Using Coulomb’s law to find the force between charged objects · Physics Playground

Example from Physics, 18.2 Coulomb's law

Using Coulomb’s law to find the force between charged objects

Suppose Coulomb measures a force of $20 \times 10^{−6} N$ between the two charged spheres when they are separated by 5.0 cm. By turning the dial at the top of the torsion balance, he approaches the spheres so that they are separated by 3.0 cm. Which force does he measure now?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Apply Coulomb’s law to the situation before and after the spheres are brought closer together. Although we do not know the charges on the spheres, we do know that they remain the same. We call these unknown but constant charges $q_{1}$ and $q_{2}$ . Because these charges appear as a product in Coulomb’s law, they form a single unknown. We thus have two equations and two unknowns, which we can solve. The first unknown is the force (which we call $F_{f}$ ) when the spheres are 3.0 cm apart, and the second is $q_{1} q_{2}$ .

Use the following notation: When the charges are 5.0 cm apart, the force is $F_{i} = 20 \times 10^{−6} N$ and $r_{i} = 5.0 cm = 0.050 m$ , where the subscript i means initial. Once the charges are brought closer together, we know $r_{f} = 3.0 cm = 0.030 m$ , where the subscript f means final.
Solution
Coulomb’s law applied to the spheres in their initial positions gives

$F_{i} = \frac{k q_{1} q_{2}}{r_{i}^{2.}} .$
18.8

Coulomb’s law applied to the spheres in their final positions gives

$F_{f} = \frac{k q_{1} q_{2}}{r_{f}^{2.}} .$
18.9

Dividing the second equation by the first and solving for the final force $F_{f}$ leads to

$\begin{matrix} \frac{F_{f}}{F_{i}} & = & \frac{k q_{1} q_{2} / r_{f}^{2}}{k q_{1} q_{2} / r_{i}^{2}} . \\ = & \frac{r_{i}^{2}}{r_{f}^{2}} \\ F_{f} & = & F_{i} \frac{r_{i}^{2}}{r_{f}^{2}} . \end{matrix}$
18.10

Inserting the known quantities yields

$\begin{matrix} F_{f} & = & F_{i} \frac{r_{i}^{2}}{r_{f}^{2}} \\ = & (20 \times 10^{−6} N) \frac{{(0.050 m)}^{2}}{{(0.030 m)}^{2}} \\ = & 56 \times 10^{−6} N \\ = & 5.6 \times 10^{−5} N. \end{matrix}$
18.11

The force acts along the line joining the centers of the spheres. Because the same type of charge is on each sphere, the force is repulsive.

0 of 2 steps open

How did it go?

Strategy

Solution