Answer:
2E
Explanation:
The electric field inside a charged, nonconducting sphere is given by
[tex]E(r)=\frac{kQr}{R^3}[/tex] (1)
where
k is the Coulomb's constant
Q is the total charge on the sphere
r is the distance from the centre of the sphere
R is the radius of the sphere
In this problem, the magnitude of the electric field at r = R/4 is E, so we can write:
[tex]E=\frac{kQ(\frac{R}{4})}{R^3}=\frac{1}{4}\frac{kQ}{R^2}[/tex]
Then, we want to calculate the magnitude of the electric field at a distance of
[tex]r=\frac{R}{2}[/tex]
Substituting into (1), we find:
[tex]E'=\frac{kQ(\frac{R}{2})}{R^3}=\frac{1}{2}\frac{kQ}{R^2}=2E[/tex]
Therefore, the electric field at R/2 is twice the electric field at R/4.
