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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