JEE PYQ: Electrostatics Question 50
Question 50 - 2019 (09 Jan Shift 1)
Three charges $+Q$, $q$, $+Q$ are placed respectively, at distance, $d/2$ and $d$ from the origin, on the $x$-axis. If the net force experienced by $+Q$, placed at $x = 0$, is zero, then value of $q$ is:
(1) $-Q/4$
(2) $+Q/2$
(3) $+Q/4$
(4) $-Q/2$
Show Answer
Answer: (1)
Solution
Force on $+Q$ at $x = 0$: due to $q$ at $d/2$ and $+Q$ at $d$. $\frac{kQq}{(d/2)^2} + \frac{kQQ}{d^2} = 0$. $\frac{4kQq}{d^2} = -\frac{kQ^2}{d^2}$. $q = -Q/4$.
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