Four charges each of `-Q`, are placed at the four corners of a square. Find out the value of a charge q placed at the centre of the square, such that all the charges would be at equilibrium.

Explanation

Let each side of the square =a [Fig.]

`therefore AC = sqrt(a^(2) + a^(2)) =sqrt(2)a` and `AO = (AC)/2 = (sqrt(2)a)/2 = a/sqrt(2)`
Force on the charge `-Q` at A due to the charge `-Q` at B.
`F=1/(4pi epsilon_(0)).Q^(2)/a^(2)`, along `vec(BA)`
Similarly, the force on `-Q` at A due to `-Q` at D will also be `F_(1)`, but along `vec(DA)`.
The resultant of these two forces will be,
`F_(1) = sqrt(F^(2) + F^(2)) =sqrt(2)F = 1/(4pi epsilon_(0)). (sqrt(2)Q^(2))/a^(2)`, along `vec(CA)`
Again, force on `-Q` at A due to the charge `-Q` at C.
`F_(2) =1/(4pi epsilon_(0)).Q^(2)/(sqrt(2)a)^(2) =1/(4pi epsilon_(0)).Q^(2)/(2a^(2))`, along `vec(CA)`
Now, force on -Q at A due to the charge q at O.
`F_(3) = 1/(4pi epsilon_(0)).(qQ)/(a/sqrt(2))^(2) =1/(4pi epsilon_(0)).(2qQ)/a^(2)`, along `vec(CA)`
At equilibrium, the resultant of these three forces along `vec(CA)` must be zero, i.e.,
`F_(1) + F_(2) + F_(3) =0` or `1/(4pi epsilon_(0))[((sqrt(2)Q^(2))/a^(2) + Q^(2)/(2a^(2)) -(2qQ)/a^(2))]=0`
or `1/(4pi epsilon_(0)).Q/a^(2)[Q(sqrt(2) + 1/2) - 2q]=0`
or `Q(sqrt(2) + 1/2) -2q=0`
or `2q = Q(2sqrt(2)+1)/2` or, `q=Q/4(2sqrt(2)+1)`