Question

$ABCD$ is a square of side $0.2m$. Charges of $2\times {10}^{-9},4\times {10}^{-9},8\times {10}^{-9}$ coulomb are placed at the corners A, B and C respectively. Calculate the work required to transfer a charge of $2\times {10}^{-9}$ coulomb from corner D to the centre of the square.

Answer 1

$AC=BD=\sqrt{(0.2{)}^2+(0.2{)}^2}$

$=\sqrt{2}\times 0.2=0.28m$

$\therefore AO=BO=CO=\frac{0.28m}{2}=0.14m$

$V$ i.e, Potential at $O;V_o=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1}{AO}+\frac{q_2}{BO}+\frac{q_3}{CO}]$

$V$ i.e, Potential at $D;V_D=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1}{AD}+\frac{q_2}{BD}+\frac{q_3}{CD}]$

Work done $=q[V_o-V_D]$

$=\frac{2\times {10}^{-9}}{4\pi {\epsilon }_0}[\frac{2\times {10}^{-9}}{0.14}+\frac{4\times {10}^{-9}}{0.14}+\frac{8\times {10}^{-9}}{0.14}-\frac{2\times {10}^{-9}}{0.2}-\frac{4\times {10}^{-9}}{0.28}-\frac{8\times {10}^{-9}}{0.2}]$

On solving, we get

$W=2\times {10}^{-9}\times 2\times {10}^9[\frac{14\times {10}^{-9}}{0.14}-\frac{10\times {10}^{-9}}{0.2}-\frac{4\times {10}^{-9}}{0.28}]$

$=4\times [{10}^{-7}-0.5\times {10}^{-7}-0.14\times {10}^{-7}]$

$=5.40\times {10}^{-7}$ Joules.