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sohan singh singh Grade: 12
        

http://www.goiit.com/upload/2010/7/20/ba6c2b3e37a0223d86bc3ca07df884db_1439300.jpg

6 years ago

Answers : (3)

Chetan Mandayam Nayakar
312 Points
										

Just as there is a shell theorem for spherical shells, there is an analogous "ring theorem" for circular disks.


Answer= ∫ρr2l/(4*epsilon-nought*(l2+r2)3/2) dl = ρr/(4*epsilon-nought)


 

6 years ago
Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
										

Dear Sohan,


Ans:- You have not said anything about the type of distribution i.e whether is it linear, surface or vol density, for simplicity let's take it linear density


now at any point inside or the cylinder the intensity is 0.


So, let's consider that point at a infinitesimally small distance outside from the surface.


Let E is the field due to an infinite cylinder. Then from gauss's law we get,


EƒdA=ρl/€


or E 2∏rl=ρl/€


or E=ρ/2∏r€


But the given cylinder was not infinite rather it was semi infinite. hence the required field is half the calculated field


Hence  E=ρ/4∏€r (ans) 


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All the best Sohan!!!


 



Regards,


Askiitians Experts


SOUMYAJIT IIT_KHARAGPUR

6 years ago
sohan singh singh
38 Points
										

SIR it is volume density.Sir please sir help me again.

6 years ago
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