System of Charges at Equal Distances
Three charges (all q=10 C) are placed at the edge of an equilateral triangle of side 2 m. Find the net potential energy of the system.

Explanation

Given, charge, `q=10C(q_(1)=q_(2)=q)`
Each side of equilateral triangle, r=2m
Potential energy (PE) = ?
Potential energy between two charges is given by
`PE=(kq_(1)q_(2))/(r)" "(because" r = distance between "q_(1)" and "q_(2))`
`:.` PE of system will be three imes the potential energy between the two charges as the equal charge is placed at the vertices of equilateral triangle.
So, `"PE"_("net")=(3xxkqq)/(r)=(3kq^(2))/(r)=(3xx9xx10^(9)xx10xx10)/(2)`
`=135xx10^(10)J`