Question

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

• A. $9J$
• B. $\frac{81}{20}J$
• C. $\frac{9}{25}J$
• D. $\frac{9}{4}J$

Answer 1

The correct option is A $9J$

R.E.F image

Work done in displacing charge of blue from B to C

$W=5\times {10}^{-6}(V_C-V_B)$

$V_B=9\times {10}^9\times \frac{100\times {10}^{-1}}{0.4}=\frac{9}{6}\times {10}^{6}v$

$V_C=9\times {10}^9\times \frac{100\times {10}^{-5}}{0.5}=\frac{9}{5}\times {10}^{6}v$

$W=5\times {10}^{-6}\times (\frac{9}{5}\times {10}^6-\frac{9}{4}\times {10}^{-6})$

$W=\frac{9}{u}J$