity Can you tell? is proportional to the number of lines of force per unit area of the surface held Lines of force are imaginary, can they perpendicular to the field. have any practical use? 198 10.7 Electric Flux: imagine a small charge +q present at a point O As discussed previously, the number of inside closed surface. Imagine an infinitesimal lines of forceper unit area is the intensity of the area dA of the given irregular closed surface. electric field E. Number of lines of force... E =Area enclosing the lines of force - (10.14) Fig. 10.15: Gauss' law. The magnitude of electric field intensity at point P on dS due to charge +q at point O is, Fig. 10.14: Flux through area S. 1 q E= Number of lines of force = (E).(Area) 4 0  r 2  When the area is inclined at an angle θ with the The direction of E is away from point O. direction of electric field, Fig. 10.14, the electric Let θ be the angle subtended by normal drawn flux can be calculated as follows.  to area dS and the direction of E. Electric flux, Let the angle between electric field E dφ, passing through area dS, = Ecosθ dS and area vector d S be θ, then the electric flux q passing through area dS is given by  cos  dS   4 0 r 2 dφ = (component of dS along E ).(area of d S )  q   dS cos  dφ = E (dS cos θ)     4 0   r 2  dφ = EdS cos θ    q  dφ = E.d S --- (10.15) d    d --- (10.17)  4 0  Total flux through the entire surface        d   E.d S  E.S --- (10.16) where, d = dS cos is the solid angle s r2 The SI unit of electric flux can be calculated subtended by area dS at point O. using,   Total electric flux, φE, crossing the given   E  S =(V/m) m2 =Vm closed surface can be obtained by integrating 10.8 Gauss' Law: Eq. (10.17) over its area. Thus, Karl Friedrich Gauss (1777-1855) one   q q  E   d   E.ds   4 0  d  d of the greatest mathematician of all times, 4 0 s formulated a law expressing the relationship between the electric charge and its electric field But  d  4  solid angle subtended by which is called the Gauss’ law. Gauss' law is entire closed surface at point O analongous to Coulomb’s law in the sense that it too expresses the relationship between electric q Total flux  ( 4 ) field and electric charge. Gauss' law provides 4 0    equivalent method for finding electric intensity.  E   E ds   q /  0 --- (10.18) It relates values of field at a closed surface and s the total charges enclosed by that surface. This is true for every electric charge Consider a closed surface of any shape enclosed by a given closed surface. which encloses number of positive electric Total flux due to charge q1, over the given charges (Fig. 10.15). To prove Gauss’ theorem, closed surface = + q1/ε0 199  Total flux due to charge q2, over the given Example 10.8: A charge of 5.0 C is kept at the closed surface = + q2/ε0 centre of a sphere of radius 1 m. What is the flux Total flux due to charge qn, over the given passing through the sphere? How will this value closed surface = + qn/ε0 change if the radius of the sphere is doubled? Positive sign in Eq. (10.18) indicates that Solution: Flux per unit area is given by Eq. the flux is directed outwards, away form the 10.16. charge. If the charge is negative, the flux will According to Gauss law,  the total flux   be is directed inwards as shown in Fig 10.16 through the sphere    E.ds , where the (b). If a charge is outside the closed surface the integration is over the surface of the sphere. As net flux through it will be zero Fig 10.16 (c). the electric field is same all over the sphere i.e. | E | = constant and the direction of E as well as that of ds is along the radius, we get flux =  = | E | 4 R 2 q 5.0 C E  9  109  4 0 r 2 (1.0 m)2 E  9  109  5  4.55  1010 NC-1 Fig. 10.16 (a): Flux due to positive charge.      ES  flux = 4.5 1010  4 (1)2  5.65  1011 Vm Thus the total flux is independent of radius. E ∝ 1/r2, and area ∝ r2. This can also be seen Fig. 10.16 (b): Flux due to negative charge. from Gauss' law, where the net flux crossing a closed surface is equal to q/ε0 where q is the net charge inside the closed surface. As the charge inside the sphere is unchanged, the flux passing through a sphere of any radius is the same. Thus, if the radius of the sphere is increased by a factor of 2, the net flux passing through Fig. 10.16 (c): Flux due to charge outside a its surface remains unchanged. As shown in closed surface is zero. Fig. 10.17, same number of lines of force cross According to the superposition principle, both the surfaces. The total flux is independent the total flux φ due to all charges enclosed of shape of the closed surface because Eq. 10.18 within the given closed surface is does not involve any radius. q1 q2 q3 qn i=n qi Q E = + + + ---- + =  0 0 0  0 i=1  0  0 Statement of Gauss' law The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0 where Q is the total charge within the surface. Fig. 10.17: Flux is independent of the shape Gauss' law is applicable to any closed and size of closed surface. surface of regular or irregular shape. 200   Do you know ? dipole moment p. Gaussian surface Dipole moment is defined as   All the lines of force originating from p = q( 2l ) --- (10.19) a point charge penetrate an imaginary A dipole moment is a vector whose three dimensional surface. The total flux magnitude is q (2l) and the direction is from ΦE = q/ε0. The same number of lines of force the negative to the positive charge. The unit of will cross the surface of any shape. The dipole moment is Columb-meter (Cm) or Debye total flux through both the surfaces is the (D). 1D = 3.33×10-30 Cm. If two charges +e and same.  Calculating flux involves calculating -e are separated by 1.0A0, the dipole moment is  E  ds , hence it is convenient to consider a 1.6×10-29 Cm or 4.8 D. For example, a water regular surface surrounding the given charge molecule has a permanent dipole moment of distribution. A surface enclosing the given charge distribution and symmetric about it is Natural dipole: a Gaussian surface. The water molecule is non-linear, i.e., For example. if we have a point charge the two hydrogen atoms and one oxygen atom the Gaussian surface will be a sphere. If the are not in a straight line. The two hydrogen- charge distribution is linear, the Gaussian oxygen bonds in water molecule are at an surface would be a cylinder with the charges angle of 105°. The positive charge of a water distributed along its axis. Gaussian surface molecule is effectively concentrated on the offers convenience    of calculating the hydrogen side and the negative charge on integral  E  ds. the oxygen side of the molecule. Thus, the Remember that a Gaussian surface positive and negative charges of the water is purely imaginary and does not exist molecule are inherently separated by a small physically. distance. This separation of positive and negative charges gives rise to the permanent 10.9 Electric Dipole: dipole moment of a water molecule. A pair of equal and opposite charges Molecules of water, separated by a finite distance is called an electric ammonia, sulphur di- dipole. It is shown in Fig. (10.18). oxide, sodium chloride etc. have an inherent separation of centers of positive and negative charges. Such molecules are called polar molecules. Polar molecules are the molecules in Fig. 10.18: Electric dipole. x-y axial line, which the center of positive charge and P-Q equatorial line. the negative charge is naturally separated. Line joining the two charges is called the Molecules such as H2, Cl2, CO2 CH4 and dipole axis. A line passing through the dipole many others have their positive and negative axis is called axial line. A line passing through charges effectively centered at the same the centre of the dipole and perpendicular to the point and are called non-polar molecules. axial line is called the equatorial line as shown Non-polar molecules are the in Fig. 10.18. molecules in which the center of positive Strength of a dipole is measured in terms charge and the negative charge is one and the same. They do not have a permanent of a quantity called the dipole moment. Let  electric dipole. When an external electric q be the magnitude of each charge and 2l be field is applied to such molecules the centers the distance from negative charge  to positive of positive and negative charge are displaced charge. Then the product q ( 2l ) is called the and a dipole is induced. 201 6.172×10-30 Cm or 1.85 D. Its direction is from Example 10.9: An electric dipole of length 2.0 oxygen to hydrogen. See box on Natural dipole. cm is placed with its axis making an angle of 10.9.1 Couple Acting on an Electric Dipole in 30° with a uniform electric field of 105 N/C. as a Uniform Electric Field: shown in figure. If it experiences a torque of Consider an electric dipole placed in a 10 3 Nm, calculate the magnitude of charge on uniform electric field E. The axis of electric the dipole. dipole makes an angle θ with the direction of electric field as shown in Fig. 10.19 a. P Solution: Given   10 3, Nm, E  105 N / C , 2l  2.0  102 m,   300 Fig. 10.19 (a): Dipole in uniform electric field.   qE 2l sin  1 10 3Nm  q105 N / C 2.0  102 m   2 10 3 q =  3  102 C 103 = 1.73  102 C Fig. 10.19 (b): Couple acting on a dipole. 10.9.2 Electric Intensity at a Point due to an Figure 10.19. b shows the couple acting on Electric Dipole: an electric dipole in uniform electric field. Case 1 : At a point on the axis of a dipole. The force acting on charge - q at A is Consider an electric dipole consisting of   F A = - qE in the direction opposite to that two charges -q and +q separated by a distance  of E and the 2l as shown in Fig. 10.20. Let P be a point at a  force acting on charge +q at B  is FB = + qE in the direction of E. Since F A distance r from the centre C of the dipole. The = - F B , the two equal and opposite forces electric intensity E a at P due to the dipole is the separated by a distance form a couple. Moment vector sum of the field due to the charge - q at of A and + q at B.  the  couple  is called torque and is defined by   d  F where, d is the perpendicular distance between the two equal and opposite forces....Magnitude of Torque = Magnitude of force × Perpendicular distance   ... Torque on the dipole =   BP  qE Fig. 10.20: Electric field of a dipole along... τ = its axis.  qE2lsinθ  --- (10.20) Electric field intensity at P due to the but p  q  2l... τ = pEsinθ --- (10.21) charge -q at A     1 ( q )   In vector form   p  E --- (10.22) = EA   PDPD 2PQ If θ =90° sin θ =1, then τ = pE 4 0 ( r + l ) When the axis of electric dipole is  where PD is unit vector directed along PD perpendicular to uniform electric field, torque Electric intensity at P due to charge +q at B of the couple acting on the electric dipole is  1 q   maximum, i.e., τ = pE. It θ = 0 then τ = 0, this is = EB   PQ PQ 2PQ the minimum torque on the dipole. Torque tends 4 0 ( r - l ) to align the dipole axis along the direction of  is a unit vector directed along PQ where PQ. electric field. 202     is the unit vector directed along PQ  The magnitude of E B is greater than that of where  PQ E A. (Because BP < AB) or BP      Resultant field E a at P on the axis, due to Electric field at PE Aisand the,Esum B of E A and E B the dipole is       ∴ E eq = EA + EB Ea = E B + E A  Consider ∆ ACP The magnitude of E a is given by (AP)2 = (PC)2 + (AC)2 = r2 + l2 = (BP)2  1  q q   | Ea |   1 q  2  EA  4 0  (r  l ) (r  l )  2 4 0 ( r  l 2 ) 2 --- (10.25)  q  r 2 + l 2  2lr - r 2  2lr  l 2   1 q | Ea | =   EB  --- (10.26) 4 0  (r  l ) 2 2 2  4 0 (r  l 2 ) 2    2(2lq )r EA = EB | Ea | = 4 0 (r 2  l 2 ) 2   The resultant of fields E A and E B acting But 2lq = p, the dipole moment at point P can be calculated by resolving these   1 2 pr vectors E A and E B along the equatorial line | Ea | = 4 0 (r 2  l 2 ) 2 and along a direction perpendicular to it. --- (10.23)  E a , is directed along PQ, which is the  direction of the dipole moment p i.e. from the Fig. 10.21 (b): negative to the positive charge, parallel to the Components of axis. If r>> l, l2 can be neglected compared to r2, the field at point  1 2p P. Ea = --- (10.24) 4 0 r 3 The field will Consider Fig. 10.21 (b). Let the y-axis  be along the direction of the coincide with the equator of the dipole x-axis dipole moment p. will be parallel to dipole axis, as shown. The Case 2: At a point on the equatorial line. As origin is at point P. shown in Fig. 10.21 (a) The y-components of EA and EB are EAsinθ and EBsinθ respectively. They are equal in magnitude but opposite in direction and cancel Fig. 10.21 (a): each other. There is no contribution from them Electric field of a r towards the resultant. dipole at a point on The x-components of EA and EB are EAcosθ the equatorial line. and EBcosθ respectively. They are of equal magnitude and are in the same direction  Electric field at point P due to charge -q at ∴| Eeq | E A cos   E B cos  --- (10.27)  1 ( q )  A is: E A  PA By using Eq. 10.25 and 10.26 4 0 ( A P) 2    Eeq = 2E A cos where PA is the unit vector direction along PA.  q  l  2 2  Similarly, Electric field at P due to charge + q at  4 0 (r  l )  r  l 2 2 2  1 (q)   B is: E B  PQ 2ql 4 0 (BP) 2  4 0 (r 2  l 2 )3/ 2 203  2 If r>>l then l is very small compared to r2 (b) Surface charge density (σ)  1 p 1 p Suppose a charge q is uniformly distributed Eeq   --- (10.28) 4 0 (r ) 2 3/ 2 4 0 r 3 over a surface of area A. As shown in Fig. 10.23,  then the surface charge density σ is defined as The direction  of this field is along - p (anti- q  --- (10.30) parallel to p ) as shown in Fig. 10.21 (c). A Fig. 10.21 (c): SI unit of σ is (C/m2) Electric field at For example, charge distributed uniformly point P is anti- on a thin disc or a synthetic cloth. If the charge parallel to p. is not distributed uniformally over the surface Comparing Eq. 10.28 and 10.24 we find of a conductor, then charge dq on small area that the electric intensity at an axial point is element dA can be written as dq = σ dA. twice that at a point on the equatorial position, lying at the same distance from the centre of the dipole. 10.10 Continuous Charge Distribution: A system of charges can be considered as Fig. 10.23: Surface charge. a continuous charge distribution, if the charges (c) Volume charge density (ρ) are located very close together, compared Suppose a charge q is uniformly distributed to their distances from the point where the throughout a volume V, then the volume charge intensity of electric field is to be found out. density ρ is defined as the charge per unit The charge distribution is continuous volume. in the sense that, a system of closely spaced q charges is equivalent to a total charge which  --- (10.31) V is continuously distributed along a line or a S.I. unit of ρ is (C/m3) surface or a volume. To find the electric field For example, charge on a solid plastic due to continuous charge distribution, we define sphere or a solid plastic cube. following terms for different types of charge distribution. If the charge is not distributed uniformaly over the volume of a material, then charge dq (a) Linear charge density (λ). over small volume element dV can be written As shown in Fig. 10.22 charge q is as dq = ρ dV. uniformly distributed along a liner conductor of length l. The linear charge density λ is defined as, q  --- (10.29) l Fig. 10.24:Volume charge. SI unit of λ is (C/m). Electric field due to a continuous charge For example, charge distributed uniformly distribution can be calculated by adding electric on a straight thin rod or a thin nylon thread. If fields due to all these small charges. the charge is not distributed uniformly over the length of thin conductor then charge dq on small element of length dl can be written as dq = λdl Can you tell? The surface charge density of Earth is σ = -1.33 nC/m2. That is about 8.3×109 Fig. 10.22: Linear charge. electrons per square meter. If that is the case why don't we feel it? 204  Do you know ? iii. One can get static shock if charge transferred is large. Static charge can be useful iv. Dust or dirt particles gathered on Static charges can be created whenever computer or TV screens can catch static there is a friction between an insulator and charges and can be troublesome. other object. For example, when an insulator Precautions against static charge like rubber or ebonite is rubbed against i. Home appliances should be grounded. a cloth, the friction between them causes ii. Avoid using rubber soled footwear. electrons to be transferred from one to the iii. Keep your surroundings humid. (dry air other. This property of insulators is used can retain static charges). in many applications such as Photocopier, Inkjet printer, Panting metal panels, Electrostatic precipitation/separators etc. Internet my friend Static charge can be harmful 1. h t t p s : / / w w w. p h y s i c s c l a s s r o o m. i. When charge transferred from one body com>class to other is very large sparking can take 2. https://courses.lumenlearing.com>elect place. For example lightning in sky. 3. https://www.khanacademy.org>science. ii. Sparking can be dangerous while 4. https://www.topper.com>guides>physics refuelling your vehicle. Ex erc ises Exercises 1. Choose the correct option. (A) 1:4 (B) 1:2 i. A positively charged glass rod is brought (C) 1:1 (D) 1:16 close to a metallic rod isolated from iv. Two charges of 1.0 C each are placed ground. The charge on the side of the one meter apart in free space. The force metallic rod away from the glass rod will between them will be be (A) 1.0 N (B) 9×109 N (A) same as that on the glass rod and equal (C) 9×10-9 N (D) 10 N in quantity v. Two point charges of +5 µC are so (B) opposite to that on the glass of and placed that they experience a force of equal in quantity 80×10-3 N. They are then moved apart, (C) same as that on the glass rod but lesser so that the force is now 2.0×10-3 N. The in quantity distance between them is now (D) same as that on the glass rod but more (A) 1/4 the previous distance in quantity (B) double the previous distance ii. An electron is placed between two (C) four times the previous distance parallel plates connected to a battery. If (D) half the previous distance the battery is switched on, the electron vi. A metallic sphere A isolated from ground will is charged to +50 µC. This sphere is (A) be attracted to the +ve plate brought in contact with other isolated (B) be attracted to the -ve plate metallic sphere B of half the radius of (C) remain stationary sphere A. The charge on the two sphere (D) will move parallel to the plates will be now in the ratio iii. A charge of + 7 µC is placed at the centre (A) 1:2 (B) 2:1 of two concentric spheres with radius 2.0 (C) 4:1 (D) 1:1 cm and 4.0 cm respectively. The ratio of vii. Which of the following produces uniform the flux through them will be electric field? 205  (A) point charge (B) linear charge iii. Four charges of + 6×10-8 C each are (C) two parallel plates placed at the corners of a square whose (D) charge distributed an circular any sides a are 3 cm each. Calculate the viii. Two point charges of A = +5.0 µC and resultant force on each charge and show B = -5.0 µC are separated by 5.0 cm. its direction on a digram drawn to scale. A point charge C = 1.0 µC is placed [Ans: 6.89×10-2 N] at 3.0 cm away from the centre on the iv. The  electric field in a region is given by perpendicular bisector of the line joining E = 5.0 kN/C. Calculate the electric flux the two point charges. The charge at C Through a square of side 10.0 cm in the will experience a force directed towards following cases (A) point A (a) the square is along the XY plane (B) point B [Ans: = 5.0×10-2 Vm] (C) a direction parallel to line AB (b) The square is along XZ plane (D) a direction along the perpendicular [Ans: Zero] bisector (c) The normal to the square makes an angle of 450 with the Z axis. 2. Answer the following questions. [Ans: 3.5×10-2 Vm] i. What is the magnitude of charge on an v. Three equal charges of 10×10-8 C electron? respectively, each located at the corners ii. State the law of conservation of charge. of a right triangle whose sides are 15 cm, iii. Define a unit charge. 20 cm and 25 cm respectively. Find the iv. Two parallel plates have a potential force exerted on the charge located at the difference of 10V between them. If the 90° angle. plates are 0.5 mm apart, what will be the [Ans: 4.59.×10-3 N] strength of electric charge. vi. A potential difference of 5000 volt is v. What is uniform electric field? applied between two parallel plates 5 cm vi. If two lines of force intersect of one a part a small oil drop having a charge of point. What does it mean? 9.6 ×10-19 C falls between the plates. Find vii. State the units of linear charge density. (a) electric field intensity between the viii. What is the unit of dipole moment? plates and (b) the force on the oil drop. ix. What is relative permittivity? [Ans: (a) 1.0.×105 N/C 3. Solve numerical examples. (b) 9.6.×10-14 N] i. Two small spheres 18 cm apart have vii. Calculate the electric field due to a charge equal negative charges and repel each of -8.0×10-8 C at a distance of 5.0 cm other with the force of 6×10-3 N. Find the from it. total charge on both spheres. [Ans: -2.88×10-2 N/C] [Ans: q = 2.938×10-7 C] ii. A charge +q exerts a force of magnitude *** -0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q. [Ans: q = 0.8333 µC] 206