Question

Calculate the work done in moving a charge $-2\times {10}^{-9}C$ from A to B via C.

Answer 1

We know that, potential is path independent.
Now, $V_A=\frac{kq}{r_A}=\frac{9\times {10}^9\times 8\times {10}^{-3}}{3\times {10}^{-2}}$
$=24\times {10}^{8}V$
$V_B=\frac{kq}{r_B}=\frac{9\times {10}^9\times 8\times {10}^{-3}}{4\times {10}^{-2}}$
$=18\times {10}^{8}V$
$\therefore \Delta W=Q(V_B-V_A)=-2\times {10}^{-9}\times (18-24)\times {10}^8$
$=12\times {10}^{-1}=1.2J$