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

Explanation

(a) Let a square ABCD charges are placed as shown

Now forces on charge Q (at point A) due to other charge are `vec(F)_(Q Q), vec(F)_(Q q)` and `F_(Q q)` respectively shown in figure.
`F_("net")` on `Q = vec(F)_(Q Q) + vec(F)_(Q q) + vec(F)_(Q q)`
But `F_("net") = 0`, So, `SigmaF_(x) = 0`
`Sigma F_(x) = -F_(Q Q) cos 45^(@) - F_(Q q)`
`rArr (KQ^(2))/((sqrt(2)a)^(2)).(1)/(sqrt(2)) + (KQ q)/(a^(2)) = 0`
`rArr q = - (Q)/(2sqrt(2))`
(b) For resultant force on each charge to be zero : From previous data, force on charge Q is zero when `q = -(Q)/(2sqrt(2))` if for this value of charge q, force on q is zero then and only then the value of q exists for which the resultant force on each charge is zero.
Force on q : -
Forces on charges (at point D) due to other three charges are `vec(F)_(q Q), vec(F)_(q Q)` respectively shown in figure.

Net force on charge q : -
`vec(F)_("net") = vec(F)_(q q) + vec(F)_(q Q)` But `vec(F)_("net") = 0`
So, `Sigma F_(x) = 0`
`Sigma F_(x) = -(Kq^(2))/((sqrt(2)a)^(2)).(1)/(sqrt(2)) - (KQq)/((a)^(2))`
`rArr q = -2sqrt(2)Q`
But from previous condition, `q = - (Q)/(2sqrt(2))`
So, no value of q makes the resultant force on each charge zero.