Question

A conducting sphere of radius R has surphase charge density. The electric potential on surface is

Answer 1

Answer:

Explanation:

you know that the charge will be distributed on the surface of the conducting sphere.

Exploiting the spherical symmetry with Gauss’s Law, for r≥R,

EA=E(4πr2)E=q/ϵ0=1/4πϵ0qr2

V=1/4πϵ0qr

At r = R, just sub. r = R into the above equations. Note that the surface charge density can be introduced to give E=σϵ0

For r≤R,

E=0

Since the electric field within the conductor is 0, the whole conductor must be at the same potential (equipotential). The electric potential within the conductor will be:

V=1/4πϵ0qR

Answer 2

★ Question Given :

• ➦ A conducting sphere of radius R has surphase charge density. The electric potential on surface is ?

★ Required Solution :

• ➦ Electric potential on the surface of conducting sphere of radius R

✪ According to Question :

• ➝ V = 1 / 4πε0 q / R

• ➝ q = 4πR²σ

• ➝ V = σ R / ε0

★ Therefore :

• ➦ A conducting sphere of radius R has surphase charge density. The electric potential on surface is V = σ R / ε0.

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