Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point C is

A
$\frac{q}{4 {πε}_{0} a^{2}}$
B
$\frac{\sqrt{2} q}{4 {πε}_{0} a^{2}}$
C
$\frac{\sqrt{3} q}{4 {πε}_{0} a^{2}}$
D
$\frac{q}{2 {πε}_{0} a^{2}}$

Solution:

$|E_{A}| = |E_{B}| = k \cdot \frac{q}{a^{2}}$

$So, E_{net} = \sqrt{E_{A}^{2} + E_{B}^{2} + 2 E_{A} E_{B} \cos 60^{\circ}} = \frac{\sqrt{3} k \cdot q}{a^{2}} \Rightarrow E_{net} = \frac{\sqrt{3} q}{4 {πε}_{0} a^{2}}$

Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point C is

Solution:

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