Three charges q,q and `-q` are kept in the three vertices of an equilateral triangle of side l. Find out the resultant force on each of the charges.

Explanation

Due to symmetry, the magnitude of the force between each pair of charges would be the same, given by
`F =1/(4pi epsilon_(0))q^(2)/l^(2)`

Force acting on charge q at A due to the charges at points B and C,
`F_(A) = sqrt(F_(BA)^(2) + F_(BC)^(2) + 2F_(BA). F_(BC) cos 120^(@))`
`=sqrt(F_(BA)^(2) + F_(BC)^(2) -F_(BA). F_(BC))`
`=F_(BA) [therefore F_(BA) = F_(BC)]`
`=1/(4pi epsilon_(0)). q^(2)/r^(2)`, along `bar(BE)`
Force acting on charge `-q` at C due to the charges at point A and B,
`F_(C) = sqrt(F_(CA)^(2) + F_(CB)^(2) + 2F_(CA). F_(CB). cos 60^(@))`
`=sqrt(3)F_(CA) [therefore f_(CA) = F_(CB)]`
`therefore = sqrt(3).1/(4pi epsilon_(0)). q^(2)/l^(2)`, along `vec(CF)`

The forces `F_(A), F_(B)` and `F_(C)` are shown in the fig.
Let `vec(F)` be the resultant of the forces `vecF_(A)` and `vecF_(B)`
`therefore F=sqrt(F_(A)^(2) + F_(B)^(2) + 2F_(A)F_(B)cos 60^(@)) = sqrt(3)F_(A) = sqrt(3). 1/(4pi epsilon_(0)). q^(2)/l^(2)`
Therefore, F and `F_(C)` have the same magnitude, and they are oppositely directed. So, they cancel each other.
`therefore vecF_(A) + vecF_(B) + vecF_(C) = vecF + vecF_(C) = -vecF_(C) + vecF_(C) =0`