Question

Three charges \(Q(-2 q)\) and \((-2 q)\) are placed at the vertices of an isosceles right angled triangle as shown in figure. The net electrostatic potential energy is zero if \(Q\) is equal to

• A. \(\sqrt{2}q\)
• B. \(\frac{\mathrm{q}}{2}\)
• C. \(\frac{\mathrm{q}}{\sqrt{2}}\)
• D. \(\frac{q}{2\sqrt{2}}\)

Hint:

Add the pair energies of the three charges and set the sum to zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The potential energy of the system is the sum over all pairs, \(U=\sum\dfrac{kq_iq_j}{r_{ij}}\).

Step 2: Distances
Q is at the right angle with the two charges \(-2q\) at distance \(l\) each. The two \(-2q\) charges are \(\sqrt2\,l\) apart.

Step 3: Total energy
\(U=k\left[\dfrac{Q(-2q)}{l}+\dfrac{Q(-2q)}{l}+\dfrac{(-2q)(-2q)}{\sqrt2\,l}\right]=\dfrac{k}{l}\left[-4Qq+2\sqrt2\,q^{2}\right]\).

Step 4: Set U = 0
\(4Qq=2\sqrt2q^{2}\), so \(Q=\dfrac{\sqrt2\,q}{2}=\dfrac{q}{\sqrt2}\).

Final Answer:
Q must equal q divided by root 2. \[ \boxed{\text{(C) }\dfrac{q}{\sqrt{2}}} \]