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sohan singh singh Grade: 12

7 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)


7 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,


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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Askiitians Experts


7 years ago
sohan singh singh
38 Points

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

7 years ago
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