electric field due to remaining part  =  field due to complete plate - field due to removed disk

i m taking  sigma/epsilon = k

field due to complete plate = k/2

field due to disk = -x/2k(R²+x²)^1/2               

here R is radius of aperture (disk) , x is perpendicular distance ...

field due to remaining = k/2root2 (given) = K/2 + x/2k(R²+x²)^1/2

now after solving we get

R = [2root2 -1 /3-2root2]^1/2 x

 x = a , this is the required ans

this can be solved if we calculate the net dipole moment & then

using formula , E = kP/x³                            (electric field due to dipole at equitorial position)

from figure it is clear that two dipoles are at 60^o & other one is at 120^o to one of these dipoles ......

dipoles are vector so we can apply rule of vector addition ...

resultant of dipoles which are alinged at 60 with each other is (p²+p²+2p²cos60)^1/2 = (root3) p

now this resultant will be perpendicular to the other remaining one dipole ...

net resultant = (P² + (root3p)²)^1/2 = 2p

now , electric field = kP/R² = k2P/R³

this is the required ans

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