Question

Three point charges of 0.1 C each are placed at the corners of an equilateral triangle with side L =1 m. If this system is supplied energy at the rate of 1 kW, how much time will be required to move one of the charges to the midpoint of the line joining the other two?

Answer 1

As the potential energy of two point charges separated by a distance, r is given by $U=\left(q_1{q}_2/4\pi {\epsilon }_{0}r\right)$, the initial and final potential energies of the system will be

$(Us{)}_I=3\times \frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r}=3\times 9\times {10}^9\times \frac{(0.1{)}^2}{1}=2.7\times {10}^{8}J$

$(Us{)}_F=2\times \frac{1}{4\pi {\epsilon }_0}\frac{q^2}{\left(r/2\right)}+\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r}=\times \frac{q^2}{4\pi {\epsilon }_{0}r}=4.5\times {10}^{8}J$

So work done in changing the configuration of the system

$W=(Us{)}_F-(Us{)}_I=1.8\times {10}^{8}J$

Now, as energy is supplied at the rate of 1 kW,

i.e., ${10}^{3}J{s}^{-1}$, the time required to do work W is given by

$t=\frac{W}{P}=\frac{1.8\times {10}^8}{{10}^3}=1.8\times {10}^{5}s=50h$