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(C) The work done $W$ in moving a charge $q$ from point $B$ to point $C$ is given by $W = q(V_C - V_B)$,where $V_C$ and $V_B$ are the electric potenti

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$\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^2}{R}\left(\frac{\sqrt{2}-1}{2}\right)$

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(C) The work done $W$ in moving a charge $q$ from point $B$ to point $C$ is given by $W = q(V_C - V_B)$,where $V_C$ and $V_B$ are the electric potentials at points $C$ and $B$ respectively.The potential at any point is the sum of potentials due to the three charges: one at $O$ (charge $q$),one at $A$ (charge $q$),and one at $B$ (charge $q$).Potential at $B$ $(V_B)$: The distance from $O$ to $B$ is $R$,from $A$ to $B$ is $2R$,and the charge at $B$ is the one being moved,so we consider the potential due to the other two charges: $V_B = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{R} + \frac{q}{2R} \right) = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{2R}$.Potential at $C$ $(V_C)$: The distance from $O$ to $C$ is $R$,from $A$ to $C$ is $\sqrt{R^2 + R^2} = R\sqrt{2}$,and from $B$ to $C$ is $R\sqrt{2}$. The charge at $C$ is the one being moved,so we consider the potential due to the other two charges: $V_C = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{R} + \frac{q}{R\sqrt{2}} \right) = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R} \left( 1 + \frac{1}{\sqrt{2}} \right)$.Work done $W = q(V_C - V_B) = q \cdot \frac{1}{4 \pi \varepsilon_{0}} \left[ \frac{q}{R} (1 + \frac{1}{\sqrt{2}}) - \frac{3q}{2R} \right] = \frac{q^2}{4 \pi \varepsilon_{0} R} \left[ 1 + \frac{1}{\sqrt{2}} - 1.5 \right] = \frac{q^2}{4 \pi \varepsilon_{0} R} \left[ \frac{\sqrt{2}}{2} - 0.5 \right] = \frac{q^2}{4 \pi \varepsilon_{0} R} \left( \frac{\sqrt{2}-1}{2} \right)$.

$A$ charge $q$ is placed at the centre $O$ of a circle of radius $R$. Two other charges $q$ and $q$ are placed at the ends of the diameter $AB$ of the circle. The work done to move the charge at point $B$ along the circumference of the circle to a point $C$ as shown in the figure is

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