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

Options

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

Answer

B

Explanation

Consider the equilibrium of charge -Q at A.
`F_(B)=(1)/(4pivarepsilon_(0))(Q_(2))/(a^(2))`
`F_(D)=(1)/(4pivarepsilon_(0))(Q_(2))/(a^(2))`
`F_(BD)=sqrt(2)F_(B)=(1)/(4pivarepsilon_(0))(sqrt(2)Q^(2))/(a^(2))`
`F_(C)=(1)/(4pivarepsilon_(0))(Q^(2))/((sqrt2a)^(2))=(1)/(4pivarepsilon_(0))(Q^(2))/(2a^(2))`
`F_(O)=(1)/(4pivarepsilon_(0))(qQ)/(((sqrt2a)/(2))^(2))=(1)/(4pivarepsilon_(0))(2qQ)/(a^(2))`
For equilibrium, `F_(BD)+F_(C)=F_(O)`
`(1)/(4pivarepsilon_(0))(sqrt(2)Q^(2))/(a^(2))+(1)/(4pivarepsilon_(0))(Q^(2))/(2a^(2))=(1)/(4pivarepsilon_(0))(2qQ)/(a^(2))`
or `sqrt(2)Q+(Q)/(2)=2q`
or `2q=(Q)/(2)(2sqrt(2)+1)`
or `q=(Q)/(4)(1+2sqrt(2))`