Four charges, each equal to `-Q`, are placed at the corners of a square of side a and a charge `+q` is placed at its centre. If the system is in equilibrium, the value of q is ………………

Options

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

Answer

A

Explanation

`OA=OC=r=(a)/(sqrt(2))`
`F_(1)=F_(2)=(kQ^(2))/(a^(2))F_(3)=(kQ^(2))/(2a^(2))F_(4)=(-2kQq)/(a^(2))`,
Resultant of `F_(1) and F_(2)=F=sqrt(2)F_(1)=sqrt(2)(kQ^(2))/(a^(2))`
F and `F_(3)` act along AP `F.=F+F_(3)=sqrt(2)(kQ^(2))/(a^(2))+(kQ^(2))/(2a^(2))=(2kQq)/(a^(2))`
`2q=sqrt(2)Q+(Q)/(2)`
`q=(Q)/(4)+(Q)/(sqrt(2))=(Q)/(4)[1+2sqrt(2)]`