When an uncharged conducting ball of radius $R$ is placed in an external uniform electric field, a s

Question

When an uncharged conducting ball of radius $R$ is placed in an external uniform electric field, a surface charge density $\sigma = \sigma_0 \cos 	heta$ is induced on the ball's surface (here $\sigma_0$ is a positive constant, $	heta$ is the polar angle). What is the magnitude of the resultant electric force acting solely on the induced charge of positive sign (which resides on the hemisphere $0 \le 	heta \le \pi/2$)?

Professor PC Thomas Classes · Answer

The electrostatic pressure acting outwards on any surface element of a conductor is $P = \frac{\sigma^2}{2\epsilon_0}$, regardless of the presence of an external field (the external field's effect is already accounted for in the local total $\sigma$). The resultant force on the positive charges ($0 \le 	heta \le \pi/2$) is directed along the $z$-axis (symmetry axis). $F_z = \int P \cos	heta \, dA = \int_0^{\pi/2} \frac{\sigma^2}{2\epsilon_0} \cos	heta (2\pi R^2 \sin	heta \, d	heta)$. Substituting $\sigma = \sigma_0 \cos	heta$, we get $F_z = \frac{2\pi R^2 \sigma_0^2}{2\epsilon_0} \int_0^{\pi/2} \cos^3	heta \sin	heta \, d	heta$. Let $u = \cos	heta$, then $du = -\sin	heta \, d	heta$. The integral becomes $\int_0^1 u^3 \, du = [\frac{u^4}{4}]_0^1 = \frac{1}{4}$. Thus, the resultant force is $F_z = \frac{\pi R^2 \sigma_0^2}{\epsilon_0} 	imes \frac{1}{4} = \frac{\pi R^2 \sigma_0^2}{4\epsilon_0}$.