A charge Q accumulates on the hollow metallic dome,of radius R,of a Van de Graaff generator.A probe measures the electric field strength at various points outside the sphere surface.By what factor will the electric field value at the 2R distance be changed if the charge value were increased to (5/3)Q?

Question

Quizplus · Accepted Answer

The electric field produced by a point charge Q at a distance r is given by:
E = kQ/r^2
where k is Coulomb's constant.

For a uniformly charged hollow sphere like the Van de Graaff generator dome, the electric field outside the sphere is the same as that produced by a point charge Q located at the center of the sphere (using the shell theorem).

So, if the charge on the dome is increased to (5/3)Q, the electric field at a distance 2R from the center of the sphere will be:
E' = k(5/3)Q/(2R)^2
E' = (5/3)[kQ/(2R)^2]
E' = (5/3)E
 
Hence, the electric field value at the 2R distance will be changed by a factor of 5/3.

A Charge Q Accumulates on the Hollow Metallic Dome,of Radius