# 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:

(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} $

Answer

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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**

Here,

$ AC=BC=2a $

As D and E are midpoints of BC and AC (given).

$ \therefore $ $ AE=EC=a $

And,

$ 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.**

**Note**

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.

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