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  4. Two point charges 100 μ C and 5 μ C are placed at points A and B respectively with A B=40 cm. The work done by external force in displacing the charge 5 μ C from B to C, where B C=30 cm, angle A B C=π / 2 and 1 / 4 π ε0=9 × 109 Nm 2 / C 2

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Q. Two point charges $100\, \mu C$ and $5\, \mu C$ are placed at points $A$ and $B$ respectively with $A B=40\, cm$. The work done by external force in displacing the charge $5\, \mu C$ from $B$ to $C$, where $B C=30 \,cm$, angle $A B C=\pi / 2$ and $1 / 4 \pi \varepsilon_{0}=9 \times 10^{9} Nm ^{2} / C ^{2}$

Electrostatic Potential and Capacitance

Solution:

Work done in displacing charge of $5 \,\mu u C$ from $B$ to $C$ is
$W=5 \times 10^{-6}\left(V_{C}-V_{B}\right)$ where
image
$V_{B}=9 \times 10^{9} \times \frac{100 \times 10^{-6}}{0.4}=\frac{9}{4} \times 10^{6} V$
and $V_{C}=9 \times 10^{9} \times \frac{100 \times 10^{-6}}{0.5}=\frac{9}{5} \times 10^{6} V$
So $\quad W=5 \times 10^{-6} \times\left(\frac{9}{5} \times 10^{6}-\frac{9}{4} \times 10^{6}\right)=-\frac{9}{4} J$