A charges Q is placed at each of the two opposite corners of a square. A charge q is placed to each of the other two corners. If the resultant force on each charge q is zero, then

Answer

`q = -2 sqrt(2)Q`

Explanation

Let each side of square be x Fig.
`:.` Diagonal, `BD = sqrt(x^(2) + x^(2)) = x sqrt(2)`
`F_(1) = F_(2) = (Qq)/(4pi in_(0) x^(2))`

`F_(3) = (qq)/(4pi in_(0) (x sqrt(2))^(2)) = (q^(2))/(2xx 4pi in_(0) x^(2))`
As `F_(1) and F_(2) are _|_` to each other, their resultant force
`= sqrt(F_(1)^(2) + F_(2)^(2)) = sqrt(F_(1)^(2) + F_(1)^(2)) = F_(1) sqrt(2)`
As net force on q is zero, therefore,
`F_(1) sqrt(2) = -F_(3)`
`(Qq sqrt(2))/(4pi in_(0) x^(2)) = (-q^(2))/(2xx 4pi in_(0) x^(2))`
`:. q = -2 sqrt(2)Q`