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Two point charges $100\,\mu \,C$ and $5\,\mu \,C$ are placed at points $A$ and $B$ respectively with $AB = 40\,cm$. The work done by external force in displacing the charge $5\,\mu \,C$ from $B$ to $C$, where $BC = 30\,cm$, angle $ABC = \frac{\pi }{2}$ and $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}/{C^2}$.........$J$
$9$
$\frac{{81}}{{20}}$
$\frac{9}{{25}}$
$-2.25$
Solution
(d) Work done in displacing charge of $5\, µC$ 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 $W = 5 \times {10^{ – 6}} \times \left( {\frac{9}{5} \times {{10}^6} – \frac{9}{4} \times {{10}^6}} \right) = – \frac{9}{4}\,J$
