Question

Two point charges 100μc and 5μc are placed at points A and B respectively with AB=40 cm . The work done by external force in displacing the charge 5μc from B to C , where BC=30 cm, angle $ABC=\frac{\pi }{2}and\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}N{M}^2/c^2$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Work done in displacing charge of $5\mu C$ from B to C is
$W=5\times {10}^{-6}(V_C-V_B)$ where

$V_B=9\times {10}^9\times \frac{100\times {10}^{-6}}{0.4}=\frac{9}{4}\times {10}^{6}V$
and $V_C=9\times {10}^9\times \frac{100\times {10}^{-6}}{0.5}=\frac{9}{5}\times {10}^{6}V$
So $W=5\times {10}^{-}6\times (\frac{9}{5}\times {10}^6-\frac{9}{4}\times {10}^6)=-\frac{9}{4}J$