Consider a sphere of radius R which carries a uniform charge density $\text{'}\rho \text{'}$ . If a sphere of radius $\left(\frac{R}{2}\right)$  is carved out of it, as shown. The ratio  $\frac{\left|E_A\right|}{\left|E_B\right|}$ of magnitude of electric field ${\overrightarrow{E}}_A$  and ${\overrightarrow{E}}_B$, respectively at points A and B due to the remaining portion is

Consider the smaller sphere having negative charge density.

$$\begin{gathered} \left|E_A\right|=O+\frac{K\rho .\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(R/2\right)}^2}=K\rho \frac{4}{3}\pi \left(\frac{R}{2}\right) \\ \left|E_B\right|=\frac{K\rho \frac{4}{3}\pi R^3}{R^2}-\frac{K\rho \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{3R}{2}\right)}^2} \\ =K\rho \frac{4}{3}\pi R-K\rho \frac{4}{3}\pi \frac{R}{18} \\ =K\rho \frac{4}{3}\pi \left(\frac{17R}{18}\right) \\ \therefore \frac{E_A}{E_B}=\frac{K\rho \frac{4}{3}\pi \left(R/2\right)}{K\rho \frac{4}{3}\pi \frac{17R}{18}} \\ \frac{E_A}{E_B}=\frac{18}{34} \end{gathered}$$