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 midpoints of BC and CA, as shown in the 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

The correct option is C zero

Here, $AC=BC=2a$

$D$ and $E$ are the midpoints of $BC$ and $AC$.

$\therefore AE=EC=a$ and $BD=DC=a$

In $\Delta ADC,(AD{)}^2=(AC{)}^2-(DC{)}^2$

$=(2a{)}^2-(a{)}^2=4a^2-a^2=3a^2$

$AD=a\sqrt{3}$

Similarly, potential at point $D$ due to the given charge configuration is

$V_D=\frac{1}{4\pi {ϵ}_0}[\frac{q}{BD}+\frac{q}{DC}+\frac{q}{AD}]$

$=\frac{q}{4\pi {ϵ}_0}[\frac{1}{a}+\frac{1}{a}+\frac{1}{\sqrt{3a}}]=\frac{q}{4\pi {ϵ}_{0}a}[2+\frac{1}{\sqrt{3}}]$...........(i)

Potential at point $E$ due to the given charge configuration is

$V_E=\frac{1}{4\pi {ϵ}_0}[\frac{1}{AE}+\frac{q}{EC}+\frac{q}{BE}]$

$=\frac{1}{4\pi {ϵ}_0}[\frac{1}{a}+\frac{1}{a}+\frac{1}{a\sqrt{3}}]=\frac{q}{4\pi {ϵ}_{0}a}[2+\frac{1}{\sqrt{3}}]$...........(2)

From the (i) and (ii), it is clear that

$V_D=V_E$

The work done in taking a charge $Q$ from $D$ to $E$ is

$W=Q(V_E-V_D)=0$ $(\because V_D=V_E)$