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Three charges each have $ +q $ charge, are placed at the corner of an isosceles triangle ABC of sides BC and AC, 2a.D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is:
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(A) $ \dfrac{eqQ}{8\pi {{\in }_{0}}a} $
(B) $ \dfrac{qQ}{4\pi {{\in }_{0}}a} $
(C) Zero
(D) $ \dfrac{3qQ}{4\pi {{\in }_{0}}a} $

Last updated date: 13th Jun 2024
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Hint: To find out the work done is taking charge Q from one point to another, we will find electrostatic potential at those points using
 $ V=\dfrac{q}{4\pi {{\in }_{0}}r} $
 $ V $ is the potential generated between the charges
 $ q $ is the charge due to which our test charge is affected
r is the distance between the charges.

Complete step by step solution
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 $ AC=BC=2a $
As D and E are midpoints of BC and AC (given).
 $ \therefore $ $ AE=EC=a $
 $ BD=DC=a $
In $ \Delta ADC, $
  $ \begin{align}
  & {{(AD)}^{2}}={{(AC)}^{2}}-{{(DC)}^{2}} \\
 & ={{(2a)}^{2}}-{{(a)}^{2}}={{(4a)}^{2}}-{{(a)}^{2}}={{(a)}^{2}} \\
 & AD=a\sqrt{3} \\
\end{align} $
Similarly, potential at points D due to the given charge distribution is
 $ {{V}_{D}}=\dfrac{1}{4\pi {{\in }_{0}}}[\dfrac{q}{BD}+\dfrac{q}{DC}+\dfrac{q}{AD}] $
 $ \begin{align}
  & =\dfrac{q}{4\pi {{\in }_{0}}}\left[ \dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{\sqrt{3}a} \right] \\
 & =\dfrac{q}{4\pi {{\in }_{0}}a}\left[ 2+\dfrac{1}{\sqrt{3}} \right] \\
\end{align} $ ...........................(1)
Potential at point E due to given charge configuration is
 $ {{V}_{E}}=\dfrac{1}{4\pi {{\in }_{0}}}\left[ \dfrac{1}{q}+\dfrac{1}{q}+\dfrac{1}{a\sqrt{3}} \right] $
 $ =\dfrac{q}{4\pi {{\in }_{0}}a}\left[ 2+\dfrac{1}{\sqrt{3}} \right] $ ...........................(2)
From (1) and (2), it is clear that
The work done in taking a charge Q from D to E is
 $ W=Q({{V}_{E}}-{{V}_{D}})=0 $ $ (\because {{V}_{D}}={{V}_{E}}) $
Therefore, option (C) is correct.

Electric potential due to single charge is spherically symmetric. It should be clearly borne in mind that due to single charge,
 $ F\propto \dfrac{1}{{{r}^{2}}} $ ; $ E\propto \dfrac{1}{{{r}^{2}}} $ but $ V\propto \dfrac{1}{r} $ ,
where r is the distance from the charge.