A solid metal ball of radius $R = 1.5 ext{ cm}$ has a total uniform charge $q = 10 \mu ext{C}$. F

Question

A solid metal ball of radius $R = 1.5 	ext{ cm}$ has a total uniform charge $q = 10 \mu	ext{C}$. Find the modulus of the resultant electrostatic force (in N) acting on the charge located on exactly one half (a hemisphere) of the ball.

Professor PC Thomas Classes · Answer

The force per unit area (electrostatic pressure) on the surface of the conducting ball is $P = \frac{\sigma^2}{2\epsilon_0}$. The resultant force on a hemisphere is obtained by integrating the component of this pressure parallel to the symmetry axis of the hemisphere. Force $F = \int P \cos	heta \, dA = P \int \cos	heta \, dA$. The integral $\int \cos	heta \, dA$ over a hemisphere is exactly its projected cross-sectional area, which is $\pi R^2$. Thus, $F = P(\pi R^2) = \frac{\sigma^2}{2\epsilon_0} \pi R^2$. Since $\sigma = \frac{q}{4\pi R^2}$, we have $F = \frac{(q/4\pi R^2)^2}{2\epsilon_0} \pi R^2 = \frac{q^2}{32\pi\epsilon_0 R^2} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{8 R^2}$. Substituting the given values: $q = 10^{-5} 	ext{ C}$, $R = 1.5 	imes 10^{-2} 	ext{ m}$, and $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$. We get $F = (9 	imes 10^9) \frac{(10^{-5})^2}{8(1.5 	imes 10^{-2})^2} = \frac{9 	imes 10^9 	imes 10^{-10}}{8 	imes 2.25 	imes 10^{-4}} = \frac{0.9}{18 	imes 10^{-4}} = \frac{9000}{18} = 500 	ext{ N}$.