Four equal charges `Q` are placed at the four corners of a square of each side is 'a'. Work done in removing a charge `-Q` from its centre to infinity is

Options

A. `0`
B. `(sqrt(2) Q^(2))/(4 pi epsilon_(0) a)`
C. `(sqrt(2) Q^(2))/( pi epsilon_(0) a)`
D. `(Q^(2))/(2 pi epsilon_(0) a)`

Answer

2

Explanation

`U_(1) = (1)/(4 pi in_(0)) .(Q xx -Q)/(a//sqrt(2)) , U_(2) = 0`
`W_(1 rarr 2) = U_(2) - U_(1) = (sqrt(2) Q^(2)) /(4 pi in_(0)) a`