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Grade: 11
        
A disc of radius a/4 having a uniformly distributed charge 6C is placed in the xy plane with centre at (-a / 2, 0, 0 ) A rod of length a carrying a uniformly distributed charge 8c is placed on the x-axis from x = a / 4 to x =5a / 42 . Two point Charges -7 c & 3c are placed at (a / 4, -a/ 4 , 0) and ( - 3a / 4, 3a / 4, 0 ) respectively consider a cubical surface formed by 6 surface x=+- a / 2, y=+-a / 2, z =+- a / 2 . The electric flux through the cubicle surface is 
(a) -2C/E?
Please explain in detail
4 years ago

Answers : (1)

Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
							Hi Ashu,
flux=charge enclosed/epsilon
Here if we try to get the charge enclosed then
3C will not be enclosed
Half of the disc will be in the cube
so


Thanks & Regards
Arun Kumar
IIT Delhi
Askiitians Faculty
4 years ago
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