Question

If three equal charges q are placed at the vertices of an equilateral triangle of side l, what is the net potential energy of the system?

Answer 1

Step 1. Given data:

The magnitude of three equal charges = $q$

Side of the triangle = $l$

Net potential energy =?

Step 2. Formula used:

Net potential energy $U=\frac{q_1.q_2}{4{\mathrm{\pi \epsilon }}_{0}l}$

Step 3. Solution:

Where $U_{AB}$= $\left(\frac{q^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{l}}\right)$ Potential energy along AB length of the triangle.

$U_{BC}$=$\left(\frac{q^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{l}}\right)$ Potential energy along BC length of the triangle.

$U_{CA}$= $\left(\frac{q^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{l}}\right)$Potential energy along the CA length of the triangle.

$U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q^2}{I}+\frac{q^2}{I}+\frac{q^2}{I}\right)$……..As all the charges are equal

$U=\frac{3q^2}{4{\mathrm{\pi \epsilon }}_{0}l}$

Hence, the net potential energy of the system will be $\frac{3q^2}{4{\mathrm{\pi \epsilon }}_{0}l}$,