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

The correct option is C $1.8N$
Diagram

$F_A=$ force on $C$ due to charge placed at $A$
$F_B=$ force on $C$ due to charge placed at $B$
$F_A=\frac{\left(q_A{q}_C\right)}{\left(4\pi {ϵ}_0{r}^2\right)},F_B=\frac{\left(q_C{q}_B\right)}{\left(4\pi {ϵ}_0{r}^2\right)}$
Given $q_A=+1\times {10}^{-6}C,q_B=-1\times {10}^{-6}C,q_C=+2\times {10}^{-6}C,r=0.1m$
$\left|F_A\right|=9\times {10}^9\times \frac{({10}^{-6}\times 2\times {10}^{-6})}{(10\times {10}^{-2}{)}^2}=1.8N$
$\left|F_B\right|=9\times {10}^9\times \frac{({10}^{-6}\times 2\times {10}^{-6})}{(10\times {10}^{-2}{)}^2}=1.8N$
Net force on C:
$F_{net}=\sqrt{(F_A{)}^2+(F_B{)}^2+2\left|F_A\right|\left|F_B\right|\cos {120}^{\circ })}$
$=1.8N$

Hence, option (b) is correct.