Question

A hollow charged metal sphere has radius R. If the potential difference between the surface and a point at distance 3R from centre is V, then the electric field intensity at a distance 5R from surface is​

Answer 1

On the surface of a hollow metal sphere, the charge Q is uniformly distributed. This will result in equal potential and the electric field outside the sphere.

Look at the solution below.

Solution:

The potential difference is given below.

V = 1/4π ε0 ( Q/r - Q / 3r)

V = 2/3 1/4π ε0 Q/r

Now the electric field is

E = 1/4π ε0 (Q/3r)^2 = 1/9 1/4π ε0 Q/r^2

E = 1/9 3/2V  ÷  / r

E = 1/6 v/r

Thus the value of electric field intensity is E = 1/6 v/r

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