Field inside a conductor is zero

[madah12]

Homework Statement

I know that the field inside a conductor is zero but what if there was a charged object in the cylinder I mean in some examples I see infinite charged wire inside an infinite hollow conduction cylinder with radius a
but if we take a cylindrical gaussian surface with r <a Qinside = linear density of the wire times l which isn't zero right? so why is the field zero?

Homework Equations

The Attempt at a Solution

 

[madah12]

and if you could help me please do it today because tomorrow is my quiz and I can't continue studying unless I understand this point.

 

[Doc Al]

The electrostatic field within the conducting material is zero. But if you can certainly have a non-zero field within the cavity of a hollow conductor if there's charge inside the cavity.

 

[madah12]

ok another question in this case of a wire inside a hollow conduction cylinder is E at the surface of the cylinder equal only to sigma/epsilon naught or is it (lamba * l + sigma *A)=E*A? because the book says the field on the surface of the cylinder is sigma/epsilon

 

[Doc Al]

ok another question in this case of a wire inside a hollow conduction cylinder is E at the surface of the cylinder equal only to sigma/epsilon naught or is it (lamba * l + sigma *A)=E*A? because the book says the field on the surface of the cylinder is sigma/epsilon

What does sigma represent? The surface charge on the outside of the cylinder? If so, then sigma/epsilon is correct.

In applying Gauss's law, don't forget to include all charge, including that on the inner surface of the cylinder.

 

[madah12]

yes sigma is the outer surface density ,so I understand so you are saying that because on the inner surface there is q induced that is equal and opposite to the one of the wire?
so even outside the surface of the cylinder Q only equals sigma outer *outer surface area?

 

[Doc Al]

yes sigma is the outer surface density ,so I understand so you are saying that because on the inner surface there is q induced that is equal and opposite to the one of the wire?
so even outside the surface of the cylinder Q only equals sigma outer *outer surface area?

That's right. And if the cylinder has no net charge of its own, the field outside its surface will equal that of the line charge.

 

[madah12]

http://physics.kuniv.edu.kw/phys102/08-09-S-ms1.pdf
look at number 5 how can the field be zero if Q=sigma *a

 

[Doc Al]

http://physics.kuniv.edu.kw/phys102/08-09-S-ms1.pdf
look at number 5 how can the field be zero if Q=sigma *a

In that problem statement, sigma represents the total charge per unit area on the cylinder, not just the charge on its outer surface. (The problem is confusingly worded.) Think of sigma as the charge on the outer surface before the line charge is introduced. Once you include the induced charge due to the line charge, then the outer surface will have zero sigma.

 

[madah12]

so sometimes they say the linear charge of a cylinder do you know what that means?

 

[Doc Al]

so sometimes they say the linear charge of a cylinder do you know what that means?

I imagine that means the total charge per unit length. (As opposed to the surface charge, which is per unit area.)