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):
[tex]E=k\frac{Q}{R^2}[/tex]
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:
[tex]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[/tex]
hence, the new electric field E' is 3/16 times the previous electric field E
