Question

When an uncharged conducting ball of radius r is placed in an external uniform electric field, a surface charge density $\sigma = \sigma_0 \cos \theta$ is induced on the ball's surface where $\sigma_0$ is a constant, $\theta$ is a polar angle measured from a direction parallel to external electric field. Find the magnitude of the resultant electric force acting on an induced charge of the same sign on one half of hemisphere.

• A. $\frac{\pi \sigma_0^2 r^2}{4 \epsilon_0}$

Answer 1

Answer: (A)

$\frac{\pi \sigma_0^2 r^2}{4 \epsilon_0}$

Answer 2

The force on a surface charge element $dq$ on a conductor is $d\vec{F} = \frac{\sigma^2}{2\epsilon_0} dA \hat{n}$, directed outward normal. For a sphere, $dA = r^2 \sin\theta d\theta d\phi$ and $\hat{n} = \sin\theta \cos\phi \hat{i} + \sin\theta \sin\phi \hat{j} + \cos\theta \hat{k}$. Substituting $\sigma = \sigma_0 \cos\theta$ and integrating over the upper hemisphere ($0 \le \theta \le \pi/2$, $0 \le \phi \le 2\pi$), the $x$ and $y$ components of the force cancel due to symmetry. The $z$-component is calculated as:

$F_z = \int_0^{2\pi} \int_0^{\pi/2} \frac{\sigma_0^2 r^2}{2\epsilon_0} \cos^3\theta \sin\theta d\theta d\phi = \frac{\sigma_0^2 r^2}{2\epsilon_0} (2\pi) \int_0^{\pi/2} \cos^3\theta \sin\theta d\theta$.

The integral $\int_0^{\pi/2} \cos^3\theta \sin\theta d\theta = \frac{1}{4}$.

Thus, $F_z = \frac{\sigma_0^2 r^2}{2\epsilon_0} (2\pi) (\frac{1}{4}) = \frac{\pi \sigma_0^2 r^2}{4\epsilon_0}$. This is the magnitude of the resultant force.