# MIT problem set problem, I have problems with the solutions

## Homework Statement

Wow that was three problems in my title. Anyways please go to this link

http://stuff.mit.edu/afs/athena/course/8/8.02-esg/Spring03/www/8.02pset2sol3.pdf

and go to problem 4: 24.28

I have a question to the solutions

a) They constructed a bigger cylinder for their Gaussian surface. Now my question is, is the thin cylindrical shell itself a charge and that is why we must construct our own surface?

What happens if the $$\vec{E}$$ field is 36kN/C at 7.00cm?

b) Why is it 0? Why can't you just replace 19cm with 4.00cm?

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a) They constructed a bigger cylinder for their Gaussian surface. Now my question is, is the thin cylindrical shell itself a charge and that is why we must construct our own surface?
I cant understand what are you asking

What happens if the $$\vec{E}$$ field is 36kN/C at 7.00cm?
It depends on the side of surface. Inside the material its 0, just outside ... follow formula used in (a), inside its 0 again

b) Why is it 0? Why can't you just replace 19cm with 4.00cm?
Because according to gauss law, flux or even E is directly disproportional to charge in-closed by the surface.

I cant understand what are you asking
Does it matter if it is even a thin cylindrical shell? Can it be a solid?

It depends on the side of surface. Inside the material its 0, just outside ... follow formula used in (a), inside its 0 again
Yes, I plug in 4, I get a number

Because according to gauss law, flux or even E is directly disproportional to charge in-closed by the surface.
But it doesn't mean it can't right?

Does it matter if it is even a thin cylindrical shell? Can it be a solid?
No, even if its solid, you need to use a Gaussian surface ... Gaussian surface i just a hypothetical surface ... it may even coincide with your cylinder's outer surface!!!

Yes, I plug in 4, I get a number
You can use a formula made for 1 thing for something else but it will give a wrong result. same is happening with you!!

But it doesn't mean it can't right?
No there cant be any field .... you know that charges on conductor resides on surfaces ...most probably on outer to reduce its potential energy
so draw the figure of cylindrical shell we're talking about ... draw charge on outer surface ... and you see there is no charge inside to provide electric field!!! the outer charge provides field to outer area. This concept in electrostatic shielding

$$(3.60x10^4) = \frac{2(8.99x10^9)\frac{Q}{2.4}}{4.00cm}$$

I don't understand

Wait, is it because the radial E-field outside the surface of the shell?

Look at the pic
If charge is + or -, E will be outside the area,
you think there will be field inside too. can you tell me why?

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