Question

Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smaller sphere. If the electric field just outside of the smaller sphere is E, then the electric field just outside of the larger sphere is: 9E, 1/9E, 1/3E, E, 3E?

Answer 1

Answer:

1/9 E

Explanation:

The electric field produced by a charged sphere outside the sphere is equal to that produced by a single point charge:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

In this problem, we have a sphere of radius R (smaller sphere) with a charge Q that produces an electric field of magnitude E at r=R.

For the larger sphere,

R' = 3R

Therefore, the electric field at r=R' will be

$E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E}{9}$

So, the electric field just outside the larger sphere will be 1/9 E.