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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 60o & other one is at 120o 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
i have solved some of ur previously asked questions , give ur responce to them also...
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