Question

Three charges Q + q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, it the value of Q is:

• A. -2q
• B. $\frac{-q}{1+\sqrt{2}}$
• C. $\frac{-\sqrt{2}q}{\sqrt{2}+1}$
• D. +q

Answer 1

Answer: (C)

$\frac{-\sqrt{2}q}{\sqrt{2}+1}$

Answer 2

Let the vertices of the right-angled isosceles triangle be A, B, and C, with the right angle at A. Let the charges be placed at these vertices.

Case 1: Charge Q at A, +q at B, +q at C. Let AB = AC = a, BC = $a\sqrt{2}$. Potential energy $U = \frac{kQq}{a} + \frac{kQq}{a} + \frac{kq^2}{a\sqrt{2}} = \frac{k}{a}(2Qq + \frac{q^2}{\sqrt{2}})$. Setting $U=0$ gives $Q = -\frac{q}{2\sqrt{2}}$.

Case 2: Charge +q at A, Q at B, +q at C. Let AB = AC = a, BC = $a\sqrt{2}$. Potential energy $U = \frac{k(+q)Q}{a} + \frac{k(+q)(+q)}{a} + \frac{kQ(+q)}{a\sqrt{2}} = \frac{k}{a}(Qq + q^2 + \frac{Qq}{\sqrt{2}})$. Setting $U=0$ gives $Qq + q^2 + \frac{Qq}{\sqrt{2}} = 0$. $Q(1+\frac{1}{\sqrt{2}}) = -q$. $Q = -\frac{q}{1+1/\sqrt{2}} = -\frac{\sqrt{2}q}{\sqrt{2}+1}$. This matches option (3).

Since option (3) is a provided choice, it is highly probable that the intended configuration is Case 2 (or Case 3 by symmetry), despite the figure showing Q at the right angle.