Question

A solid conducting sphere of radius 3.0 cm has a charge of 30 nC distributed uniformly over its surface. Let A be a point 1.0 cm from the center of the sphere, S be a point on the surface of the sphere, and B be a point 5.0 cm from the center of the sphere. What are the electric potential differences VAVBV_A - V_B?

Solution

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Answered 5 months ago
Answered 5 months ago
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On the surface we have:

VS=k q/R=(9e9)(30e9)0.03=9000 VV_S = k\ q/R = \dfrac{(9e9)(30e-9)}{0.03} = 9000 \ V

The electric potential inside a solid conducting sphere is equal to the electric potential on the surface:

VA=VS=9000 VV_A = V_S = 9000 \ V

Then, outside the sphere we have:

VB=k q/r=(9e9)(30e9)/(0.05)=5400 VV_B = k\ q/r = (9e9)*(30e-9)/(0.05) = 5400 \ V

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