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 on 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(D)_(Q q)` and `vec(F)_(Q q)` respectively as shown in figure.
`F_("net")` on `Q=vec(F)_(Q. Q)+vec(F)_(Qq)+vec(F)_(Qq)` (at point A)
But `F_("net")=0`
So, `SigmaF_(x)=0`
`SigmaF_(x)=-F_(Q Q) cos 45^(@)-F_(Q q)`
`implies(KQ^(2))/((sqrt(2)a)^(2)). 1/sqrt(2)+(KQq)/a^(2)=0 implies" "q=-Q/(2sqrt(2))`
(b) For resultant force on each charge o 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 charge q (at point D) due to other three charges are `vec(F)_(q Q), vec(F)_(q q)` and `vec(F)_(q Q)` respectively as shown in figure.
Net force on charge q :
`vec("net")=vec(F)_(q q)+vec(F)_(q Q)+vec(F)_(q Q)` But `vec(F)_("net")=0`
So, `SigmaF_(x)=0`
`SigmaF_(x)=-(Kq^(2))/((sqrt(2)a)^(2)). 1/sqrt(2)-(KQq)/((a)^(2))" "implies" "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.