Consider the charges q, q and -q placed at vertices of an equilateral triangle as shown it figure. What is the force on each charge ?

Explanation

Situation given in the statement is depicted in the following figure.
Here `|q_(1)| = |q_(2)| = |q_(3)|=q`

Here since AABC is an equilateral triangle, we have AB = BC = CA = I
Magnitudes of mutual Coulombian forces are equal to F.
`F_(12) = F_(21) = F_(23) = F_(13) = F_(31) = F = (kq^(2))/l^(2)`....(1)
Resultant Coulombian force on `q_(1)` is:
`vecF_(1) = vecF_(12) + vecF_(13)`
`therefore F_(1) = sqrt(F^(2) + F^(2) + 2F F cos (120^(@))) (therefore F_(12) = F_(13) =F)`
`=sqrt(2F^(2) + 2F^(2) (-1/2))` (Here, `theta = 120^(@)`)
`=sqrt(F^(2))`
`therefore F_(1) = F` (in the direction parallel to `bar(BC)`).....(2)
Resultant Coulombian force on `q_(3)` is,
`vecF_(3) = vecF_(31) + vecF_(32)`
`therefore F_(3) = sqrt(F^(2) + F^(2) + 2F F cos 60^(@)) (therefore F_(31) = F_(32) = F)`
`therefore F_(3) = sqrt(2F^(2) + 2F^(2)(1/2))` (Here `theta = 60^(@)`)
`=sqrt(3F^(2))`
`therefore F_(3) = sqrt(3F)` (in the direction of `vec(CD)`)
where D is the midpoint of `bar(AB)` ) (or along the direction bisecting `angle(ACB)`)
`therefore vecF_(B) = {(F cos 60^(@))(-hati) + (F sin 60^(@)hatj)} + Fcos 60^(@) hati + F sin 60^(@) hatj + sqrt(3)F(-hatj)`
`=-F/2hati + sqrt(3)/2 Fhatj + F/2hati + sqrt(3)/2 Fhatj - sqrt(3)Fhatj`
`=sqrt(3)Fhatj - sqrt(3)Fhatj`
`therefore vecF_(B) = vec0`
`therefore vecF_(B) =0`
Here, if given point charges have equal mass then resultant gravitational force on this isolated system will also be zero.
Another alternative Method (Aliter) :
Here, resultant Coulombian force acting on whole system is,
`vecF_(B) = vecF_(1) + vecF_(2) + vecF_(3)`
`therefore =(vecF_(12) + vecF_(13) + vecF_(21) + vecF_(23) + vecF_(31) + vecF_(32))`
`=(vecF_(12) + vecF_(21)) + (vecF_(13) + vecF_(31)) + (vecF_(23) + vecF_(32))`
`=vec0 + vec0 + vec0`
`=vec0` (`therefore` According to Newton.s third law)
`vecF_(12) =-vecF_(21).vecF_(13) =-vecF_(31)` and `vecF_(23) =-vecF_(32)`