Question

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

Answer 1

The correct option is A $\frac{-\sqrt{2}q}{\sqrt{2}+1}$

$U=K[\frac{q^2}{a}+\frac{Qq}{a}+\frac{Qq}{a\sqrt{2}}]=0$
$\Rightarrow q=-Q[1+\frac{1}{\sqrt{2}}]$
$\Rightarrow Q=\frac{-q\sqrt{2}}{\sqrt{2}+1}$