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Charges $+q$ and $-q$ are placed at points A and B respectively which are a distance $2L$ apart, C is the midpoint between A and B. The work done in moving a charge $+Q$ along the semicircle CRD is
(The semicircle CRD is centred at B and has radius $L$; D lies on the line AB produced, at a distance $L$ beyond B.)
(The semicircle CRD is centred at B and has radius $L$; D lies on the line AB produced, at a distance $L$ beyond B.)
A
$\dfrac{qQ}{2\pi\varepsilon_0 L}$
B
$\dfrac{qQ}{6\pi\varepsilon_0 L}$
C
$-\dfrac{qQ}{6\pi\varepsilon_0 L}$
D
$\dfrac{qQ}{4\pi\varepsilon_0 L}$
Detailed Solution
The work done in moving a charge in an electrostatic field depends only on the end points: $W = Q(V_D - V_C)$.
C is equidistant ($L$) from $+q$ and $-q$, so the potential at C is $V_C = \dfrac{Kq}{L} - \dfrac{Kq}{L} = 0$, where $K = \dfrac{1}{4\pi\varepsilon_0}$.
D is at a distance $3L$ from A ($+q$) and $L$ from B ($-q$).
Potential at D: $V_D = K\left(\dfrac{-q}{L}\right) + \dfrac{Kq}{3L} = -\dfrac{2}{3}\dfrac{Kq}{L}$
Potential difference: $V_D - V_C = -\dfrac{2}{3}\dfrac{Kq}{L} = \dfrac{1}{4\pi\varepsilon_0}\left(-\dfrac{2}{3}\cdot\dfrac{q}{L}\right)$
Work done $= Q(V_D - V_C) = -\dfrac{2}{3} \times \dfrac{1}{4\pi\varepsilon_0}\dfrac{qQ}{L}$
$W = -\dfrac{qQ}{6\pi\varepsilon_0 L}$