Gauss' Law: Charge enclosed zero = E field zero?

AI Thread Summary
In the discussion about Gauss' Law, a cylindrical shell with a uniform charge distribution is analyzed to determine the electric field at a point inside the shell. The chosen Gaussian surface, which is inside the shell, encloses no charge, leading to zero electric flux and enclosed charge. However, it is clarified that a zero flux does not automatically imply a zero electric field; the symmetry of the charge distribution must also be considered. The participants conclude that due to the symmetry of the cylindrical shell, the electric field at any point inside the shell is indeed zero. This reinforces the importance of symmetry in applying Gauss' Law effectively.
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Homework Statement


A cylindrical shell of radius R and length H has its charge uniformly distributed on its curved surface

Find the electric field at a point P from the axis, a distance r away, measured radially outward from the midpoint of the shell such that R>r

Homework Equations



φ = ∫E⋅dA = Qenc / ∈o

The Attempt at a Solution



I constructed a gaussian surface (cylinder) inside the larger cylinder of radius R.

I realize, my chosen gaussian surface encloses no charge ∴ φ & Qenc are both zero.

I have been told I am not allowed to simply jump and say the E field must also then be zero if the flux is zero, but for this case, since the charge on the outer cylinder is uniformly distributed, does that also mean the net electric field at point P must be zero?

If not, please shed some light on this!

Thank you
 
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AKJ1 said:

Homework Statement


A cylindrical shell of radius R and length H has its charge uniformly distributed on its curved surface

Find the electric field at a point P from the axis, a distance r away, measured radially outward from the midpoint of the shell such that R>r

Homework Equations



φ = ∫E⋅dA = Qenc / ∈o

The Attempt at a Solution



I constructed a gaussian surface (cylinder) inside the larger cylinder of radius R.

I realize, my chosen gaussian surface encloses no charge ∴ φ & Qenc are both zero.

I have been told I am not allowed to simply jump and say the E field must also then be zero if the flux is zero, but for this case, since the charge on the outer cylinder is uniformly distributed, does that also mean the net electric field at point P must be zero?

If not, please shed some light on this!

Thank you
Is there sufficient symmetry to conclude that the field is zero everywhere inside to charge distribution?
 
SammyS said:
Is there sufficient symmetry to conclude that the field is zero everywhere inside to charge distribution?

Yes.

I think that answers my question, so then, we should be examining at the symmetry of the chosen surface with respect to the original surface.
 
AKJ1 said:
Yes.

I think that answers my question, so then, we should be examining at the symmetry of the chosen surface with respect to the original surface.
It also has a lot to do with what symmetry implies regarding the electric field and whether you an find a surface so the field is normal to the surface over regions where the field's magnitude is constant and in all other regions the field is parallel to the surface.
 
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