Solution:

Ans. (i) Finding the magnitude of the resultant force on charge $q$

(ii) Finding the work done

(i) Force on charge $q$ due to the charge $-4 q$

$$ \overrightarrow{F_{1}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{4 q^{2}}{l^{2}}\right), \text { along } A B $$

Force on the charge $q$, due to the charge 2

$$ \vec{F}{2}=\frac{1}{4 \pi \varepsilon{0}}\left(\frac{2 q^{2}}{l^{2}}\right), \text { along } C A $$

The forces $F_{1}$ and $F_{2}$ are inclined to each other at an angle of $120^{\circ}$

Hence, resultant electric force on charge $q$

$$ \begin{aligned} F & =\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta} \ & =\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos 120^{\circ}} \ & =\sqrt{F_{1}^{2}+F_{2}^{2}-F_{1} F_{2}} \ & =\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{l^{2}}\right) \sqrt{16+4-8} \end{aligned} $$

$$ =\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{2 \sqrt{3} q^{2}}{l^{2}}\right) $$

(ii) Net P.E. of the system

$$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{l}[-4+2-8] \ & =\frac{(-10)}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{l} \ \therefore \quad \text { Work done } & =\frac{10 q^{2}}{4 p \varepsilon_{0} l}=\frac{5 q^{2}}{2 \pi \varepsilon_{0} l} \end{aligned} $$

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Detailed Answer :

(ii) The amount of work done to separate the charges at infinite distance is equal to the (-ve) potential energy of the given system. Now we know that the potential energy of three (03) charges at the corners of an equilateral triangle $A B C$ of side $l$ is given by

$$ U_{P E}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right] $$

given $q_{1}=q, q_{2}=-4 q, q_{3}=2 q, r_{12}=r_{13}=r_{23}=l$

$$ \begin{aligned} U_{P E} & =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q(-4 q)}{l}+\frac{q \cdot 2 q}{l}+\frac{2 q(-4 q)}{l}\right] \ & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{l}[-4+2-8] \ U_{P E} & =-\frac{10 q^{2}}{4 \pi \varepsilon_{0} l} \end{aligned} $$

Therefore work done, $W=-U_{P E}=-\left(\frac{-10 q^{2}}{4 \pi \varepsilon_{0} l}\right)$

$$ \Rightarrow \quad W=+\frac{5 q^{2}}{2 \pi \varepsilon_{0} l} $$