Q.43. A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential is half the potential at its centre?
(a) R
(b) R/2
(c) R/3
(d) 2R

Dear Student,

Here it is not mentioned that, wether the sphere is a conductor or non conductor. I am considering the sphere is a conducting sphere.

Please find below the solution to the asked query:

The potential at the centre of the solid conducting sphere is,

V=14πε0QR
Q is the charge on the sphere and R is the radius of the sphere.

Let the distance from the surface of sphere is x, where the potential is half of the potential at the centre. Therefore,

V'=14πε0QR+x  V2=14πε0QR+x  14πε0Q2R=14πε0QR+x 12R=1R+x  2R=R+x x=R

 

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