Two metal spheres, each of radius 4.0 cm, have a center-to-center separation of

Question

Two metal spheres, each of radius 4.0 cm, have a center-to-center separation of 1.4 m. Sphere 1 has charge +2.60 10-8 C. Sphere 2 has charge of 3.90 10-8C. Assume that the separation is large enough for us to assume that the charge on each sphere is uniformly distributed (the spheres do not affect each other). Take V = 0 at infinity. (a) Calculate the potential at the point halfway between the centers. V

(b) Calculate the potential on the surface of sphere 1. V

(c) Calculate the potential on the surface of sphere 2.

Explanation / Answer

here,

q1 = 2.60 * 10^-8 C
q2 = -3.90 * 10^-8 C
seperation distance, d = 1.4 m
Radius of each sphere, r = 4 cm = 0.04 m

Potential is given as:
V = K*q/s

Where, q is charge
k = columb's constant
r distance to point

Part A:
Distance to midpoint, s = d/2 = 1.4/ = 0.7 m

V = v sphere 1 + V sphere 2
V = k/s (q1 + q2)
V = (9*10^9/0.7)*(2.60 * 10^-8 + (-3.90 * 10^-8))
V = -167.143 V

Part B:
distance to surface ,
for sphere 1, s1 = R = 0.04
for Sphere 2, s2 = d - R = 1.4 - 0.04 = 1.36 m

Net potential on surface of sphere 1 :
V1 = k*(q1/s1 + q2/s2)
V1 = 9*10^9((2.60 * 10^-8)/0.04 + (-3.90 * 10^-8)/1.36)
V1 = 5591.912 V

Part C:
Potential at surface of sphere 2,
V2 = k*(q1/s2 + q2/s1)
V1 = 9*10^9((2.60 * 10^-8)/1.360 + (-3.90 * 10^-8)/0.04)
V1 = -8602.941 V

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