Question

Two point charges $100\mu C$ and $5\mu C$ are placed at points $A$ 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 $ABC=\pi /2$ and $1/4\pi {\epsilon }_0=9\times 10^{9}N{m}^2/C^2$

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

Answer 1

Answer: (D)

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

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