Question

Three charges Q, \(+q\) and \(+q\) are placed at the vertices of the right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if Q is equal to [a \(=\) length of two equal sides of a triangle]

• A. \(\frac{-q}{1+\sqrt{2}}\)
• B. \(\frac{-\sqrt{2}q}{1+\sqrt{2}}\)
• C. \(\frac{-q}{1+\sqrt{3}}\)
• D. \(-q\)

Hint:

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

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Q is at the top vertex. One \(+q\) is at the right-angle corner (distance \(a\) from Q). The other \(+q\) is at the far corner, at \(a\) from the first \(+q\) and \(\sqrt{2}a\) from Q.

Step 2: Total energy
\(U=\dfrac{1}{4\pi\varepsilon_{0}}\left[\dfrac{Qq}{a}+\dfrac{Qq}{\sqrt{2}a}+\dfrac{q^{2}}{a}\right]\)

Step 3: Set U to zero
\(Q\left(1+\dfrac{1}{\sqrt{2}}\right)=-q\)
\(Q=\dfrac{-q}{1+1/\sqrt{2}}=\dfrac{-\sqrt{2}\,q}{\sqrt{2}+1}\)

Final Answer:
Q equals \(\frac{-\sqrt{2}q}{1+\sqrt{2}}\). \[ \boxed{\text{(B) }\dfrac{-\sqrt{2}\,q}{1+\sqrt{2}}} \]