Q 12-02-184JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
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
Answer: (B) $\dfrac{-\sqrt2\,q}{\sqrt2 + 1}$
Let the equal sides be $a$. $Q$ is at distance $a$ from one $+q$ and $\sqrt2a$ from the other; the two $+q$ charges are $a$ apart.
$$U = k\left[\frac{Qq}{a} + \frac{Qq}{\sqrt2a} + \frac{q^2}{a}\right] = 0$$
$$Q\left(1 + \frac1{\sqrt2}\right) = -q \Rightarrow Q = \frac{-\sqrt2\,q}{\sqrt2 + 1}$$
Solution by Sreeraj P, M.Sc Physics