1. Tardigrade
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  4. Two point charges 100 μ C and 5 μ C are placed at points A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 μ C from B to C, where BC = 30 cm, angle ABC = π/2 and 1/4 π ε0 = 9 × 109 Nm2/C2

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Q. 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 $ABC = \pi/2$ 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 $5\, \mu C$ from $B$ to $C$ is
$W= 5 \times 10^{-6}( V_c - V_B)$
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 $W = 5\times 10^{-6} \times (\frac{9}{5} \times 10^6 - \frac{9}{4} \times 10^6)$
$ = -\frac{9}{4} \,J$