# (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

By coulomb's law force between two charges is
$F= \frac{1}{4\pi \varepsilon _{0}}\: \frac{q_{1}q_{2}}{r^{2}}$
where q1 and q2 are charges separated by distance r
Here, to find the force on the charge at A by charge at B and C
Force acting on charge q by -4q is
$F_{1}= \frac{1}{4\pi \varepsilon _{0}}\:\frac{4q\times q}{l^{2}}= \frac{1}{4\pi \varepsilon _{0}}\, \cdot \frac{4q}{l^{2}}$
Force acting on charge q by 2q is
$F_{2}= \frac{1}{4\pi \varepsilon _{0}}\, \cdot \frac{2q^{2}}{l^{2}}$

For the above diagram, the resultant force acting on charge at
A is,

$F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\: \cos \left ( 120 \right )}$
here, $F_{1}=2F_{2}$   and  $\cos 120^{\circ}= \frac{-1}{2}$
$F_{net}=\sqrt{\left ( 2F_{2} \right )^{2}+F_{2}^{2}-\frac{4F_{2}}{2}}= \sqrt{4F_{2}^{2}+F_{2}^{2}-2F_{2}^{2}}$
$F_{net}=\sqrt{3}F_{2}$
So-net  force on q is

, $F= \sqrt{3}\times \frac{1}{4\pi \varepsilon _{0}}\times \frac{2q^{2}}{l^{2}}$
b) The amount of the work done to separate the charges at infinite distance is equal to the potential energy of the system.
The potential energy between two charges separated by a distance 'r' is
$\cup = \frac{1}{4\pi \varepsilon _{0}}\, \frac{q_{1}q_{2}}{r}$
Here the potential energy of the system is,
$\cup = \frac{1}{4\pi \varepsilon _{0}\times l}\left [ \left ( q\times -4q \right ) +\left ( q\times 2q \right )+\left ( -4q\times 2q \right )\right ]$
$\cup = \frac{1}{4\pi \varepsilon _{0}\times l}\left [-4q^{2}+2q^{2}-8q^{2}\right ]$
$\cup = \frac{1}{4\pi \varepsilon _{0}\times l}\times \left ( -10q^{2} \right )$
therefore,

work done is

$\frac{5q^{2}}{2\pi \varepsilon _{0}l}$

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