Electric charges of $1\mu C,-1\mu C$ and $2\mu C$ are placed in air at the corners $A,B$ and $C$ of an equilateral triangle having length of each side $10cm$ respectively. The resultant force on the charge at $C$ is

The correct option is B $1.8N$


Let $F_A$ be force on $C$ due to charge placed at $A$
$F_A=9\times {10}^9\times \frac{2\times {10}^{-6}\times {10}^{-6}}{(10\times {10}^{-2}{)}^2}=1.8N$
Here $F_A$ is repulsive in nature, as both $A$ and $B$ have positive charges. Hence, $F_A$ acts away from $A$

Let $F_B$ be force on $C$ due to charge placed at $B$
$F_B=9\times {10}^9\times \frac{{10}^{-6}\times 2\times {10}^{-6}}{(0.1{)}^2}=1.8N$
Here $F_B$ is attractive in nature, as $C$ and $B$ have charges of opposite polarities. Hence, $F_B$ acts towards $B$.

As angle between $F_A,F_B$ is ${120}^{\circ }$, net force on $C$ is
$F_{net}=\sqrt{(F_A{)}^2+(F_B{)}^2+2F_A{F}_B\cos {120}^{\circ }}$
$F_{net}=\sqrt{(1.8{)}^2+(1.8{)}^2+2\left(1.8\right)\left(1.8\right)\cos {120}^{\circ }}=1.8N$