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Q.

If 3 charges are placed at the vertices of equilateral triangle of charge ‘q’ each. What is the net potential energy, if the side of equilateral $Δ$ is 1 cm

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a

$\frac{1}{4 {πε}_{0}} \frac{4 q^{2}}{l}$

b

$\frac{1}{4 {πε}_{0}} \frac{q^{2}}{l}$

c

$\frac{1}{4 {πε}_{0}} \frac{3 q^{2}}{l}$

d

$\frac{1}{4 {πε}_{0}} \frac{2 q^{2}}{l}$

answer is C.

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Detailed Solution

$U = \frac{1}{4 {πε}_{0}} . \frac{Q_{1} Q_{2}}{r};$ net potential energy

$U_{net} = 3 \times \frac{1}{4 {πε}_{0}} . \frac{q^{2}}{l}$

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