Blog: Two similar charges, when placed 2 cm apart, repel each other with a force of 44.1 N. Find the magnitude of either charge.

Hello!

This problem requires using Coulomb's law, one of the basic electrostatic laws. It states that for point charges $\displaystyle{q}_{{1}}$ and $\displaystyle{q}_{{2}},$ which are stationary relative to each other, the electrostatic force is:

$\displaystyle{F}={k}\cdot\frac{{{q}_{{1}}\cdot{q}_{{2}}}}{{{r}}^{{2}}}.$

Here $\displaystyle{r}$ is a distance between these points and $\displaystyle{k}$ is a constant which depends on the electrostatic properties of a medium in which the charges are reside. For air it is about $\displaystyle{9}\cdot{{10}}^{{9}}{N}\cdot\frac{{{m}}^{{2}}}{{{C}}^{{2}}}.$

It is given that...

Hello!

$\displaystyle{F}={k}\cdot\frac{{{q}_{{1}}\cdot{q}_{{2}}}}{{{r}}^{{2}}}.$

It is given that both charges are equal: $\displaystyle{q}_{{1}}={q}_{{2}}={q}.$ Then:

$\displaystyle{{q}}^{{2}}=\frac{{{F}\cdot{{r}}^{{2}}}}{{k}}$ and $\displaystyle{\left|{q}\right|}={r}\sqrt{{\frac{{F}}{{k}}}}.$

Numerically it is:

$\displaystyle{0.02}\cdot\sqrt{{\frac{{44.1}}{{{9}\cdot{{10}}^{{9}}}}}}={2}\cdot{{10}}^{{-{2}}}\cdot\sqrt{{\frac{{4.9}}{{{10}}^{{9}}}}}={2}\cdot{{10}}^{{-{2}}}\cdot\sqrt{{\frac{{49}}{{{10}}^{{{10}}}}}}=$

$\displaystyle={14}\cdot{{10}}^{{-{2}}}\cdot{{10}}^{{-{5}}}={1.4}\cdot{{10}}^{{-{6}}}{\left({C}\right)}.$

$\displaystyle{C}$ is for the Coulomb, the Si unit for electric charge. Note that we transformed $\displaystyle{2}{c}{m}$ into $\displaystyle{0.02}{m}$ to use the proper units.

So the answer is: the magnitude of each charge is $\displaystyle{1.4}\cdot{{10}}^{{-{6}}}{C}.$ The sign of this charge is still unknown, charges may be both positive or both negative.