Two point charges $100 \mu C$ and $5 \mu C$ are placed at points $A$ and B respectively with $AB = 40 \text{ cm}$ . The work done by external force in displacing the charge $5 \mu C$ from B to $C_0$ , where $BC = 30 \text{ cm}$ , angle $ABC = \frac{\pi}{2}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^{9} \, \mathrm{Nm^{2} / C^{2}}$
Text Solution
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Work done in displacing charge of 5 μ μ C from B to C is
$W = 5 \times 10^{-6} (V_C - V_B)$ where
$V_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{-6}}{0.4} = \frac{9}{4} \times 10^{6} \mathrm{V}$
and $V_c = 9 \times 10^9 \times \frac{100 \times 10^{-6}}{0.5} = \frac{9}{5} \times 10^6 \mathrm{V}$
So $W = 5 \times 10^{-6} \times \left( \frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{6} \right) = - \frac{9}{4} J$
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