Question
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
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
A.
$$\frac{{qQ}}{{2\pi {\varepsilon _0}L}}$$
B.
$$\frac{{qQ}}{{3\pi {\varepsilon _0}L}}$$
C.
$$ - \frac{{qQ}}{{6\pi {\varepsilon _0}L}}$$
D.
$$\frac{{qQ}}{{4\pi {\varepsilon _0}L}}$$
Answer :
$$ - \frac{{qQ}}{{6\pi {\varepsilon _0}L}}$$
Solution :
Potential at $$C = {V_C} = 0$$
Potential at $$D = {V_D}$$
$$\eqalign{ & = k\left( {\frac{{ - q}}{L}} \right) + \frac{{kq}}{{3L}} \cr & = - \frac{2}{3}\frac{{kq}}{L} \cr} $$
Potential difference
$$\eqalign{ & {V_D} - {V_C} = \frac{{ - 2}}{3}\frac{{kq}}{L} = \frac{1}{{4\pi { \in _0}}}\left( { - \frac{2}{3}.\frac{q}{L}} \right) \cr & \Rightarrow {\text{work}}\,{\text{done}} = Q\left( {{V_D} - {V_C}} \right) \cr & = - \frac{2}{3} \times \frac{1}{{4\pi {\varepsilon _0}}}\frac{{qQ}}{L} = \frac{{ - qQ}}{{6\pi {\varepsilon _0}L}} \cr} $$
Potential at $$C = {V_C} = 0$$
Potential at $$D = {V_D}$$
$$\eqalign{ & = k\left( {\frac{{ - q}}{L}} \right) + \frac{{kq}}{{3L}} \cr & = - \frac{2}{3}\frac{{kq}}{L} \cr} $$
Potential difference
$$\eqalign{ & {V_D} - {V_C} = \frac{{ - 2}}{3}\frac{{kq}}{L} = \frac{1}{{4\pi { \in _0}}}\left( { - \frac{2}{3}.\frac{q}{L}} \right) \cr & \Rightarrow {\text{work}}\,{\text{done}} = Q\left( {{V_D} - {V_C}} \right) \cr & = - \frac{2}{3} \times \frac{1}{{4\pi {\varepsilon _0}}}\frac{{qQ}}{L} = \frac{{ - qQ}}{{6\pi {\varepsilon _0}L}} \cr} $$