Q. Two charges $-Q$ and $-9Q$ are placed at a separation of $3m.$ Where should charge q be placed so that net force on it is zero ?

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A

$2m$ from $-Q$

B

$0.75m$ from $-Q$

C

$1m$ from $-Q$

D

$3m$ from $-Q$

Solution:

$k\frac{Q q}{x^{2}}=k\frac{q Q q}{\left(\right. 3 - x \left(\left.\right)^{2}}$
$\frac{1}{x}=\frac{3}{3 - x}$
$3-x=3x$
$4x=3$
$x=\frac{3}{4}=0.75m$

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