CHAPTER3 Electric Potential 3.1-3.50 Introduction 31 Work Done to Move a Charge in an Electrostatic Field 3.1 Potential and Potential Difference 32 Potential 32 Continuous Charges 33 Finding Potential Difference 34 Potential Difference in a Uniform Electric Field 34 Equipotential Surface 36 Concept Application Exercise 3.1 38 Finding Electric Field from Electric Potential 39 Concept Application Exercise 3.2 312 Electric Potential Energy 3.12 Potential Energy of a System of Three Charges 3.13 Potential Energy of Charges in an External Electric Field 3.14 Potential Energy of Single Charge in an External Field 3.14 Potential Energy of a System of Two Charges in an External Field 3.14 Concept Application Exercise 3.3 B17 Electric Potential of Some Continuous Charge Distributions 3.18 ‘Charged Conducting Sphere 3.18 Non-Conducting Solid Sphere 3.20 Outside the Sphere 3.20 Inside the Sphere 3.20 Uniform Line of Charge 3.21 A Ring of Charge 3.22 Charged Disk 3.23 Energy for a Continuous Distribution of Charge 3.23Self-Energy of a Charged Spherical Shell Sphere Self-Energy of a Uniformly Charged Sphere Self and Interaction Energy of Two Spheres Electric Field and Potential Due to Induced Charges Earthing of a Conductor Charge Distribution on a Conductor Surface (Uniqueness Theorem) Circular Motion of a Charged Particle in Electric Field Concept Application Exercise 3.4 Potential Due to an Electric Dipole Work Done in Rotating an Electric Dipole in a Uniform Electric Field Potential Energy of an Electric Dipole in a Uniform Electric Field Concept Application Exercise 3.5 Van De Graaff Generator Exercis Single Correct Answer Type Multiple Correct Answers Type Linked Comprehension Type Matrix Match Type Numerical Value Type s Archives Answers Key 3.23 3.23 3.24 3.24 3.25 3.26 3.27 3.28 3.29 3.29 3.29 3.31 3.32 3.33 3.33 3.38 3.41 3.44 3.46 3.47 349INTRODUCTION The concept of potential energy was introduced in mechanies in connection with conservative forces such as the gravitational force and the elastic force exerted by a spring. By using the law of conservation of energy, we can avoid working directly with forces when solving problems in mechanics. The concept of potential energy is also of great value in the study of electricity. Because electrostatic force is conservative, the electrostatic phenomena can be conveniently described in terms of electric potential energy. With this idea, we can define a scalar quantity known as electric potential. Since the electric potential at any point in an electric field is a scalar quantity, it can be used to describe electrostatic phenomena more simply than by relying only on electric field and electric forces. ‘The potential is characteristic of the field only, independent of a charged test particle that might be placed in the field. Potential energy is characteristic of the charge-field system due to an interaction between the field and a charged particle placed in the field. WORK DONE TO MOVE A CHARGE IN AN ELECTROSTATIC FIELD Let us consider an arbitrary electric field due to any charged object. If we move a test charge +g from position 1 to position 2 (see figure) in this electrostatic field, at each position of the test charge, it will experience an electrostatic force F, p move the test charge slowly, we must pull it against the electric force (field) witha force F,=-q E, opposite to the electric field arge along the dashed line from 1 to 2 is, Wog= ff Fey dt= f° (-gB)-dt = ‘The ext get des werk W=-4 [Ein wansponing ne ts 1 charge q slowly from position | to position 2 in the static electric field. Find the work done by an extemal agent in slowly shifting a charge q = | UC in the electric field E=10°7 Vm"! from the point P(1, 2) to 0 (3, 4). [SGD The work done by the external agent against the electric field is =-gJeei- si say (10 (10° )(3 = 1) = = 2 «10°F Accharge particleg=—10 uC is carried along OP and PQ and then back to O along QO as shown in figure, in an electric field E=(x+2y)i+2a7. Find the work done by anexternal 5} ‘agent in (a) each path and a) (b) the round trip. 012, 2) [SAD Work done by an external agent in an electric field (@) W =~] fi di=—q(f Ede + JE,dv) In Path 0 — P: Since the displacement is along the x-axis for the path OP, Wan =-af El where y=0 or Wy jp=+20x 10°F In Path PQ: Since the displacement is along y-axis for path PQ, 2 afEd, -a2h, xdy) whe ==2(-10% 10) 2d = 80x 105 In Path Q — 0: Since the displacement dl = dxi + dyj along OO, Wo-s0=—4f Exde + Eyd,) 10x10) (x + 29), Weg -afier + 2y) de + xdy =-a[ce+2ndr+ 2f ray]3.2_Electrostatics and Current Electricity here the equation of the path QO is given as Wo so=—a[ fete 2d + 2ft a] -~«(3ER-40"1) =-100x 10° (b) Then, the total work done in the round trip is W= Wop t Wyo t Wog=0 Intheabove example, Edi and static. ILLUSTRATION 3.3 : £20 A uniform electric field is present in region 2, whereas itis dropped to zero abruptly. 3 in regions 1 and 3. Is this electric field conservative? ). Hence, the field is conservative BD The closed integral of £ in the loop abeda is Since E.Ld 1 fromd'o a and b toc. féar=[éai=o Since £0 in region 3, | E-d1'= | EdlcosO =[z dl =Eh(#0) Since § E-d1'#0, the given fiel non-conservative. POTENTIAL AND POTENTIAL DIFFERENCE POTENTIAL We know that in slowly bringing a point charge from point I to point 2, the work done by the external agent is independent of the path followed by the charge, which can be given by 2 ~qfE-dt 1 If we choose the initial point at infinity and the final point at P, the above expression will be e Woy=—a) E-dl ‘Wecan see that IV, will be different for different values ofthe test charge. This means that even though you perform the same work in bringing a test charge from infinity to any point P following different paths, you will have to do different works in bringing different test charges. However, if we take the ratio I7,,/q, that is, work done per unit test charge, we will get a constant quar which can be defined as “potential at the point P” denoted by V. Itis a constant quantity for a given point. Hence, itis a point function (like electric field) that may vary with positions. The “potential 7” characterizes an electric field as a scalar function in addition to “field strength E” as a vector function. Hence, potential field is a scalar field. Infinity 1 mh The potential at any point P is given by Vv fe & Which is equal to the external ‘work done per unit positive charge in shifting slowly from infinity to this point. The potential at any point P is given by Wego P) _ q F v -[é-a? Then, the field expression for potential is, The potential at any observation point P of a static electric field is defined as the work done by the external agent (or negative of work done by the electrostatic field) in slowly bringing a unit positive point charge from infinity to the observation point. The potential at a point is more if the external agent does more ‘work to shift the charge from infinity to the given point and vice versa. In this way, we define potential as the extemal work done by unit charge or roughly potential energy per unit charge,Electric Potential 3.3 Potential Due to a Point Charge The definition of a potential is given by the expression From co e e V=JE-dl=JE-dlcoso a fee an 2 =! aaa * P -2 ~ ane yr if If r30,V 30. Hence, varies hyperbotically. If Q is positive, V is positive and viee versa, ‘The electric field in a region is given by E=4i, Write x an expression for the potential in the region assuming the potential at infinity to be zero. r, Jt [SAD As £ = 41x", potential in the region is te fA va=-fE-d Si} je-d=-{(4))-an x 27 A a rae] J 2b ar Potential Due to Collection of Charges Discrete Charges Following the principle of superposition of potentials as t described in the last section, let us find the potential V due to a collection of discrete point charges 4), dy. dys at apoint P. As we know, the potential at | Thepotentalat Pduetothe system ras of point charges is givenas the sum point oftheir individual potentials tP, vay, _ 1 > Yip, Be where V.= y =—4— and Ane gy, ris the magnitude of the position vector P relative to g,.Then 4, La © AE) oy 7 If the positions of the charges are given from a fixed origin O (see figure) put = 7 —7 to obtain Ane) Continuous CHARGES For continuous charge distribution 1 pdg yoy ina, ! r where r is the position of the point p from elementary charge dg. Put dg (for volume charge) (surface charge) (line charge) yo J PAY Cor volume charge distribution) ana) = J (For surface charge distribution) ane, aa! fA for tine charge dist ane, where p, 6, and 2 may be a function of r ‘Three charges g, = 1 UC, g, = ~2 HC, and g, = —I WC are placed at 4(0, 0, 0), B-1, 2, 3,) and C(2, -1, 1). Find the potential of the system of three charges at P(1, ~2,—1).. [SAD Ir §,, is the position of P from the charge, its potential 10° x10" x9 9x10° -2j-b- G7 +07+0k)| V6 A are ipa 7 p 2X1 XOX __ 5 aghy = b)-Ci+2} +36) 2 <10V \G-2j-H-@i-j+b) Vo Then, V,=EV,=V,+V,+V,=3x10-V