Question

A sphere of radius R has charge Q.The electric field strength at distance r > R is E intial. What is the ratio E of the final to initial electric field strengths if R is halved?

Answer 1

Answer:

Efinal / Einitial = 1

Explanation:

Applying Gauss´s Law to a closed spherical surface encircling the sphere of radius R, it can be showed that the electric field, at a distance r> R, behaves like all the charge of the sphere (which is distributed over the surface) were concentrated in the center, so the sphere produces the same electric field as the one created by a point charge located at that point.

The expression for the field E initial is as follows:

$E initial = \frac{K*Q}{r^{2}}$

So, at a distance r> R, as the center of the sphere is located at the same point, the electric field will be the same as before.

The expression for E final is as follows:

$E final = \frac{K*Q}{r^{2}}$

So the ratio $\frac{Ef}{Ei}$ is equal to one.