It is requird to hold four equal point charges `+q` in equilibrium at the corners of a square. Find the point charge that will do this, if placed at the centre of the square.

Explanation

Let ABCD be a square of side a, and Q be the charge placed at the centre.
`r=(asqrt(2))/(2)=(a)/(sqrt(2))`
`vecF_(BA)=(kq^(2))/(a^(2))hati,vecF_(BC)=(-kq^(2))/(a^(2))hatj`
`vecF_(BD)=(kq^(2))/((asqrt(2))^(2))(cos45^(@)hati-sin45^(@)hatj)`
`vecF_(BQ)=(kQq)/((a//sqrt(2))^(2))(cos45^(@)hati-sin45^(@)hatj)`
Here `hatiandhatj` have usual meaning
Net force on the charge at B is
`vec(F_(R))=((kq^(2))/(a^(2))+(kq^(2))/((asqrt(2))^(2))cos45^(@)+(kQq)/((a//sqrt(2))^(2))cos45^(@))hati`
`-((kq^(2))/(a^(2))+(kq^(2))/((asqrt(2))^(2))sin45^(@)+(kQq)/((a//sqrt(2))^(2))sin45^(@))hatj=F_(x)hati+F_(y)hatj`
For charge, q to be in equilibrium at B, the net force on it must be zero.
`therefore F_(x)=0&F_(y)=0`
`implies k[(q^(2))/(a^(2))+(q^(2))/((asqrt(2))^(2)).(1)/(sqrt(2))+(Qq)/((a//sqrt(2))^(2)).(1)/(sqrt(2))]=0`
`therefore Q=-(q)/(4)(1+2sqrt(2))`
Similarly, `F_(y)=0," if "Q=-(q)/(4)(1+2sqrt(2))`