Question

Two charges $2.0\times {10}^{-6}$ are placed a ta separation of 10 cm. Where should a third charge be placed such that it experienes no net orce due to these charges?

Answer 1

$q_1=2.0\times {10}^{-6}C{q}_2=1.0\times {10}^{-6}C$

$r=10cm=0.1m{F}_{1q}=\frac{9\times {10}^9\times 1.0\times {10}^{-6}\times q}{x^2}{F}_{2q}=\frac{9\times {10}^9\times {10}^{-6}\times q}{(10-x{)}^2}{f}_{2q}=\frac{9\times {10}^9\times {10}^{-6}\times q}{(10-x{)}^2}{F}_{1q}-F_{2q}=0$
So,
$\frac{9\times {10}^9\times {10}^{-6}\times q}{x^2}=\frac{9\times {10}^9\times {10}^{-6}\times q}{(10-x{)}^2}{x}^2=2(10-x{)}^{2}x=5.9cm$
From the larger cahrge to the line joining the charge in the side to the smaller charge.