Question

PQR is a right angled triangle with a point charge $q=+3\times {10}^{-10}C$ at Q. How much work is required to be done to carry a unit positive charge from R to P? PQ = 0.6 m and QR = 0.8 m.

Answer 1

Initial potential energy $U_1=\frac{kq{q}_1}{RQ}$ where $k=9\times {10}^{9}N{m}^2/C^2$

$⟹U_1=\frac{9\times {10}^9\times 3\times {10}^{-10}\times 1}{0.8}=\frac{27}{8}J$

Finally,

$U_2=\frac{kq{q}_1}{QP}=\frac{9\times {10}^9\times 3\times {10}^{-10}\times 1}{0.6}=\frac{27}{6}J$

Work done $W=U_2-U_1=\frac{27}{6}-\frac{27}{8}$

$⟹W=1.125J$