Question

The charges of $2\mu c,4\mu c$ and $6\mu c$ are placed at the corner $A,BandC$ respectively of a square $ABCD$ of side $0.2m$ calculate the work done to transfer a charge of $2\mu c$ from $D$ to the center of the square.

Answer 1

Work done on $2\mu c$ to transfer it from $D$ to $O$ is equals to the energy difference at the $2$ positions on the charge.

Energy=$q\times potential$

$E_d=q\left(V_d\right)=2\times {10}^{-6}\times [V_A+V_B+V_C]=2\times {10}^{-6}\times [\frac{K\times 2{\times 10}^{-6}}{0.2}+\frac{K\times 4{\times 10}^{-6}}{\left(\frac{\sqrt{2}}{5}\right)}+\frac{K\times 6{\times 10}^{-6}}{0.2}]=97.45\times {10}^{-6}J{E}_O=2\times {10}^{-6}\times \left[\frac{(K\times 2{\times 10}^{-6})+(K\times 4{\times 10}^{-6})+(K\times 6{\times 10}^{-6})}{\left(\frac{1}{5\sqrt{2}}\right)}\right]E_O=152.73\times {10}^{-2}JW.D=(152.3-97.45)\times {10}^{-2}JW.D=55.28\times {10}^{-2}J$