Four charges equal to `-Q` are placed at the four corners fo a square and a charge `q` is at its centre, If the system is in equilibrium the value of `q` si

Options

A. `(-Q)/(4)(1+2sqrt2)`
B. `(Q)/(4)(1+2sqrt2)`
C. `(-Q)/(2)(1+2sqrt2)`
D. `(Q)/(2)(1+2sqrt2)`

Answer

B

Explanation

The system is in equilibrium which means that the force experienced by each charge is zero.

`F_(CD)+F_(CA)cos 45^(@)+F_(CO)cos 45^(@)=0`
`rArr" "(1)/(4piepsilon_(0)).((-Q)(-Q))/(a^(2))+(1)/(4piepsilon_(0))((-Q)(-Q))/((sqrt2a)^(2))xx(1)/(sqrt2)+(1)/(4piepsilon_(0))((-Q)q)/((sqrt2a//2)^(2))xx(1)/(sqrt2)=0`
`rArr Q+(Q)/(2sqrt2)-sqrt2q=0 rArr 2sqrt2Q+Q-4q=0`
`q=(Q)/(4)(2sqrt2+1)`