Question

(a) Three point charges q, - 4q and 2q are placed at the vertices of an equilateral triangle ABC of side 'l' as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.
(b) Find out the amount of the work done to separate the charges at infinite distance.

Answer 1

(a) Electric force at A due to charge 2q
$F_c=\frac{1}{4\pi {ϵ}_0}\frac{q\times 2q}{l^2}\text{along}\underset{CA}{\to }$

Electric force at A due to charge (-4q),

$F_B=\frac{1}{4\pi {ϵ}_0}\frac{q\times (-4q)}{l^2}\text{along}\underset{AB}{\to }$


Resultant force, F = $\sqrt{F_B^a+F_C^2+2F_B{F}_{C}cos{120}^{\circ }}$
$=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{l^2}\sqrt{(-4{)}^2+(2{)}^2+2(-4)\left(2\right)(-\frac{1}{2})}$
$=\frac{1}{4\pi {ϵ}_0}\frac{2\sqrt{7}{q}^2}{l^2}$
(b) Work done to seperate the charges to infinity :
Initial potential energy,
$U_i=\frac{1}{4\pi {ϵ}_0}[\frac{(-4q)q}{l}+\frac{(-4q)\left(2q\right)}{l}+\frac{\left(q\right)\left(2q\right)}{l}]$

=$\frac{1}{4\pi {ϵ}_0}\frac{q^2}{l}[-4-8+2]=\frac{1}{4\pi {ϵ}_0}\left(\frac{-10q^2}{l}\right)$
Final potential energy, $U_f=0$
Thus, work done
= $U_f-U_i=0-\left(\frac{-10q^2}{4\pi {ϵ}_{0}l}\right)=\frac{10q^2}{4\pi {ϵ}_{0}l}$