Physics - Electrostatics Question with Solution | TestHub

Let the charge $+e$ be placed at vertex C of the equilateral triangle ABC, while $+4e$ and $+2e$ are placed at vertices A and B respectively.

The electric intensity (electric field) at C due to the charge $+4e$ at A is:
$E_1=\frac{k(4e)}{a^2}$

The electric intensity at C due to the charge $+2e$ at B is:
$E_2=\frac{k(2e)}{a^2}$

Since ABC is an equilateral triangle, the angle between the electric field vectors ${\vec{E}}_1$ and ${\vec{E}}_2$ is $60^{\circ }$.

The net electric intensity at C is given by the vector sum of ${\vec{E}}_1$ and ${\vec{E}}_2$:
$E_{net}=\sqrt{E_1^2+E_2^2+2E_1{E}_2\cos 60^{\circ }}$

Substituting the values of $E_1$, $E_2$, and $\cos 60^{\circ }=\frac{1}{2}$:
$E_{net}=\sqrt{{\left(\frac{4ke}{a^2}\right)}^2+{\left(\frac{2ke}{a^2}\right)}^2+2\left(\frac{4ke}{a^2}\right)\left(\frac{2ke}{a^2}\right)\left(\frac{1}{2}\right)}$

$E_{net}=\frac{ke}{a^2}\sqrt{16+4+8}$

$E_{net}=\frac{ke}{a^2}\sqrt{28}$

$E_{net}=\frac{2\sqrt{7}ke}{a^2}$

Answer: $\frac{2\sqrt{7}ke}{a^2}$