# Charges of +2microC, +3microC and -8microC are placed in air at the vertices of an equilateral triangle of ide 10cm.What is the magnitude of the force acting on the -8microC due to the other two charges?

Mar 14, 2018

Let charge $2 \mu C , 3 \mu C , - 8 \mu C$ are placed at point $A , B , C$ of the triangle shown.

So,net force on $- 8 \mu C$ due to $2 \mu C$ will act along $C A$

and the value is ${F}_{1} = \frac{9 \cdot {10}^{9} \cdot \left(2 \cdot {10}^{-} 6\right) \cdot \left(- 8\right) \cdot {10}^{-} 6}{\frac{10}{100}} ^ 2 = - 14.4 N$

And due to $3 \mu C$ it will be along $C B$ i.e ${F}_{2} = \frac{9 \cdot {10}^{9} \cdot \left(3 \cdot {10}^{-} 6\right) \left(- 8\right) \cdot {10}^{-} 6}{\frac{10}{100}} ^ 2 = - 21.6 N$

So,two forces of ${F}_{1}$ and ${F}_{2}$ are acting on the charge $- 8 \mu C$ with an angle of ${60}^{\circ}$ in between,so the nect force will be, $F = \sqrt{{F}_{1}^{2} + {F}_{2}^{2} + 2 {F}_{1} {F}_{2} \cos 60} = 31.37 N$

Making an angle of ${\tan}^{-} 1 \left(\frac{14.4 \sin 60}{21.6 + 14.4 \cos 60}\right) = {29.4}^{\circ}$ with ${F}_{2}$