Answer to Question #127392 in Physics for Jaydee Galvez

Question

Two Identical charges(mass= 1.0 x 10^6kg) are at rest on the surface of a hemispherical bowl of radius R=0.25m. 3 forces act om each particle: Normal force, weight force and repulsive force they exert on each other. The normal force acting on either charge (caused by the bowl's surface) is 60° to the horizontal. Find the charge on each particle.

Answer 1

$N\sin60°=mg$

$N\cos60°=k\frac{q^2}{r^2}\to \frac{mg}{\sin60°}\cdot\cos60°=k\frac{q^2}{r^2}$

$r=2\cdot R\cdot\cos60°$

$q^2=\frac{mg}{k\sin60°}\cdot\cos60°\cdot(2R\cos60°)^2$

$q=\sqrt{\frac{mg}{k\sin60°}\cdot\cos60°\cdot(2R\cos60°)^2}=$

$=\sqrt{\frac{1\cdot 10^{6}\cdot9.81}{9\cdot 10^9 \cdot \sin60°}\cdot\cos60°\cdot(2\cdot 0.25\cdot \cos60°)^2}=6.3\cdot10^{-3}$ $C$