Question

Two positive point charges are$5.0\mathrm{cm}$ apart. If the electric potential energy is$72\mathrm{\mu J}$ , what is the magnitude of the force between the two charges?

Short answer

The force between the two charges has a magnitude of$1.4\times {10}^{-3}\mathrm{N}$.

Step-by-step solution

Formula for magnitude of force

Given information:

Distance between two positive charges is $5.0cm$.

Electric potential energy is $72\mu \mathrm{J}$.

Find:

The objective is to find magnitude of the force between two charges.

The potential energy is calculated as follows:

$\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{\mathrm{r}}..1$

when the force's magnitude is

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}.......2$

${\mathrm{Q}}_1$-Charge of point $1$

${\mathrm{Q}}_2$-Charge of point $2$

$\mathrm{r}$-Distance

$\mathrm{U}$-Electric potential energy

Calculation for magnitude of force

when the force's magnitude is

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}$

Equation$2$divided by equation$1$,

we obtain,

$\frac{\mathrm{F}}{\mathrm{U}}=\frac{\frac{1}{{\mathrm{r}}^2}}{\frac{1}{\mathrm{r}}}=\frac{1}{\mathrm{r}}$

This results in

$\mathrm{F}=\frac{1}{\mathrm{r}}\mathrm{U}$

$=\frac{72\mathrm{\mu J}}{5.0\mathrm{cm}}$

localid="1649412963669" $=1.4\times {10}^{-3}\mathrm{N}$