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. By what factor will the electric field value at the 2R distance be changed if the charge value were increased to (4/3)Q?

Answer 1

Answer:

The factor must be 4/3

Explanation:

the solution is in the attached Word file

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Answer 2

Answer:

E'=(3/16)E

Explanation:

The electric field generated by a Van de Graff generator can be taken as the electric field generated by a hollow shell (for distance out of the shell):

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

Q: electric charge

K: Coulomb's constant

R: distance in which E is measured.

If the distance is increased to R'=2R, and the charge to Q'=(4/3)Q the new electric field is:

$E'=k\frac{Q'}{R'^2}=k\frac{(3/4)Q}{(2R)^2}=\frac{3}{4}k\frac{Q}{4R^2}=\frac{3}{16}k\frac{Q}{R^2}=\frac{3}{16}E$

hence, the new electric field E' is 3/16 times the previous electric field E