Three point charge $q, - 4 q, 2 q$ are placed at the vertices of equilateral $Δ$ ABC of side $^{'} l^{'}$ as shown in figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge $q$ .

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Solution

$F_{1} = (\frac{k q 2 q}{l^{2}}) = 2 ((\frac{k q^{2}}{l^{2}})) = 2 F$
$F_{2} = (\frac{k q 2 q}{l^{2}}) = 4 ((\frac{k q^{2}}{l^{2}})) = 4 F$
$∴ r e s u l t a n t = \sqrt{(F_{1})^{2} + (F_{2})^{2} + 2 F_{1} F_{2} c o s θ}$
$= \sqrt{4 F^{2} + 16 F^{2} + 16 F^{2} c o s 120^{\circ}}$
$= \sqrt{20 F^{2} + 16 F^{2} \times (\frac{- 1}{2})}$
$= \sqrt{12 F^{2}} = 2 \sqrt{3} F = 2 \sqrt{3} (\frac{K q^{2}}{l^{2}})$

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