Electrostatics JEE (Main) PDF

ELECTROSTATICS CLASS: XII / XIII | COURSE: ANOOP (EP), AKHIL (EF), AJAY (ER), ABHYAAS (ED) Target: JEE (Main)  Foreword by Founder Dear Aspirant,...

ELECTROSTATICS CLASS: XII / XIII | COURSE: ANOOP (EP), AKHIL (EF), AJAY (ER), ABHYAAS (ED) Target: JEE (Main)  Foreword by Founder Dear Aspirant, Welcome aboard! You are about to begin one of the most exciting stages in your life and every single member of ‘Team Resonance’ is committed to making it an enriching experience for you. Now that you have become a RESONite and have chosen one of the leaders in the coaching industry, you can be rest assured that your career shall reach the potential which it truly deserves. About the Sheets: The study material at Resonance is developed meticulously as per the updated curriculum of JEE (Main) & JEE (Advanced). Sheets, which you get, are divided into two parts - the Theory and the Exercises. Theoretical part of the sheets is broadly written in systematic structure, which a faculty delivers in the classrooms at Resonance. In other words, these are the lectures in the theoretical format or the written format. There will be some difference in the lectures and the actual sheets, as a faculty always improvises his/her lectures to add more value in the classroom. In a broader sense, however, theory more or less can be termed as the written format of the class lectures. The next part, the Exercises given after the topics follow a definite flow or pattern. Resonance provides three levels of Exercises and a super-special class of questions called High Level Problems (HLP) in the sheets. Brief of Exercise 1: It is the easiest one and the questions are as per the sequence / topics to increase the learning pace of the students. It contains multiple types of questions. The Motive: To let the students gain confidence & increase fluency of problem solving. Point to Remember: The questions are framed to make the student understand the fundamental concepts of the topics. Brief of Exercise 2: In this part, the level of the questions increases a step further. These Exercises contain numerical questions also. The Motive: These Exercises tests the analytical ability and the ability to handle unknown situations. Point to Remember: Students may get stuck in solving questions of Exercises 2, it’s natural as the questions are intended to test enhanced abilities and to better cement the learning. Brief of Exercise 3: These contain Previous Years Questions (last 10 years) of JEE Main & JEE Advanced. The Motive: To give exposure to the students of the varied types of questions asked in the competitive examinations. Point to Remember: These structured questions provide the ability to students to carefully select the best approach to problem solving. Brief about HLP: HLP is not compulsory for JEE (Main) or JEE (Advanced). These types of questions are a class apart. The Motive: To cater to students who have deep interest in the subject, like for those students who have an interest in research. Point to Remember: These questions require a super specialized approach, are time consuming and may be laboriously lengthy. Conclusion Students must note that the theoretical portion of the sheet is provided as an add-on so that the students can complete and enhance their notes which they would have made while attending the lectures. The sheets are more of backup for students in case students miss making complete notes in the class but sheets are not a substitute for the notes. The Exercises cements the learning aspect and solidifies the application of theoretical concepts. They are a must if students seriously intend to go past the examination barrier to achieve their desired dreams. I hope you will make full use of the enriched content of sheet. Why Sheets With best wishes, R.K. Verma (RKV Sir) Managing Director Head - Academics & Senior Faculty (Physics) RKV Sir’s Video Resonance Eduventures Limited STUDENT'S INFORMATION ACADEMIC SESSION: 20__ - 20__ | DIVISION:____________________ Name: Reso Roll No.: Target Exam / Year: Class: Course Name / Code: Phase / Batch Code: Study Centre: Campus: Contact No.: Email:  CONTENTS Page Number S. No. Particulars From To 1. Weightage Analysis : (JEE (Main + Advanced) 2. THEORY 01 86 2.1. Theory with Examples (Solved + Unsolved) 01 86 3. EXERCISES 87 116 3.1 EXERCISE 1 87 97 3.1.1 Part I : Multiple Choice Questions (MCQs) : Only One Correct Answer 87 95 3.1.2 Part II : Numerical Value Questions (NVQs) 95 97 3.2 EXERCISE 2 97 108 3.2.1 Part I : Multiple Choice Questions (MCQs) : Only One Correct Answer 97 105 3.2.2 Part II : Numerical Value Questions (NVQs) 105 108 3.3 EXERCISE 3 108 116 3.3.1 JEE (MAIN) (Previous Years) Problems 108 116 3.4. ANSWER KEY 117 118 4. SUMMARY 119 122 TOTAL NO. OF PAGES SHEET 124 Yearwise count of ELECTROSTATICS In LAST 05 YEARS (2020 - 2024) Exam Year 2024 2023 2022 2021 2020 Grand Total Total Papers 10 24 22 26 16 98 Total Questions from Topic 16 33 26 32 23 130 MCQs 16 18 21 21 21 97 NVQs 0 15 5 11 2 33 * Exam Year 2024 is counted as session 1 in JEE (Main). © Copyright reserved. All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to enrolled student of RESONANCE only. Any sale/resale of this material is punishable under law, subject to Kota Jurisdiction only. Electrostatics ELECTROSTATICS  1. INTRODUCTION The branch of physics which deals with electric effect of static charge is called electrostatics. 2. ELECTRIC CHARGE Charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some of naturally charged particles are electron, proton,  -particle etc. Charge is a derived physical quantity. Charge is measured in coulomb in S.. unit. In practice we use mC (10–3C), C (10–6C), nC(10–9C) etc. C.G.S. unit of charge = electrostatic unit = esu. 1 coulomb = 3 × 109 esu of charge Dimensional formula of charge = [MºLºT11] 2.1. Properties of Charge (i) Charge is a scalar quantity : It adds algebraically and represents excess, or deficiency of electrons. (ii) Charge is of two types : (i) Positive charge and (ii) Negative charge Charging a body implies transfer of charge (electrons) from one body to another. Positively charged body means loss of electrons, i.e., deficiency of electrons. Negatively charged body means excess of electrons. This also shows that mass of a negatively charged body > mass of a positively charged identical body. (iii) Charge is conserved : In an isolated system, total charge (sum of positive and negative) remains constant whatever change takes place in that system. (iv) Charge is quantized : Charge on any body always exists in integral multiples of a fundamental unit of electric charge. This unit is equal to the magnitude of charge on electron (1e = 1.6 × 10– 19 coulomb). So charge on anybody Q = ± ne, where n is an integer and e is the charge of the electron. Millikan's oil drop experiment proved the quantization of charge or atomicity of charge. 1 2 Note : Recently, the existence of particles of charge ± e and ± e has been postulated. 3 3 These particles are called quarks but still this is not considered as the quantum of charge because these are unstable (They have very short span of life). (v) Like point charges repel each other while unlike point charges attract each other. (vi) Charge is always associated with mass, i.e., charge cannot exist without mass though mass can exist without charge. The particle such as photon or neutrino which have no (rest) mass can never have a charge. (vii) Charge is relativistically invariant: This means that charge is independent of frame of reference, i.e., charge on a body does not change whatever be its speed. This property is worth mentioning as in contrast to charge, the mass of a body depends on its speed and increases with increase in speed. (viii) A charge at rest produces only electric field around itself; a charge having uniform motion produces electric as well as magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-1 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics 2.2. Charging of a body A body can be charged by means of (a) friction, (b) conduction, (c) induction, (d) thermionic ionization or thermionic emission (e) photoelectric effect and (f) field emission. (a) Charging by Friction : When a neutral body is rubbed against other neutral body then some electrons are transferred from one body to other. The body which can hold electrons tightly, draws some electrons and the body which can not hold electrons tightly, looses some electrons. The body which draws electrons becomes negatively charged and the body which looses electrons becomes positively charged. For example : Suppose a glass rod is rubbed with a silk cloth. As the silk can hold electrons more tightly and a glass rod can hold electrons less tightly (due to their chemical properties), some electrons will leave the glass rod and gets transferred to the silk. So in the glass rod their will be deficiency of electrons, therefore it will become positively charged. And in the silk there will be some extra electrons, so it will become negatively charged (b) Charging by conduction (flow): There are three types of material in nature (i) Conductor : Conductors are the material in which the outer most electrons are very loosely bounded, so they are free to move (flow). So in a conductors, there are large number of free electrons. Ex. Metals like Cu, Ag, Fe, Al............. (ii) Insulator or Dielectric or Nonconductor : Non-conductors are the materials in which outer most electrons are very tightly bounded, so they cannot move (flow). Hence in a non- conductor there is no free electrons. Ex. plastic, rubber, wood etc. (iii) Semiconductor : Semiconductor are the materials which have free electrons but very less in number. Now lets see how the charging is done by conduction. In this method we take a charged conductor 'A' and an uncharged conductor 'B'. When both are connected some charge will flow from the charged body to the uncharged body. If both the conductors are identical & kept at large distance, if connected to each other, then charge will be divided equally in both the conductors otherwise they will flow till their electric potential becomes same. Its detailed study will be done in last section of this chapter. (c) Charging by Induction : To understand this, lets have introduction to induction. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-2 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics We have studied that there are lot of free electrons in the conductors. When a charge particle +Q is brought near a neutral conductor. Due to attraction of +Q charge, many electrons (–ve charges) come closer and accumulate on the closer surface. On the other hand a positive charge (deficiency of electrons) appears on the other surface. The flow of charge continues till there is resultant force on free electrons of the conductor becomes zero. This phenomena is called induction, and charges produced are called induced charges. A body can be charged by induction in the following two ways : Method I : Step 1. Take an isolated neutral conductor.. Step 2. Bring a charged rod near to it. Due to the charged rod, charges will induce on the conductor. Step 3. Connect another neutral conductor with it. Due to attraction of the rod, some free electrons will move from the right conductor to the left conductor and due to deficiency of electrons positive charges will appear on right conductor and on the left conductor there will be excess of electrons due to transfer from right conductor.. Step 4. Now disconnect the connecting wire and remove the rod. The first conductor will be negatively charged and the second conductor will be positively charged. Method II : Step 1. Take an isolated neutral conductor.. Step 2. Bring a charged rod near to it. Due to the charged rod, charges will induce on the conductor. Step 3. Connect the conductor to the earth (this process is called grounding or earthling). Due to attraction of the rod, some free electrons will move from earth to the conductor, so in the conductor there will be excess of electrons due to transfer from the earth, so net charge on conductor will be negative. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-3 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Step 4. Now disconnect the connecting wire. Conductor becomes negatively charge. (d) Thermionic emission : When the metal is heated at a high temperature then some electrons of metals are ejected and the metal becomes positively charged. (e) Photoelectric effect : When light of sufficiently high frequency is incident on metal surface then some electrons gains energy from light and come out of the metal surface and remaining metal becomes positively charged. (f) Field emission : When electric field of large magnitude is applied near the metal surface then some electrons come out from the metal surface and hence the metal gets positively charged. SE SOLVED EXAMPLE Example 1. If a charged body is placed near a neutral conductor, will it attract the conductor or repel it? Solution : Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-4 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics If a charged body (+ve) is placed leftside near a neutral conductor, (–ve) charge will induce at left surface and (+ve) charge will induce at right surface. Due to positively charged body –ve induced charge will feel attraction and the +ve induced charge will feel repulsion. But as the –ve induced charge is nearer, so the attractive force will be greater than the repulsive force. So the net force on the conductor due to positively charged body will be attractive. Similarly we can prove for negatively charged body also. From the above example we can conclude that. "A charged body can attract a neutral body." If there is attraction between two bodies then one of them may be neutral. But if there is repulsion between two bodies, both must be charged (similarly charged). So "repulsion is the sure test of electrification". Example 2. A positively charged body 'A' attracts a body 'B' then charge on body 'B' may be: (A) positive (B) negative (C) zero (D) can't say Answer. B, C Example 3. Five styrofoam balls A, B, C, D and E are used in an experiment. Several experiments are performed on the balls and the following observations are made (i) Ball A repels C and attracts B. (ii) Ball D attracts B and has no effect on E. (iii) A negatively charged rod attracts both A and E. For your information, an electrically neutral styrofoam ball is very sensitive to charge induction, and gets attracted considerably, if placed nearby a charged body. What are the charges, if any, on each ball ? A B C D E (A) + – + 0 + (B) + – + + 0 (C) + – + 0 0 (D) – + – 0 0 Answer. C Solution : From (i), As A repels C, so both A and C must be charged similarly. Either both are +ve or both are –ve. As A also attract B, so charge on B should be opposite of A or B may be uncharged conductor. From (ii) As D has no effect on E, so both D and E should be uncharged, and as B attracts uncharged D, so B must be charged and D must be on uncharged conductor. From (iii) a –ve charged rod attract the charged ball A, so A must be +ve, and from exp. (i) C must also be +ve and B must be –ve. Example 4. Charge conservation is always valid. Is it also true for mass? Solution : No, mass conservation is not always. In some nuclear reactions, some mass is lost and it is converted into energy. Example 5. What are the differences between charging by induction and charging by conduction ? Solution : Major differences between two methods of charging are as follows : (i) In induction, two bodies are close to each other but do not touch each other while in conduction they touch each other. (or they are connected by a metallic wire) (ii) In induction, total charge of a body remains unchanged while in conduction it changes. (iii) In induction, induced charge is always opposite in nature to that of source charge while in conduction charge on two bodies finally is of same nature Example 6. If a glass rod is rubbed with silk it acquires a positive charge because : (A) protons are added to it (B) protons are removed from it (C) electrons are added to it (D) electrons are removed from it. Answer. D Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-5 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics  3. COULOMB’S LAW (INVERSE SQUARE LAW) On the basis of experiments Coulomb established the following law known as Coulomb's law. The magnitude of electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. 1 qq Kq1q2 i.e. F  q1q2 and F 2  F  12 2  F= r r r2 Important points regarding Coulomb's law : (i) It is applicable only for point charges. 1 (ii) The constant of proportionality K in SI units in vacuum is expressed as and in any other 4 0 1 medium expressed as. If charges are dipped in a medium then electrostatic force on one 4 1 q1q2 charge is. 0 and  are called permittivity of vacuum and absolute permittivity of 4 0 r r 2 the medium respectively. The ratio /0 = r is called relative permittivity of the medium, which is a dimensionless quantity. (iii) The value of relative permittivity r is constant for medium and can have values between 1 to . For vacuum, by definition it is equal to 1. For air it is nearly equal to 1 and may be taken to be equal to 1 for calculations. For metals the value of r is  and for water is 81. The material in which more charge can induce r will be higher. 1 (iv) The value of = 9 × 109 Nm2 C–2  0 = 8.855 × 10–12 C2/Nm2. 4 0 Dimensional formula of  is M–1 L–3 T4 A2 (v) The force acting on one point charge due to the other point charge is always along the line joining these two charges. It is equal in magnitude and opposite in direction on two charges, irrespective of the medium, in which they lie. (vi) The force is conservative in nature i.e., work done by electrostatic force in moving a point charge along a close loop of any shape is zero. (vii) Since the force is a central force, in the absence of any other external force, angular momentum of one particle w.r.t. the other particle (in two particle system) is conserved, (viii) In vector form formula can be given as below.  1 q1q2  1 q1q2 F =  3r =  r̂ (q1 & q2 are to be substituted with sign.) 4 0 r | r | 4 0 r | r |2  here r is position vector of the test charge (on which force is to be calculated) with respect to the source charge (due to which force is to be calculated). SE SOLVED EXAMPLE Example 7. Find out the electrostatics force between two point charges placed in air (each of + 1C) if they are separated by 1m. kq q 9  109  1  1 Sol. Fe = 12 2 = = 9×109 N r 12 From the above result we can say that 1 C charge is too large to realize. In nature, charge is usually of the order of C Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-6 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 8. Two particles having charges q1 and q2 when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled then what will be the force between the particles: Answer. 16 F kq q Solution : F = 12 2 r r If q’1 = 2q1, q’2 = 2q2 r’ = , 2 kq'1 q'2 k(2q1 )(2q2 ) then F’ = 2 = 2 r' r   2 16kq1q2 F’ = r2 F’ = 16F Example 9. A particle of mass m carrying charge q1 is revolving around a fixed charge –q2 in a circular path of radius r. Calculate the period of revolution and its speed also. 1 q1q2 2 4 2mr Solution : = mr = 4 0 r 2 T2 (40 )r 2 (4 2mr)  0mr T2 = or T = 4r q1q2 q1q2 and also we can say that q1q2 mv 2 q1q2 =  V= 40 r 2 r 4 0mr Example 10. A point charge qA = + 100 µc is placed at point A (1, 0, 2) m and an another point charge qB = +200µc is placed at point B (4, 4, 2) m. Find : (i) Magnitude of Electrostatic interaction force acting between them   (ii) Find FA (force on A due to B) and FB (force on B due to A) in vector form Solution : (i) kqA qB (9  109 )(100  10 6 )(200  10 6 ) Value of F : F  = = 7.2 N r2 (4  1)2  (4  0)2  (2  2)2  kq q  (9  109 )(100  10 6 )(200  106 ) Force on B FB = A 3B r = (4  1) ˆi  (4  0) ˆj  (2  2)kˆ    |r| (4  1)2  (4  0)2  (2  2)2 3 4  = 7.2  ˆi  ˆj N 5 5    3 4  Similarly FA = 7.2   ˆi  ˆj N  5 5  Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-7 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics   Action( FA ) and Reaction ( FB ) are equal but in opposite direction.  4. PRINCIPLE OF SUPERPOSITION The electrostatic force is a two body interaction, i.e., electrical force between two point charges is independent of presence or absence of other charges and so the principle of superposition is valid, i.e., force on charged particle due to number of point charges is the resultant of forces due to individual point charges, therefore, force on a point test charge due to many charges is     given by F  F1  F2  F3 ........... SE SOLVED EXAMPLE Example 11. Three equal point charges of charge +q are moving along a circle of radius R and a point charge –2q is also placed at the centre of circle as (shown in figure), if charges are revolving with constant and same speed then calculate speed Solution : mv 2 K(q)(2q) 2(Kq2 ) mv 2 F2 – 2F1 cos 30 =  – cos 30 = R R2 ( 3R)2 R kq2  1   v   2  Rm  3 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-8 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 12. Two equally charged identical small metallic spheres A and B repel each other with a force 2 × 10–5N when placed in air (neglect gravitation attraction). Another identical uncharged sphere C is touched to B and then placed at the mid-point of line joining A and B. What is the net electrostatic force on C? Solution : Let initially the charge on each sphere be q and separation between their centres be r; then according to given problem. 1 qq F= = 2 × 10–5 N 4 0 r 2 When sphere C touches B, the charge of B, q will distribute equally on B and C as sphere are identical conductors, i.e., now charges on spheres; qB = qC = (q/2) So sphere C will experience a force 1 q(q / 2)  FCA = = 2F along AB due to charge on A 4 0 (r / 2)2 1 (q / 2)(q / 2)  and FCB = 2 = F along BA due to charge on B 4 0 (r / 2) So the net force FC on C due to charges on A and B,  FC = FCA – FCB = 2F – F = 2 × 10–5 N along AB. Example 13. 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 coulomb placed at the centre of the hexagon? Solution : Method I : If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on –q at O would be zero (as the forces due to individual charges will balance each  other), i.e. FR  0   Now if f is the force due to sixth charge and F due to remaining five charges.     F + f =0 i.e. F =–f 1 qq 1 q2 or |F| = |f| = = 4 0 L2 4 0 L2   1 q2 FNet = FCO = along OD 4   L2 Method II : In the diagram we can see that force due to charge A and D are opposite to each other   FDO + FAO = 0....(i)   Similarly FBO + FEO = 0....(ii)       So FAO + FBO + FCO + FDO + FEO = FNet   1 q2 Using (i) and (ii) FNet = FCO = along OD 4   L2 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-9 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 14. A thin straight rod of length l carrying a uniformly distributed change q is located in vacuum. Find the magnitude of the electric force on a point charge 'Q' kept as shown in the figure. Solution : As the charge on the rod is not point charge, therefore, first we have to find force on charge Q due to charge over a very small part on the length of the rod. This part called element of length dy can be considered as point charge. q Charge on element dq = dy = dy  K.dq.Q K.Q.q.dy Electric force on 'Q' due to element = 2 = y y2. All forces are along the same direction  F=  dF This sum can be calculated using integration, a a  KQqdy KqQ  1 KQ.q  1 1  KQq therefore F =  y a 2 y  =   y   a =    a a    = a(a   ) Note: (1) The total charge of the rod cannot be considered to be placed at the centre of the rod as we do in mechanics for mass in many problems. KQq Note: (2) If a >> l then F = a2 behaviour of the rod is just like a point charge.  5. ELECTROSTATIC EQUILIBRIUM The point where the resultant force on a charged particle becomes zero is called equilibrium position. 5.1. Stable Equilibrium : A charge is initially in equilibrium position and is displaced by a small distance. If the charge tries to return back to the same equilibrium position then this equilibrium is called position of stable equilibrium. 5.2. Unstable Equilibrium : If charge is displaced by a small distance from its equilibrium position and the charge has no tendency to return to the same equilibrium position. Instead it goes away from the equilibrium position. 5.3. Neutral Equilibrium : If charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium. SE SOLVED EXAMPLE Example 15. Two equal positive point charges 'Q' are fixed at points B(a, 0) and A(–a, 0). Another test charge q0 is also placed at O(0, 0). Show that the equilibrium at 'O' is (i) stable for displacement along X-axis. (ii) unstable for displacement along Y-axis.     KQq0 Solution : (i) Initially FAO + FBO = 0  | FAO | = | FBO | = a2 When charge is slightly shifted towards + x axis by a small distance x,  then.  | FAO | < | FBO | Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-10 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Therefore the particle will move towards origin (its original position) hence the equilibrium is stable. (i) When charge is shifted along y axis After resolving components net force will be along y axis so the particle will not return to its original position so it is unstable equilibrium. Finally the charge will move to infinity. Example 16. Two point charges of charge q1 and q2 (both of same sign) and each of mass m are placed such that gravitation attraction between them balances the electrostatic repulsion. Are they in stable equilibrium? If not then what is the nature of equilibrium? K q1 q2 Gm2 Solution : In given example : =. We can see that irrespective of distance between them r2 r2 charges will remain in equilibrium. If now distance is increased or decreased then there is no effect in their equilibrium. Therefore it is a neutral equilibrium. Example 17. A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and a distance 2 apart. Prove that the motion of the particle will be SHM if it is displaced slightly along the line connecting them and released. Also find its time period. Solution : Let the charge q at the mid-point the displaced slightly to the left. The force on the displaced charge q due to charge q at A, 1 q2 F1 = 4 0 (  x)2 The force on the displaced charge q due to charge at B, 1 q2 F2 = 4 0 (  x)2 Net restoring force on the displaced charge q. 1 q2 1 q2 F = F2 – F1 or F = 2 – 4 0 (  x) 4 0 (  x)2 q2  1 1  q2 4x or F=  2  2  = 4 0  (  x) (  x)  4 0 (2  x 2 )2 q2 x q2 x Since  >> x,  F = or F = 0 4 0 3 We see that F  x and it is opposite to the direction of displacement. Therefore, the motion is SHM. m q2 m  0 3 T = 2 , here k = = 2  k  0 3 q2 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-11 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 18. Find out mass of the charge Q so, that it remains in equilibrium for the given configuration. Q q h q  q q   h KQq Solution :  4 Fcos = mg  4× 3/2 h = mg  2 2   2 h  2   Example 19. Two identical charged spheres are suspended by strings of equal length. Each string makes an angle  with the vertical. When suspended in a liquid of density  = 0.8 gm/cc, the angle remains the same. What is the dielectric constant of the liquid? (Density of the material of sphere is  = 1.6 gm/cc.) Solution : Initially as the forces acting on each ball are tension T, weight mg and electric force F, for its equilibrium along vertical, T cos  = mg.....(1) and along horizontal T sin  = F.....(2) Dividing Eqn. (2) by (1), we have F tan  =.....(3) mg When the balls are suspended in a liquid of density and dielectric constant K, the electric force will become (1/K) times, i.e., F' = (F/K) while weight. mg' = mg – FB = mg – Vg [as FB = Vg, where  is density of material of sphere]    m i.e., mg' = mg 1   as V       So for equilibrium of ball, F' F tan ' = =... (4) mg' Kmg[1  ( / )] According to given information ' = ; so from equations (4) and (3), we have  1.6 K= = =2 Ans. (   ) (1.6  0.8)  6. ELECTRIC FIELD Electric field is the region around charged particle or charged body in which if another charge is placed, it experiences electrostatic force.  6.1. Electric field intensity E : Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitude and direction. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-12 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics  If a test charge q0 is placed at a point in an electric field and experiences a force F due to some charges (called source charges), the electric field intensity at that point due to source charges is given   F by E  ; q0  If the E is to be determined practically then the test charge q0 should be small otherwise it will affect the charge distribution on the source which is producing the electric field and hence modify the quantity which is measured. SE SOLVED EXAMPLE Example 20. A positively charged ball hangs from a long silk thread. We wish to measure E at a point P in the same horizontal plane as that of the hanging charge. To do so, we put a positive test charge q0 at the point and measure F/q0. Will F/q0 be less than, equal to, or greater than E at the point in question? Solution : When we try to measure the electric field at point P then after placing the test charge at P it repels the source charge (suspended charge) F and the measured value of electric field Emeasured = will be less q0 than the actual value Eact that we wanted to measure.   6.2. Properties of electric field intensity E : (i) It is a vector quantity. Its direction is the same as the force experienced by positive charge. (ii) Direction of electric field due to positive charge is always away from it while due to negative charge always towards it. (iii) Its S.. unit is Newton/Coulomb. (iv) Its dimensional formula is [MLT–3A–1] (v) Electric force on a charge q placed in a region of electric field at a point where the electric field    intensity is E is given by F  qE. Electric force on point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge. (vi) It obeys the superposition principle, that is, the field intensity at a point due to a system of charges is vector sum of the field intensities due to individual point charges.     E  E1  E2  E3 +..... (vii) It is produced by source charges. The electric field will be a fixed value at a point unless we change the distribution of source charges. SE SOLVED EXAMPLE Example 21. Electrostatic force experienced by –3C charge placed at point 'P' due  to a system 'S' of fixed point charges as shown in figure is F  (21iˆ  9ˆj) µN. (i) Find out electric field intensity at point P due to S. (ii) If now 2C charge is placed and –3 C is removed at point P then force experienced by it will be     µN Solution : (i) F  qE  (21iˆ  9ˆj)µN = –3µC (E)  E = –7 î – 3 ĵ C (ii) Since the source charges are not disturbed the electric field intensity at 'P' will remain same.   F 2C = +2( E  = 2(–7 î – 3 ĵ ) = –14 î – 6 ĵ µN Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-13 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 22. Calculate the electric field intensity which would be just sufficient to balance the weight of a 2 particle of charge –10 c and mass 10 mg. (take  g = 10 ms ) Solution : As force on a charge q in an electric field E is F q = q E Fe So according to given problem   | Fq |  | W | i.e., |q|E = mg E A q mg i.e., E= = 10 N/C., in downward direction. | q| W List of formula for Electric Field Intensity due to various types of charge distribution : Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-14 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics SE SOLVED EXAMPLE Example 23. Find out electric field intensity at point A (0, 1m, 2m) due to a point charge –20C situated at point B( 2 m, 0, 1m). KQ  KQ  Solution : E=  3 r =  2 r̂  r = P.V. of A – P.V. of B (P.V. = Position vector) |r| |r|  = (– 2 î + ĵ + k̂ ) | r | = ( 2)2  (1)2  (1)2 = 2 9  109  ( 20  106 ) E= (– 2 î + ĵ + k̂ ) = – 22.5 × 103 (– 2 î + ĵ + k̂ ) N/C. 8 Example 24. Two point charges 2c and – 2c are placed at point A and B as shown in figure. Find out electric field intensity at points C and D. [All the distances are measured in meter]. Solution : Electric field at point C (EA, EB are magnitudes only and arrows represent directions) Electric field due to positive charge is away from it while due to negative charge it is towards the charge. It is clear that EB > EA.  ENet = (EB – EA) towards negative X-axis K(2c) K(2c) =  towards negative X-axis ( 2)2 (3 2)2 = 8000 (– î ) N/C Electric field at point D : Since magnitude of charges are same and also AD = BD So EA = EB   Vertical components of E A and EB cancel each other while horizontal components are in the same direction. 2.K(2c) So, Enet = 2EA cos = cos450 22 K  106 9000 = = î N/C. 2 2 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-15 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 25. Six equal point charges are placed at the corners of a regular hexagon of side ‘a’. Calculate electric field intensity at the centre of hexagon? Answer. Zero Similarly electric field due to a uniformly charged ring at the centre of ring : Note : (i) Net charge on a conductor remains only on the outer surface of a conductor. This property will be discussed in the article of the conductor. (Article no.17) (ii) On the surface of isolated spherical conductor charge is uniformly distributed.  6.3. Electric field due to a uniformly charged ring and arc. SE SOLVED EXAMPLE Example 26. Find out electric field intensity at the centre of uniformly charged semicircular ring of radius R and linear charge density . Solution :  = linear charge density. The arc is the collection of large no. of point charges. Consider a part of ring as an element of length Rd which subtends an angle d at centre of ring and it lies between  and  + d  dE = dEx î + dEy ĵ Ex =  dE x  0 (due to symmetry)   K 2K Ey =  dE y =  dE sin  = R  sin .d = R 0 0 Example 27. Find out electric field intensity at the centre of uniformly charged quarter ring of radius R and linear charge density l.  K Solution : Refer to the previous equation dE = dEx î + dEy ĵ on solving Enet =  (iˆ  ˆj) R Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-16 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics  By use of symmetry and from the formula of electric field due to half ring. Above answer can be justified. (ii) Derivation of electric field intensity at a point on the axis at a distance x from centre of uniformly charged ring of radius R and total charge Q. Consider an element of charge dq. Due to this element the electric field at the point on axis, which is at a distance x from the centre of the ring is dE. There are two component of this electric field dE dEX dEY The y-component of electric field due to all the elements will be cancelled out to each other. So net electric field intensity at the point will be only due to X-component of each element. Q K(dq) x Enet =  dE x =  dECos = R 2 x 2  R 2  X2 O KQx Enet = [R  x 2 ]3 / 2 2 dE R 2KQ E will be max when = 0, that is at x = and Emax = dx 2 3 3 R2 KQ CASE (I) if x>>R, E = 2 Hence the ring will act like a point charge x KQ x CASE (II) if x> R (ii) x > R 1/ 2  x   R2  E= [1 ] = [1 –  1  2  ] 2 0 R2 2 0  x  x 2 1 x  1 R2  R2 R2 Q = [1 – 1 + 2 + higher order terms] = 2 = 2 = 2 0 2 x 4 0 x 40 x 40 x2 i.e. behaviour of the disc is like a point charge. (ii) If x >R).   KQ  KQq0 (i) Force on the point charge + q0 due to the shell = q0 E shell = (q0)  2  r̂ = r̂  r  r2 where r̂ is unit vector along OP. From action - reaction principle, force on the shell due to the point charge will also be KQq0 ˆ Fshell = ( r) r2 Conclusion - To find the force on a hollow sphere due to outside charges, we can replace the sphere by a point charge kept at centre. Example 42. Find force acting between two shells of radius R1 and R2 which + Q1, R1 + Q2, R2 have uniformly distributed charges Q1 and Q2 respectively and r distance between their centre is r. Solution : The shells can be replaced by point charges kept at centre so force between them KQ1Q2 F= r2  6.7. Electric field due to uniformly charged solid sphere Derive an expression for electric field due to solid sphere of radius R and total charge Q which is uniformly distributed in the volume, at a point which is at a distance r from centre for given two cases. (i) r  R (ii) r  R Assume an elementry concentric shell of charge dq. Due to this shell the electric field at the point (r > R) will be Kdq dE = 2 [from above result of hollow sphere] r KQ  Enet = dE = 2 r For r < R, there will be no electric field due to shell of radius greater than r, so electric field at the point will be present only due to shells having radius less than r. KQ' E´net = 2 r Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-26 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Q 4 3 Qr 3 here Q' =  r = 3 4 3 R R 3 3 KQ´ KQr E´net = 2  3 away from the centre. r R Note : The electric field inside and outside the sphere is always in radial direction. SE SOLVED EXAMPLE Example 43. A solid non conducting sphere of radius R and uniform volume charge density  has its centre at origin. Find out electric field intensity in vector form at following positions :  R R  (i) (R/2, 0, 0) (ii)  , ,0 (iii) (R, R, 0)  2 2  Solution : (i) at (R/2, 0, 0) : Distance of point from centre = (R / 2)2  02  02 = R/2 < R, so point lies inside the sphere so   r  R E = = [ î ] 3 0 3 0 2  R R  (ii) At  , ,0 ; distance of point from centre = (R / 2)2  (R / 2)2  02 = R = R, so  2 2  point lies at the surface of sphere, therefore 4  K R3 KQ   R ˆ R ˆ   R ˆ R ˆ E = 3 r = 3 3  i j = i j R R  2 2  3 0  2 2  (iii) The point is outside the sphere 4  K R3 KQ   So E = 3 r = 3 Riˆ  Rjˆ  = 3   Riˆ  Rjˆ  r ( 2R) 6 20   Example 44. A Uniformly charged solid non-conducting sphere of uniform volume charge density  and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure : (i) Point A (ii) Point B (iii) Point C (iv) Centre of the sphere Solution : Method I: (i) For point A : We can consider the solid part of sphere to be made of large number of spherical shells which have uniformly distributed charge on its surface. Now since point A lies inside all spherical shells so electric field intensity due to all shells will be zero.  EA = 0 (ii) For point B : All the spherical shells for which point B lies inside will make electric field zero at point B. So electric field will be due to charge present from radius r to OB. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-27 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics 4  K (OB3  r 3 )  3 3  So, EB = 3 OB =  [OB  r ] OB OB3 3 0 OB3 (iii) For point C, similarly we can say that for all the shells point C lies outside the shell  K[ 34 (R3  r 3 )]  3 3  EC = OC =  R  r OC [OC]3 3 0 [OC]3 Method II: We can consider that the spherical cavity is filled with charge density  and also –, thereby making net charge density zero after combining. We can consider two concentric solid spheres one of radius R and charge density  and other of radius r and charge density –. Applying superposition principle. +      (OA) [ (OA)] (i) EA = E + E  = + =0 3 0 3 0  4     K  r 3 ()  (OB) 3  (ii) EB = E + E  = + 3 OB 3 0 (OB)   r 3     r 3   =   3  OB = 1   OB  3 0 3 0 (OB)  30  OB3  4  4     K  R3 K  r 3 ()  3  3  OC (iii) EC = E + E  = 3 + 3 OC OC  3 3  = [R  r ]OC 30 (OC)3    (iv) EO = E + E  = 0 + 0 = 0 Example 45. In above question if cavity is not concentric and centered at point P then repeat all the steps. Solution : Again assume  and – in the cavity, similar to the previous example.      [OA] ( )PA (i) EA = E + E  = + 30 30      = [ OA – PA ] = [OP] 3 0 3 0 Note : Here we can see that the electric field intensity at point P is independent of position of point P inside the cavity. Also the electric field is along the line joining the centres of the sphere and the spherical cavity.     (OB) K[ 34 r 3 ()]  (ii) EB = E + E  = + PB 3 0 [PB]3    K[ 34 R3 ]  K[ 34 r 3 ()]  (iii) EC = E + E  = OC + PC [OC]3 [PC]3    K[ 34 r 3 ()]  (iv) EO = E + E  = 0 + PO [PO]3 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-28 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Example 46. A non-conducting solid sphere has volume charge density that varies as  = 0 r, where 0 is a constant and r is distance from centre. Find out electric field intensities at following positions. (i) r < R (ii) r  R Solution : Method I: (i) for r < R The sphere can be considered to be made of large number of spherical shells. Each shell has uniform charge density on its surface. So the previous results of the spherical shell can be used. Consider a shell of radius x and thickness dx as an element. Charge on shell dq = (4x2dx)0x Kdq Electric field intensity at point P due to shell dE = 2 x Since all the shell will have electric field in same direction R r R  E  dE =  dE +  dE 0 0 r Due to shells which lie between region r < x  R, electric field at point P will be zero. r  r Kdq r K.4x2 dx0 x 4K0  x 4  0r 2 E  r 2 0 =  r2 = r2  4   = 0 4 0 r̂ 0 0 R R 2 4 K.4x dx0 x 0R (ii) r  R E=  dE =  r 2 = 4 0r 2 r̂ 0 0 Method II: (i) The sphere can be considered to be made of large number of spherical shells. Each shell has uniform charge density on its surface. So the previous results of the spherical shell can be used. we can say that all the shells for which point lies inside will make electric field zero at that point, r K (4x 2 dx)0 x   0r 2 0 so E(r  R) = = r̂ r2 4 0 (ii) similarly for r  R, all the shells will contribute in electric field, therefore R K (4x 2 dx)0 x   0R 4 0 E(r  R) = = r̂ r2 4 0r 2  7. ELECTRIC POTENTIAL In electrostatic field the electric potential (due to some source charges) at a point P is defined as the work done by external agent in taking a point unit positive charge from a reference point (generally taken at infinity) to that point P without changing its kinetic energy.. 7.1. Mathematical representation : If (W  P)ext is the work required in moving a point charge q from infinity to a point P, the electric potential of the point P is Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-29 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Wp )ext   Welc )p U Up  U Up Vp       q  K  0 q q q q Note : (i) (W  P)ext can also be called as the work done by external agent against the electric force on a unit positive charge due to the source charge. (ii) Write both W and q with proper sign. 7.2. Properties : (i) Potential is a scalar quantity, its value may be positive, negative or zero. joule (ii) S.. Unit of potential is volt = and its dimensional formula is [M1L2T–3–1]. coulmb (iii) Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point. (iv) Electric potential due to a positive charge is always positive and due to negative charge it is always negative except at infinite. (taking V = 0). (v) Potential decreases in the direction of electric field. (vi) V = V1 + V2 + V3 +....... 7.3. Use of potential : If we know the potential at some point ( in terms of numerical value or in terms of formula) then we can find out the work done by electric force when charge moves from point 'P' to  by the formula Wel )p  = qVp SE SOLVED EXAMPLE Example 47. A charge 2C is taken from infinity to a point in an electric field, without changing its velocity. If work done against electrostatic forces is –40J then find the potential at that point. Wext 40J Solution : V= = = –20 V q 2C Example 48. When charge 10 C is shifted from infinity to a point in an electric field, it is found that work done by electrostatic forces is –10 J. If the charge is doubled and taken again from infinity to the same point without accelerating it, then find the amount of work done by electric field and against electric field. Solution : W ext) p = –wel) p = wel)p  = 10 J Wext )p 10J because KE = 0  Vp = = = 1V q 10C So if now the charge is doubled and taken from infinity then w ext ) p 1=  W ext) P = 20 J  W el )  P = –20 J 20C Example 49. A charge 3C is released at rest from a point P where electric potential is 20 V then its kinetic energy when it reaches to infinite is : Solution : W el = K = Kf – 0 W el)P = qVP = 60 J So, Kf = 60 J Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-30 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics Electric Potential due to various charge distributions are given in table. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-31 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics  7.4. Potential due to a point charge : Derivation of expression for potential due to point charge Q, at a point which is Q r P at a distance r from the point charge. from definition potential r   Wext( p)   (q0E)   r r   r KQ KQ V=    =  E  dr   V = – 2 (dr)cos180º = qo qo r r   SE SOLVED EXAMPLE Example 50. Four point charges are placed at the corners of a square of side  calculate potential at the centre of square. Solution : V = 0 at 'C'. Example 51. Two point charges 2C and – 4C are situated at points (–2m, 0m) and (2 m, 0 m) respectively. Find out potential at point C. (4 m, 0 m) and. D (0 m, 5 m). Solution : Potential at point C K(2C) K( 4C) 9  109  2  10 6 9  109  4  10 6 VC = Vq1  Vq2 = + = – = –15000 V. 6 2 6 2 K(2C) K( 4C) K(2C) K( 4C) Similarly, VD = Vq1  Vq2 = + = + = – 6000 V. ( 5)2  22 ( 5)2  22 3 3  Finding potential due to continuous charges If formula of E is tought, then we take If formula of E is easy then we use r r   a small element and integrate V=  E· dr – r  V=  dv (i.e. for sphere, plate infinite wire etc.) SE SOLVED EXAMPLE Example 52. A rod of length  is uniformly charged with charge q calculate potential at point P. Solution : Take a small element of length dx, at a distance x from left end. Potential due to this small element x K(dq) k dq dV = x total potential V=  x0 x x r   q  K  dx q    = Kq log    r  dq = dx  V=  x r x  e   r  Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-32 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Electrostatics  7.5. Potential due to a ring : (i) Potential at the centre of uniformly charged ring : Potential due to the small element dq Kdq dV = R K dq Net potential V=  R K Kq V= R dq = R (ii) For non-uniformly charged ring potential at the center is Kqtotal V= R (iii) Potential due to half ring at center is : Kq V R (iv) Potential at the axis of a ring: dq Calculation of potential at a point on the axis which is a distance x from centre of uniformly charged (total charge Q) R ring of radius R. Consider an element of charge dq on the ring. x Potential at point p due to charge dq will be P K(dq) dv = R2  x 2 Net potential at point P due to all such element will be KQ V=  dv  R2  x2 SE SOLVED EXAMPLE Example 53. Figure shows two rings having charges Q and – 5 Q. Find Potential difference between A and B (VA – VB). Solution : VA = KQ +  K  5 Q  VB =  K  5Q  + K  Q R 2R  2R 2  R 2  R 2  R 2 From above we can easily find VA – VB. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected] MAINES-33 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029  Electrostatics Example 54. A point charge q0 is placed at the centre of uniformly charged ring of total charge Q and radius R. If the point charge is slightly displaced with negligible force along axis of the ring then find out its speed when it reaches to a large distance. Solution : Only electric force is acting on q0 1  W el = K = mv2 – 0  2 KQ   Now W el)c= q0 Vc = q0. R Kq0Q 1 2Kq0Q  = mv2  v= R 2 mR  7.6. Potential due to uniformly charged disc : V σ 2ε 0 2 2  R  x  x , where  is the charged density and x is the distance of the point on the axis from the center of the disc, R is the radius of disc. Finding potential due to a uniformly charged disc: A disc of radius 'R' has surface charge density (charge/area) = . We have to find potential at its axis, at point 'P' which is at a distance x from the centre. For this We can divide the disc into thin rings and lets considered a thin ring of radius r and thickness dr. Suppose charge on the small ring element = dq. Potential due to this ring at point 'P' is = Kdq dV = r 2  x2 Kdq So, net potential : Vnet = r  x2  2 dq Here,  = charge/area = d(area) So, dq =  × d (area) =   2r dr  (here d (area) = area of the small ring element =(length of ring) × (width of the ring) = (2r). (dr) r R K  2r dr  So, Vnet =  r 2  x2 r0 to integrate it let r2 + x2 = y2 2r dr = 2y dy, substituting we will get : r R 1 (2) y dy  r R Vnet   40 y  Vnet  20  y  r 0 r0 r R Vnet   20  r 2  x2 r 0   Vnet   20 R2  x 2  x   If a test charge q0 is placed at point P, then potential energy of this charge q0 due to the disc = U = q0V     U = q0   2 0  R2  x 2  x    Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : [email protected]