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, if the value of \(Q\) is :

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

Hint:

Total energy = sum over pairs of \(\frac{kq_iq_j}{r_{ij}}\). With legs \(a\), the hypotenuse is \(a\sqrt{2}\); the right-angle vertex carries \(+q\).

Answer 1

The Correct Option is D

Let the legs be \(a\); the right angle is at the bottom-left charge \(+q\), so the hypotenuse (between \(Q\) and the other \(+q\)) is \(a\sqrt{2}\).
Pairs: \(Q\) and \(q\) (bottom-left): distance \(a\); \(q\) and \(q\): distance \(a\); \(Q\) and \(q\) (bottom-right): distance \(a\sqrt{2}\).
\(U = k\left[\frac{Qq}{a} + \frac{q^2}{a} + \frac{Qq}{\sqrt{2}\,a}\right] = 0\).
\(q + Q\left(1 + \frac{1}{\sqrt{2}}\right) = 0 \Rightarrow Q = \frac{-q}{1 + \frac{1}{\sqrt{2}}} = \frac{-\sqrt{2}\,q}{\sqrt{2} + 1}\).
Answer: (4) \(\frac{-\sqrt{2}\,q}{\sqrt{2} + 1}\)