Answer
$\frac{E_1}{E_2}=0.500$
Work Step by Step
The electric field at the surface of the sphere is given as
$E_1=\frac{1}{4\pi \epsilon_{\circ}}\frac{Q}{R^2}$.........eq(1)
For $r=\frac{R}{2.00}$ and $Q=\frac{Q}{8}$, the electric field is given as:
$E_2=\frac{1}{4\pi \epsilon_{\circ}}\frac{\frac{Q}{8}}{(\frac{R}{2})^2}$.......eq(2)
Dividing eq(1) by eq(2), we obtain:
$\frac{E_1}{E_2}=\frac{1}{2}=0.500$
