Question

Two equal charges, $2.0\times {10}^{-7}$C each, are held fixed at a separation of $20$cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point $20$ cm from both the charges. How much work is done by the electric field during the process?

Answer 1

$\text{Step 1: Initial and Final Potential energy}$

When the charge $q$ is placed at A, the potential energy of the charge system will be:

$U_A=\frac{K{q}_{1}q}{r}+\frac{K{q}_{2}q}{r}$

The initial distance of the two charges from $q$ is $r=10cm$ or $0.1m$

$\therefore U_A=\frac{9\times {10}^9(2\times {10}^{-7}C{)}^2}{0.1m}+\frac{9\times {10}^9(2\times {10}^{-7}C{)}^2}{0.1m}$ $=72\times {10}^{-4}J$

When the charge $q$ is placed at B, the potential energy of the charge system will be:

$U_B=\frac{K{q}_{1}q}{r^{\prime }}+\frac{K{q}_{2}q}{r^{\prime }}$

The final distance of the two charges from $q$ is $r^{\prime }=20cm$ or $0.2m$

$U_B=\frac{2\times 9\times {10}^9\times 4\times {10}^{-14}}{0.2}=36\times {10}^{-4}J$.

$\text{Step 2: Work done}$

As electric force is conservative force, So, the Work done by Electric force in moving $q$ from A to B will be equal to the negative of change in potential energy i.e.

$W_{E.F.}=-\Delta U$

$=-(U_B-U_A)$

$=-\left(\right(36-72)\times {10}^{-4})=36\times {10}^{-4}J=3.6\times {10}^{-3}J$

$\text{Note:}$ Work done by Electric Force here is positive here, as while moving from A to B, the Electric force on third charge will be along its displacement, i.e. towards B.