Electrostatics
Beginner
Coulomb Law
Source: High school physics (Chinese)
Problem Sets:
Electrostatics
Problem
In a vacuum, two point charges, $q$ and $Q$, are separated by a distance $r = 5.0$ mm. They experience an attractive force of $F = 4.0 \times 10^{-4}$ N.
Given that $q = 1.2 \times 10^{-6}$ C, find the value of charge $Q$.
$Q = -9.26 \times 10^{-10}$ C
According to Coulomb's Law, the magnitude of the electrostatic force is $F = k \frac{|qQ|}{r^2}$. Since the force is attractive, the charges must have opposite signs. Given $q$ is positive, $Q$ must be negative.
We can solve for the magnitude of $Q$:
$$|Q| = \frac{Fr^2}{k|q|}$$Substituting the given values:
$$|Q| = \frac{(4.0 \times 10^{-4} N)(5.0 \times 10^{-3} m)^2}{(9.0 \times 10^9 N \cdot m^2/C^2)(1.2 \times 10^{-6} C)} \approx 9.26 \times 10^{-10} C$$Since $Q$ is negative, $Q = -9.26 \times 10^{-10}$ C.