Question

Three point charges of 0.1 C 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 at the midpoint of the line joining the two charges.

Answer 1

Initial electrostatic energy of system of three charges is

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

When charges A is moved to new position $A^{{}^{\prime }}$, Which is midpoint of side BC, then final electrostatic energy is

$U_f=\frac{1}{4\pi {\epsilon }_0}(\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_3{q}_1}{r_{13}})$

$=\frac{1}{4\pi {\epsilon }_0}[\frac{q^2}{A^{{}^{\prime }}B}+\frac{q^2}{BC}+q^2{A}^{{}^{\prime }}C]$

But $A^{{}^{\prime }}B=A^{{}^{\prime }}C=r/2=0.5m$. Therefore,

$U_f=9\times {10}^9[\frac{(0.1{)}^2}{0.5}+\frac{(0.1{)}^2}{1}+\frac{(0.1{)}^2}{0.5}]=4.5\times {10}^{8}J$

Work done $W=U_f-U_i$

$=4.5\times {10}^8-2.7\times {10}^8=1.8\times {10}^{8}J$

As $W=pt$

or $t=\frac{W}{p}=\frac{1.8\times {10}^8}{{10}^3}$

$=1.8\times {10}^{5}s=\frac{1.8\times {10}^5}{3600}h=50h.$