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
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.
- The electric potential at the surface of the conductor (r = R) is:
V1 = kQ/R
- The electric potential at a distance 2R from the center of the conductor (r = 2R) is:
V2 = kQ/(2R)
- 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.
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