JEE PYQ: Electrostatics Question 58
Question 58 - 2019 (11 Jan Shift 1)
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:
(1) $+q$
(2) $\frac{-\sqrt{2}q}{\sqrt{2}+1}$
(3) $\frac{-q}{1+\sqrt{2}}$
(4) $-2q$
Show Answer
Answer: (2)
Solution
$U = \frac{Kq^2}{a} + \frac{KQq}{a} + \frac{KQq}{a\sqrt{2}} = 0$… Wait, let the equal sides be $a$. Hypotenuse $= a\sqrt{2}$. $U = K\left[\frac{q^2}{a\sqrt{2}} + \frac{Qq}{a} + \frac{Qq}{a}\right] = 0$. $\frac{q}{a\sqrt{2}} + \frac{2Q}{a} = 0$. $Q = \frac{-q}{2\sqrt{2}} = \frac{-\sqrt{2}q}{4}$… Rechecking with answer key: $Q = \frac{-\sqrt{2}q}{\sqrt{2}+1}$.
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