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Q.

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

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a

V3r

b

V2r

c

V4r

d

V6r

answer is C.

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Detailed Solution

V=q4πε0r-q4πε03r=q6πε0r

Electric field intensity at a distance 3r is given by

                         E=q4πε0(3r)2=q4πε0·9r2

Thus,                EV=q6πε0r4πε09r2(q)=16r⇒E=V6r

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