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A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential is half of the potential at the centre?

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Solution: we know potential outside the sphere is V1 = kQ/x     where x is the distance from the center x>R

and potential inside the sphere V1 = kQ/2R [3-x2/R2]

for center put x=0  Vo = 3kQ/2R

condition given is V1 = Vo/2

kQ/x  = (3kQ/2R)/2

x=4R/3

distance from the surface is  =x-R

= 4R/3 -R

=R/3

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