Question

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. If the probe is initially at a distance 2R from the sphere's center and then is moved to 3R, by what factor will the electric field reading change?

Answer 1

The electric field changes by a factor 4/9

Explanation:

The electric field at a point outside a hollow charged sphere is identical to that produced by a single point charge Q, therefore its strength is given by

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

where

k is the Coulomb constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

In this problem, we have a charge Q accumulated on the dome of the generator of radius R. The electric field strength at a distance of 2R from the sphere centre is

$E=\frac{kQ}{(2R)^2}=\frac{1}{2}\frac{kQ}{R^2}$

Then, the probe is moved to a new distance of

r = 3R

From the centre of the sphere: therefore, the new electric field will be

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

So, the electric field has changed by a factor

$\frac{E'}{E}=\frac{1/9}{1/4}=\frac{4}{9}$

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