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A spherical conductor of radius R is carrying a charge of +Q. The ratio of the electric potentialscorresponding to a point on the surface of the conductor and a point at a distance2R from thecentre of the conductor are in the ratio

Question

A spherical conductor of radius R is carrying a charge of +Q. The ratio of the electric potentialscorresponding to a point on the surface of the conductor and a point at a distance2R from thecentre of the conductor are in the ratio

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Solution

The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:

V = kQ/r

where k is Coulomb's constant.

  1. The electric potential at the surface of the conductor (r = R) is:

V1 = kQ/R

  1. The electric potential at a distance 2R from the center of the conductor (r = 2R) is:

V2 = kQ/(2R)

  1. The ratio of the electric potentials V1 to V2 is:

V1/V2 = (kQ/R) / (kQ/(2R)) = 2

So, the ratio of the electric potentials corresponding to a point on the surface of the conductor and a point at a distance 2R from the center of the conductor is 2:1.

This problem has been solved

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