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1. Electric Charges and Fields
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Two charges $ + 4e$ and $ + e$ are at a distance $x$ apart. At what distance, a charge $q$ must be placed from charge $ + e$ so that it is in equilibrium
A
$x/2$
B
$2x/3$
C
$x/3$
D
$x/6$
Solution

(c) For equilibrium of $q$
|$F_1$| = |$F_2$|
Which gives ${x_2} = \frac{x}{{\sqrt {\frac{{{Q_1}}}{{{Q_2}}}} + 1}} = \frac{x}{{\sqrt {\frac{{4e}}{e}} + 1}} = \frac{x}{3}$
Standard 12
Physics
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