Correct Answer - `W=((sigma_(1)-sigma_(2))qa)/(sqrt(2) in_(0))`
Electric field near a large metallic plate is given by `E=sigma//epsilon_(0)`. In between the plates the two field will be in opposite direction. Hence,
`E_("net")=(sigma_(1)-sigma_(2))/epsilon_(0)=E_(0)` (say)
Now, `W=(q)` (potential difference)`=q(E_(0)a cos 45^(@))`
`(q)((sigma_(1)-sigma_(2))/epsilon_(0))(a/sqrt(2))=((sigma_(1)-sigma_(2))qa)/(sqrt(2)epsilon_(0))`