Three charges $q_1=1\mu C,q_2=-2\mu C$ and $q_3=3\mu C$ are placed on the vertices of an equilateral triangle of side $1.0m$. Find the net electric force acting on charge $q_1$.

 

Method $1$. In the figure,

$\left|F_1\right|=F_1=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{r^2}$

$=\text{ magnitude of force between }{q}_1\text{ and }{q}_2$

$=\frac{\left(9.0\times {10}^9\right)\left(1.0\times {10}^{-6}\right)\left(2.0\times {10}^{-6}\right)}{(1.0{)}^2}$

$=1.8\times {10}^{-2}N$

Similarly,

$\left|F_2\right|=F_2=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_3}{r^2}$

$=\text{ magnitude of force between }{q}_1\text{ and }{q}_3$

$=\frac{\left(9.0\times {10}^9\right)\left(1.0\times {10}^{-6}\right)\left(3.0\times {10}^{-6}\right)}{(1.0{)}^2}$

$=2.7\times {10}^{-2}N$

Now,

$\left|F_{\text{net }}\right|=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos 120°}$

$=\left(\sqrt{(1.8{)}^2+(2.7{)}^2+2(1.8)(2.7)\left(-\frac{1}{2}\right)}\right)\times {10}^{-2}N$

$=2.38\times {10}^{-2}N$

and

$\tan \alpha =\frac{F_2\sin 120°}{F_1+F_2\cos 120°}$

$=\frac{\left(2.7\times {10}^{-2}\right)(0.87)}{\left(1.8\times {10}^{-2}\right)+\left(2.7\times {10}^{-2}\right)\left(-\frac{1}{2}\right)}$

Or

$ct=79.2°$

Thus, the net force on charge $q_1$ is $2.38\times {10}^{-2}N$ at an angle $\alpha =79.2°$ with a line joining $q_1$ and $q_2$ as shown in the figure.

Method $2$. In this method let us assume a coordinate axes with $q_1$ at origin as shown in figure. The coordinates of $q_1,q_2$ and $q_3$ in this coordinate system are $(0,0,0),(1 m , 0,0) and (0.5 m), (0.87 m , 0)$ respectively. Now,

$F_1=\text{ force on }{q}_1\text{ due to charge }{q}_2$

$=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{{\left|r_1-r_2\right|}^3}\left(r_1-r_2\right)$

$=\frac{\left(9.0\times {10}^9\right)\left(1.0\times {10}^{-6}\right)\left(-2.0\times {10}^{-6}\right)}{(1.0{)}^3}\left[\right(0-1)\hat{i}+(0-0)\hat{j}+(0-0\left)\hat{k}\right]$ 

$=\left(1.8\times {10}^{-2}\hat{i}\right)N$

${\text{ and F}}_2=\text{ force on }{q}_1\text{ due to charge }{q}_3$

$=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_3}{{\left|r_1-r_3\right|}^3}\left(r_1-r_3\right)$

$=\frac{\left(9.0\times {10}^9\right)\left(1.0\times {10}^{-6}\right)\left(3.0\times {10}^{-6}\right)}{(1.0{)}^3}\left[\right(0-0.5)\hat{i}+(0-0.87)\hat{j}+(0-0\left)\hat{k}\right]$

$=(-1.35\hat{i}-2.349\hat{j})\times {10}^{-2}N$

$\text{ Therefore, net force on } q_1\text{ is } F=F_1+F_2$

$=(0.45\hat{i}-2.349\hat{j})\times {10}^{-2}N$