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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,
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)
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Sohan!!!

Regards,
Askiitians Experts
SOUMYAJIT IIT_KHARAGPUR
```
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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