Question

Three charges, each $+ q$, are placed at the corners of an isosceles triangle $ABC$ of sides $BC$ and $AC = 2a$. $D$ and $E$ are the mid points of $BC$ and $CA$, as shown in figure. The work done in taking a charge $Q$ from $D$ to $E$ is

• A. $\frac{qQ}{8\pi\varepsilon_{0} a}$
• B. $\frac{qQ}{4\pi\varepsilon_{0} a}$
• C. Zero
• D. $\frac{3qQ}{4\pi\varepsilon_{0} a}$

Answer 1

Answer: (C)

According to figure, $AC = BC$
Potential at $D$ = potential at $E$
i.e., $V_{D}=V_{E}=2\times\frac{1}{4\pi\varepsilon_{0}}\frac{q}{a}+\frac{1}{4\pi\varepsilon_{0}} \frac{q}{DB}$
$=\frac{2q}{4\pi\varepsilon_{0} a}+\frac{q}{4\pi\varepsilon_{0}}\times\frac{1}{2a\,sin\,60^{\circ}}$
$=\frac{q}{4\pi\varepsilon_{0} a}\left[2+\frac{1}{2\times\sqrt{3 2}}\right]$
$=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{a}\left[2+\frac{1}{\sqrt{3}}\right]$
$\therefore $ Work done in taking charge $Q$ from $D$ to $E$
$=QV_{E}-QV_{D}=Q\left(V_{E}-V_{D}\right)=Q\times0=0$