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Grade: 12
        

454_15271_EEYB11_MAT113_E1.jpg

8 years ago

Answers : (4)

vikas askiitian expert
509 Points
							

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

 

8 years ago
vikas askiitian expert
509 Points
							

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

8 years ago
vikas askiitian expert
509 Points
							

i have solved some of ur previously asked questions , give ur responce to them also...

8 years ago
SAGAR SINGH - IIT DELHI
879 Points
							

Dear student,

I will give u the tip applicable in all:

 

8 years ago
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