CHAPTER: 2 ELECTROSTATIC POTENTIAL AND CAPACITANCE

CHAPTER: 2 ELECTROSTATIC POTENTIAL AND CAPACITANCE. Define electric potential at a point. *Electric potential at a point is efine as the work one to bring a unit positive charge from infinity to that point. V = W Q ; S.I Unit : J/C Or V Note: Electric potential is a scalar quantity. 2. Derive an expression for potential at a point ue to a point charge. points is efine as the work one to bring a unit positive charge from one point to another. Potential ifference between A an B is VAB = VA VB VAB =[ q - q ] 4πε0 r 4πε0 r2 5. Derive the relation between electric fiel an potential. The work one to move a charge from A to B is, W = E ( - x ) W = F.x [Since isplacement is against force] E= q 4πε0 x 2 W = - 4πε0 q x 2 x The total work one to bring the unit positive charge from to P is, r W = W = - q r x 4πε0 x 2 = - q [- 4πε0 x ] r q = [ 4πε0 r - ] W = q 4πε0 r This is store as potential at the point P V= 4πε0 q r 3. Draw the graph showing variation of V an E with istance r. The work one to move a charge from A to B is given by, W = E ( - x ) This work one is the potential ifference between A an B i.e., V = E ( - x ) E = ( -V x ) i.e, Electric fiel intensity (E ) is the negative graient of electrostatic potential. 6. Derive an expression for the potential ue to an electric ipole. Consier an electric ipole having a ipole moment p = 2 l q. We have to fin the electric potential at a point P, istant r from the mi - point of the ipole. 4. Define electric potential ifference. * Potential ifference between two P a g e A B D U L M U K H S I T H

Electric potential at P ue to -q charge - q at A is, V- = 4πε0 r + l Cosθ Electric potential at P ue to +q charge + q at B is, V+ = 4πε0 r - l Cosθ Total Electric potential at P is, V= V+ + V- q V=[ 4πε0 r - l Cosθ + - q 4πε0 r + l Cosθ ] q r + l Cosθ - (r - l Cosθ) = [ 4πε0 r 2 - l 2 Cos 2 ] θ q = [ 2 l Cosθ 4πε0 r 2 - l 2 Cos 2 θ ] = [ p Cosθ 4πε0 r 2 - l 2 Cos 2 θ ] If r 2 >> l 2, [l 2 can be neglecte.] Cosθ V = [p 4πε0 r 2 ] Special cases:- I. Potential at a point on the axial line Put θ = 0 0, Cos 0 V = [p 4πε0 r 2 ] = [p x 4πε0 r 2 ] p V = 4πε0 r 2 II. Potential at a point on the equitorial line Put θ = 90 0, Cos 90 V = [p 4πε0 r 2 ] = [p x 0 4πε0 r 2 ] V = 0 Note: The equatorial plane of the ipole is an equipotential surface having a potential zero. 7. Define potential energy of a system of charges. * Potential energy of a system of charges is the work one to bring the charges from infinity to their present positions. (a). Bring a charge q from to a point O (no work to be one) (b). Calculate the potential at P istant r from q is V = q r 4πε0 (c). Bring a charge q2 from to B by oing work on q2 against the repulsive force, W= V q2 V = W Q qq2 W = 4πε0 r Potential energy of the system of A charges is qq2 U= 4πε0 r *Similarly to separate the charges q an - qq2 q2, Work = U = 4πε0 r P.E. of a system of three charges To assemble the charges in given positions, Work one W = W2 + W3 + W23 i.e., U = U2 + U3 + U23 qq2 U= [ + qq3 + 4πε0 r2 4πε0 r3 4πε0 q2q3 r23 ] P. E= U= For n charges, U= 4πε0 [qq2 + qq3 + q2q3 ] r2 r3 r23 i=n i= j=n qiqj j= rij j 8. Define one Electron Volt (ev). Give its relation with joule. Electron volt is a smaller unit of energy. ev is efine as the energy acquire by an electron, when it is accelerate through a p.. of V. W = qv => ev = (.6 0-9 ) J 2 P a g e A B D U L M U K H S I T H

ev =.6 0-9 J 9. What is an equipotential surface? Give examples. Write some properties of it. *It is a surface having same potential at all points. Example : Concentric spheres with a point charge at the centre are equipotential surfaces. Example 2: In a uniform electric fiel parallel planes perpenicular to the electric lines of force are equipotential surfaces. Capacitance or capacity (C):- It is the ability of capacitor to store electric charge. C = Q V ; S.I unit C or Fara (F) V Capacitance of a conuctor is the amount of charge require to raise the potential by unity.. Define one fara Ans: Capacitance of a capacitor is sai to be one fara if one coulomb of charge raises its potential by one volt. (mf = 0-3 F; μf = 0-6 F; nf= 0-9 F; pf= 0-2 F) 2. Explain the principle of a parallel plate capacitor. Properties of equipotential surfaces:-. The work one to move a charge from one point to another on an equipotential surface is zero. 2. Two equipotential surfaces will never intersect each other. (a) (b) 3. Electric lines of force(electric fiel) pass normal( ) to an equipotential surface. CAPACITORS 0. What is the use capacitor? Define capacitance. * It is a evice use to store electric charge. (c) (a). Consier a positively charge plate P. (b). If another plate P2 with no charge, is brought near to P (an place without touching), then on the insie of the plate negative charges are inuce an on the outsie positive charges are inuce. (c).if the secon plate is earthe all the positive charges, will flow to earth. 3 P a g e A B D U L M U K H S I T H

Now ue to the presence of negative charges on the plate P2, the potential (V) of P ecreases. Therefore, by equation C = Q V When potential ecreases capacitance increases. This is the principle of capacitor. 3. Derive an expression for the capacitance of a parallel plate capacitor. Consier a parallel plate air capacitor having plate area A an charge ensity σ. Charge on a plate, Q = σ A V = E V= σ ε0 We know Capacitance, C = Q V C= σ A σ = ε0 A ε0 C air = ε0 A This is the expression for capacitance of a capacitor with air as the meium between the plates. 4. What happens if a ielectric material is introuce between the plates of the capacitor? If a ielectric material is introuce between the plates of the capacitor, the capacitance becomes C ielectric = K ε0 A K- ielectric constant. *When the ielectric is introuce in the region between the plates, the capacitance increases k times C ielectric = K Cair K= εr 5. Give the expression for capacitance of a parallel plate capacitor partially fille with a ielectric slab. *When a ielectric of relative permittivity εr of thickness t is introuce partially between the plates of the capacitor. Electric fiel outsie the ielectric slab is, E0 = σ ε0 Electric fiel within the ielectric slab of thickness t (shae region) is, Ek = σ σ = K= εr ε0 εr Kε0 The potential ifference between the plates is, V= E0 (-t) + Ek (t) V= E x 4 P a g e A B D U L M U K H S I T H

V= σ (-t) + σ (t) ε0 Kε0 σ V= [(-t) + t ε0 K ] Capacitance = C = Q V Where Q= σ A C = Q V = σ A σ t ε0 [( t) + K ] C = ε 0 A t ( t) + K If the ielectric fills the entire space between the plates t = then, ε C = 0 K ε0 A A = ( ) + K 6. What are the avantages of introucing ielectric slab between the plates of a capacitor? (i). The capacitance increases εr times. (ii). Dielectric meium avois sparking between the plates. 7. Derive expressions for effective capacitance when capacitors are connecte in (i) series an (ii) parallel. (i). Series: Consier 3 capacitors of capacitances C, C2, C3 connecte in series with a voltage V. In a series circuit the charge store in each of the capacitors is the same but the voltages across them are ifferent. Applie voltage, V = V + V2 + V3. () We have C = Q V V = Q C V = Q ; V2 = Q C C ; 2 V3 = Q C 3 V= V + V2 + V3 V= Q C + Q C 2 V = Q [ C + + Q C 3 C + 2 C ] 3 V Q = C + C + 2 C 3 = Cs C + C + 2 C 3 In the case of two capacitors C C 2 Cs = C + C 2 (ii). Parallel: Consier 3 capacitors C, C2, C3 connecte in parallel with a voltage V. In a parallel circuit, the voltage is the same but the charges store in the capacitors are ifferent. Here the total charge q = q + q2 + q3. () We have C = Q V Q = CV If Cp is the effective capacitance when the three capacitors are connecte in parallel. 5 P a g e A B D U L M U K H S I T H

q = CV; q2 = C2V; q3 = C3V Cp V = V (C+ C2 + C3) Cp = (C+ C2 + C3) 8. Derive an expression for the energy store in a capacitor. * Consier a capacitor of a capacitance C, it has given a voltage V. Let at any instant the charge in the capacitor be q. Now the work one to increase the charge by an amount q is given by, w = Vq V= W Q Energy ensity (u) = Energy Volume u= ½ ε0 A E2 A u = ½ ε0 E 2 20. Derive an expression for the lost energy ue to sharing of Capacitors. Ans: Let two capacitors C an C2 having charge to potentials V an V2, connecte in parallel. Let V be the common potential. But V = q C W = q C q The total work one to increase the charge from O to Q is given by, Q Q W = W = q 0 C q 0 W = C 0 Q q q = C [ q 2 2 ] Q 0 = C [ Q 2 2-0 2 2 ] Q 2 W = 2C But Q = CV ; (CV) 2 (CV 2 ) W = = 2C 2 W = ½ CV 2, This work one is store as energy of the capacitor. U = ½ CV 2 Or U = ½ QV Or U = Q 2 2C 9. Derive an expression for energy ensity of a parallel plate capacitor. We have the expression for energy of a capacitor as, U = ½ CV 2 = ½ ( ε0 A ) (E)2 = ½ ε0 A E2 2 = ½ ε0 A E 2 When they are connecte through a conucting wire charge flows from higher potential to lower potential till both of them acquire a common potential (v). Accoring to charge conservation, Initial charge = final charge i.e., (C + C2 )V = CV + C2V2 Common potential CV + C2V2 V = C + C2 Total energy before sharing ui = ½ C V 2 + ½ C2 V2 2 Total energy after sharing uf = ½ C V 2 + ½ C2 V 2 = ½ ( C + C2 )V 2 Loss of energy = ui uf [½ C V 2 + ½ C2 V2 2 ] - [½ ( C + C2 )V 2 ] [½ C V 2 + ½ C2 V2 2 ] - [½ ( C + C2 ) ( CV + C2V2 C + C2 ) 2 ] 6 P a g e A B D U L M U K H S I T H

*Their ipole moment is zero. ½ [C C2 ( V 2 + V2 2-2 V V2)] ½ [C C2 ] C + C2 C + C2 [ V - V2 ] 2 2. What are ielectrics? Dielectrics are non-conucting substances or insulators. But they allow electric fiel to pass through them. 22. Distinguish between polar an non-polar molecules. * In certain molecules, the centres of positive charges an centres of negative charges o not coincie. These molecules are calle polar molecules. *They have permanent ipole moment. E. g: HCl, H2O, NH3, etc. *But in some other molecules the centre of positive charges an centre of negative charges coincie. These molecules are calle non-polar molecules. e. g: O2, N2, H2, CO2, etc. 23. Explain the polarization in nonpolar molecules. In the absence of external E.F, nonpolar molecules have no permanent ipole moment. In an external E.F, the positive an negative centers of the non-polar molecule are isplace in the opposite irections. Thus the molecule evelops an inuce ipole moment. Then the ielectric is sai to be polarize. The inuce ipole moments of ifferent molecules a up giving a net ipole moment of the ielectric in the presence of external electric fiel. 24. Define Polarization an electric susceptibility. Whether polar or non-polar, a ielectric evelops a net ipole moment in the presence of an external electric fiel. The ipole moment per unit volume of the ielectric is calle polarization (P ). For linear isotropic ielectrics, P = χe E χe is calle electric susceptibility of the ielectric meium. 7 P a g e A B D U L M U K H S I T H

25. Explain the main points of electrostatics of conuctors. (i) Insie a conuctor Electrostatic fiel is zero. *In the static situation, the free charge carriers are so istribute themselves that the E is zero everywhere insie. Net electric fiel intensity in the interior of a conuctor is zero. When a conuctor is place in an electrostatic fiel, the charges (free electrons) rift towars the positive plate leaving the + ve core behin. At an equilibrium, the electric fiel ue to the polarisation becomes equal to the applie fiel. So, the net electrostatic fiel insie the conuctor is zero. (ii) On the surface E is always Normal. Reason: If E were not normal to the surface, it woul have some non-zero component along the surface. Free charges on the surface of the conuctor woul then experience force an move. (iii) No charge insie the conuctor but gets istribute on the surface. Since E = 0 in the interior of the conuctor, q=0 (iv)electrostatic Potential is constant insie an on the surface Since E = 0 insie the conuctor an has no work is one in moving a small test charge within the conuctor an on its surface. That is there is no potential ifference between any two points insie or on the surface of the conuctor. (iv). Electric fiel at the surface of a charge conuctor E= σ ε0 n Proof:- Electric file at P is, E= q 4πε0 r 2 Electric fiel on the shell (r=r) E= q 4πε0 R 2 [Where q= 4πR 2 σ] 4πR 2 σ E = 4πε0 R 2 = σ ε0 i.e, E= σ n ε0 Where σ is the surface charge ensity an is a unit vector normal to the surface in the outwar irection If σ is ve, electric fiel is normal to the surface inwar. 6. Electrostatic shieling. Electric fiel is zero insie the cavity of a conuctor of any shape. By Gauss s theorem, ɸ = E s = q ε0 8 P a g e A B D U L M U K H S I T H

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