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Problem 1 : A non-conducting sphere has radius R=2.31cm and uniformly distributed charge q = +3.50 fC. Take the electric potential at the sphere’s center to be Vo = 0. What is V at radial distance (a) r =1.45cm and (b) r = R. Solution : Q = 3.5 fC Gaussian surface r R=2.31cm Let us first derive a general expression for potential at any radius r. Let us consider a Gaussian surface of radius r (r < R) as shown in the fig.. Applying Gauss theorem we get : ∫E.ds = q/Є0 where E = Electric field intensity on the Gaussian surface (it is uniform and radial to the surface) and q = Total charge enclosed within the Gaussian surface. As the charge is uniformly distributed with in the sphere, charge density ρ = Q/(4/3ΠR3) Charge enclosed with in the Gaussian surface = q = ρ(4/3Πr3) = [Q/(4/3ΠR3)] x (4/3Πr3) = Qr3/R3 Applying Gauss theorem we get : ∫E.ds = E(4Πr2) = Qr3/Є0R3 Or E = [Q/4ΠЄ0R3] r As E = -dV/dr, V = - ∫Edr = - [Q/4ΠЄ0R3]∫r dr = - [Q/4ΠЄ0R3][r2/2] Or V = - [kQ/2R3] r2 where k = 1/4ΠЄ0 = 9 x 109 N m2/C Substituting Q = 3.5 fC and R = 2.31 cm we get : V = - [9x109x3.5x10-15]/[2x(2.31x10-2)3]r2 = -1.28 r2 For r = 1.45 cm, V = -1.28 x (1.45x10-2)2 = -2.69 x 10-4 V = - 0.269 mV For r = 2.31 cm, V = -1.28 x (2.31x10-2)2 = -6.83 x 10-4 V = - 0.683 mV Problem 2 : Two particles, of charges Q1 and Q2, are separated by distance d horizontally. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x axis (other than at infinity) at which the electric potential due to the two particles is zero. Solution: E=0 V=0 Q1 Q2 P L Q d d/4 As electric field is zero at P, E = kQ1/(d/4)2 + kQ2/(3d/4)2 = 0 Solving we get : Q2 = - 9Q1 Let the point at which potential is zero be Q at a distance L from Q1. Then, Net potential at Q = V = Potential due to Q1 + Potential due to Q2 V = kQ1/L + kQ2/(d – L) = 0 or Q1/L = - Q2/(d – L) = + 9Q1/(d – L) Or 1/L = 9/(d – L) or d – L = 9L or L = d/10