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​

Answer 1

Potential at D:-

$\\ \sf\longmapsto V_D=\dfrac{Kq}{3\ell}+\dfrac{K(-q)}{\ell}$

$\\ \sf\longmapsto V_D=\dfrac{Kq}{3\ell}-\dfrac{Kq}{\ell}$

$\\ \sf\longmapsto V_D=\dfrac{Kq-3Kq}{3\ell}$

$\\ \bf\longmapsto V_D=\dfrac{-2Kq}{3\ell}$

Potential at C:-

$\\ \sf\longmapsto V_C=\dfrac{Kq}{\ell}+\dfrac{K(-q)}{\ell}$

$\\ \sf\longmapsto V_C=\dfrac{Kq}{\ell}-\dfrac{Kq}{\ell}$

$\\ \bf\longmapsto V_C=0$

We know that

$\\ \sf\longmapsto V_D-V_C=\dfrac{W_{C-D}}{Q}$

$\\ \bf\longmapsto W_{C-D}=(V_D-V_C)Q$

$\\ \sf\longmapsto W_{C-D}=\dfrac{-2Kq}{3\ell}Q$

As

$\boxed{\sf K=\dfrac{1}{4\pi \epsilon_0}}$

$\\ \sf\longmapsto W_{C-D}=\dfrac{\cancel{-2}q}{3\times \cancel{4}\pi\epsilon_0}\times \dfrac{Q}{L}$

$\\ \sf\longmapsto \underline{\boxed{\bf{W_{C-D}=\dfrac{-qQ}{6\pi\epsilon_0L}}}}$

Answer 2

$\large \bf \clubs \:  Given :-$

• Charges +q and –q are placed at points A and B respectively.

• Distance Between A & B = 2L and C is the mid point of line joining A and B.

• CRD is Senicircle ( Raidus = L )

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$\large \bf \clubs \:  To \: Find :-$

• Work Done in moving a charge + Q Along the Semi-Circle CRD

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$\large \bf \clubs \:  Main \: Concept$

• Work Done in moving any charge is independent of the path followed .

• Relation Between Potential of two points and Work Done from moving a charge Q from one point to another point : $\sf V_Y-V_X=\dfrac{W_{X \rightarrow Y}}{Q}$

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$\large \bf \clubs \:  Solution :-$

✏ Calculating Potential at C :-)

$\sf V_C = \bigg( \dfrac{kq}{L} \bigg) + \bigg( \dfrac{k( - q)}{L} \bigg) \\ \\ = \sf \cancel\dfrac{kq}{L} - \cancel \dfrac{kq}{L} \\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf V_C = 0} }}}$

✏ Calculating Potential at D :-)

$\sf V_D= \bigg( \dfrac{kq}{3L} \bigg) + \bigg( \dfrac{k( - q)}{L} \bigg) \\ \\ = \sf \dfrac{kq}{3L} - \dfrac{kq}{L} \\ \\ \sf = \dfrac{kq - 3kq}{3L} \\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf V_D = - \frac{2kq}{3L} } }}}$

Now ,

$\sf\dfrac{W_{C \rightarrow D}}{Q} = V_D - V_C \\ \\ \sf = - \dfrac{2kq}{3L} - 0 \\ \\ = - \sf\dfrac{2kq}{3L} \\ \\ :\longmapsto\sf {W_{C \rightarrow D}} = - \frac{2kqQ}{3L}$

$\bf Putting \: k = \dfrac{1}{4\pi \epsilon_{\circ}} : - \\ \\ \sf {W_{C \rightarrow D}} = - \frac{2qQ}{3L} \times \dfrac{1}{4\pi \epsilon_{\circ}} \\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf {W_{C \rightarrow D}} = - \frac{qQ}{6L\pi \epsilon_{\circ}}} }}}$

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}$

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