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

Answer 1

Answer: (C)

According to figure, $AC=BC$
Potential at $D$ = potential at $E$
i.e., $V_D=V_E=2\times \frac{1}{4\pi {\epsilon }_0}\frac{q}{a}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{DB}$
$=\frac{2q}{4\pi {\epsilon }_{0}a}+\frac{q}{4\pi {\epsilon }_0}\times \frac{1}{2asin60^{\circ }}$
$=\frac{q}{4\pi {\epsilon }_{0}a}\left[2+\frac{1}{2\times \sqrt{32}}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}\left[2+\frac{1}{\sqrt{3}}\right]$
$\therefore$ Work done in taking charge $Q$ from $D$ to $E$
$=Q{V}_E-Q{V}_D=Q\left(V_E-V_D\right)=Q\times 0=0$