Three charges, each equal to +q, are placed at the corners of an equilateral triangle ABC of side 'a' as shown in the figure. P and R are the mid-points of BC and CA. The work done in taking a charge Q from point P to R is:

Zero

The correct answer is Option (4) → Zero

Work done in moving a charge $Q$ from point $P$ to $R$ in an electrostatic field is:

$W = Q \, \big(V_R - V_P\big)$

Potentials at $P$ and $R$ are due to the three charges $+q$ at vertices $A, B, C$ of an equilateral triangle of side $a$.

Coordinates approach:

• Let $A(0, \sqrt{3}a/2)$, $B(-a/2, 0)$, $C(a/2, 0)$.
• $P$ is midpoint of $BC$: $P(0, 0)$.
• $R$ is midpoint of $CA$: $R\big(\frac{a}{4}, \frac{\sqrt{3}a}{4}\big)$.

Potential at $P$:

Distances: $PB = PC = a/2$, $PA = \sqrt{3}a/2$.

$V_P = kq \Big(\frac{1}{a/2} + \frac{1}{a/2} + \frac{1}{\sqrt{3}a/2}\Big)$

$V_P = kq \Big(\frac{2}{a/2} + \frac{1}{(\sqrt{3}/2)a}\Big)$

$V_P = kq \Big(\frac{4}{a} + \frac{2}{\sqrt{3}a}\Big)$

Potential at $R$:

Distances: $RA = RC = a/2$, $RB = \sqrt{(3a/4)^2 + (\sqrt{3}a/4)^2} = \sqrt{12a^2/16} = \frac{\sqrt{3}a}{2}$.

$V_R = kq \Big(\frac{1}{a/2} + \frac{1}{a/2} + \frac{1}{\sqrt{3}a/2}\Big)$

$V_R = kq \Big(\frac{2}{a/2} + \frac{1}{(\sqrt{3}/2)a}\Big)$

$V_R = kq \Big(\frac{4}{a} + \frac{2}{\sqrt{3}a}\Big)$

Comparison:

$V_R = V_P$

Thus, $W = Q (V_R - V_P) = 0$

Answer: Work done = 0