Point charges are placed at the vertices of a square of side 'a' as shown in fig. What should be the sign of charge q and magnitude of the ratio `|(q)/(Q)|` so that
net force on each Q is zero?
Is it possible that the entire system could be in electrostatic equilibrium?

Explanation

Consider now the forces acting on charge q placed at B. In a similar manner, as discussed in (i), for the net force on q to be zero, q and Q have to be of opposite signs. Thos is also shown in the given figures.
Now, `F_(1)=(1)/(4piepsilon_(0))(Qq)/(a^(2))` (force of Q at A on q at B)
`F_(2)=(1)/(4piepsilon_(0))(Qq)/(a^(2))` (force of Q at C on q at B)
`F_(3)=(1)/(4piepsilon_(0))(q^(2))/(2a^(2))` (force of q at D on q at B)

From we can write `vecF_(R)=vecF_(1)+vecF_(2)`
`therefore vecF_(R)=vecF_(1)+vecF_(2)`
`therefore F_(R)=sqrt(F_(1)^(2)+F_(2)^(2))=(1)/(4piepsilon_(0))(Qq)/(a^(2))sqrt(2)`
The resultant of `vecF_(1)andvecF_(2)` i.e., `vecF_(R)` is opposite to `vecF_(3)`. Net force can become zero if their magnitude are also equal, i.e.,
`(1)/(4piepsilon_(0))(Qq)/(a^(2))sqrt(2)=(1)/(4piepsilon_(0))(q^(2))/(2a^(2))or(q)/(4piepsilon_(0)a^(2))[sqrt(2)Q-(q)/(2)]=0`
or `Q=(q)/(2sqrt(2))or|(q)/(Q)|=2sqrt(2){qne0}`
Therefore, the sign of q should be negative of Q.
In this case, we need not repeat the calculation as the present situation is same as the previous one, we can directly write `|(q)/(Q)|=2sqrt(2)`. The entire system cannot be in equilibrium, since both conditions, `q=-(Q)/(2sqrt(2))` and `Q=-(q)/(2sqrt(2))` cannot be satisfied together.