A hollow metal sphere has 3 cm and 5 cm inner and outer radii, respectively. The

Question

A hollow metal sphere has 3 cm and 5 cm inner and outer radii, respectively. The surface charge density on the inside surface is -125 nC/m2. The surface charge density on the exterior surface is +125 nC/m2. What are the strength and direction of the electric field at points 2, 4, and 6 cm from the center? (Consider that the surface charge density is created by a point charge placed within the sphere.)

Can you pleases explain why this is. I have looked at other answers but do not understand how they found the charge. Someone else has already tried to answer this but they were incorrect.

Explanation / Answer

E inside,

By Gauss’ law

E(in) = qenclosed/0 = 0/0 = 0

E ouside,

By Gauss’ law

E(out) = qenclosed/0 = *A/0 = *4r2/0 = (-125*10^-9*4*3.14*(0.05)^2)/(8.85*10^-12) = -443.5 N/C

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