Physics - Electrostatics Question with Solution | TestHub

The field at the centre due to the hemispherical shell of charge Q (radius $R_1$) is $E=\frac{{\sigma }_1}{4{\epsilon }_0}=\frac{Q}{8\pi {\epsilon }_0{R}_1^2}$ along the common axis.

The force of interaction equals the force on the shell of charge 2Q in this field:

$F=2Q\times \frac{Q}{8\pi {\epsilon }_0{R}_1^2}=\frac{Q^2}{4\pi {\epsilon }_0{R}_1^2}$.

Comparing with $\frac{Q^2}{n\pi {\epsilon }_0{R}_1^2}$: n = 4.