Question

the force between two charges is 125 N. One has a positive charge of 2.0 X 10^-3 C, while the other has negative charge of 3.6 X 10^-4 C. How far apart are the two charges?

Answer 1

According to coulombs law

$f=\frac{k\times Q_{1} \times Q_{2}}{r^{2}}$ where $k=9.0\times 10^{9}Nm^{2}/c^{2}$

therefore $r^{2}= \frac{k\times Q_{1}\times Q_{2}}{f}$

substituting the given values in the equation

$r^{2}=\frac {(9.0\times 10^{9})\times(2.0\times 10^{-3})\times(3.6\times10^{-4})}{125}$

$r^{2}=51.84$

$r=\sqrt{51.84}$

.

$r=7.2m$