0
$\begingroup$

Say I have an infinitely long hollow cylinder, and on its surface I have a uniformly distributed surface charge. Now, I want to calculate the electric field of that cylinder both inside and outside. The solution in my textbook creates a Gaußian cylinder inside my cylinder of interest and then simply states that there are no charges inside, and thus the electric field is zero.

Now here is where I am confused. It doesn’t make sense that I can take any volume with no charge inside and then conclude that the electric field at that point is zero. As far as I understand Gauß's law, it only tells me the electric field coming from whatever is inside this volume. So what would be the correct way to calculate the field within my cylinder?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

It doesn’t make sense that I take any volume with no charge inside and then conclude that the electric field at that point is zero.

You are right, in general this is not true. For example, think of a Gaussian surface around a volume by, but not surrounding, a single point charge. There is no charge in the surface, but the field at any point on that surface is not $0$.

In general, Gauss's law in integral form doesn't give you the electric field. You can, however, exploit symmetries in certain geometries (points, spheres, lines, cylinders, planes) to determine the field. If you can't exploit some symmetry to pull E out of your flux integral, then you can't use Gauss's law to determine the field.

It's like saying "I have ten numbers that add to 100." If this is all you know then you can't tell me which numbers I have used to get a total of 100. But if I also tell you "each number is the same number", then you can for sure say I used ten 10s to get a total of 100.

In your case, the symmetry lets us know that $E$ is the same magnitude at all points on the surface and would have to have the same radial direction if there was a field. These two facts let us conclude that then $E\propto Q_\text{enclosed}=0$

$\endgroup$
3
  • $\begingroup$ okay that makes sense thank you. So it does indeed work generally? Consider another cylinder around the first one that also has uniformely distibuted charges on its surface. Now the electric field between these two cylinders is only dependent on the inner one, no matter how many charges I put onto the outer one. Here's where my intuition says goodnight ^^ $\endgroup$
    – Axodarap
    Commented Jun 5, 2023 at 7:28
  • $\begingroup$ @Axodarap When the symmetry is right, and paired with the inverse square law, the field inside does end up going to $0$; it is the perfect balance of "less" stronger fields (from closer charges) and "more" weaker fields (from charges farther away). Draw vectors from an interior point in question to points on the outer cylinder and maybe you can see how it comes together. $\endgroup$ Commented Jun 5, 2023 at 13:36
  • $\begingroup$ ah my bad, it's of course the same principle as within the first cylinder. Thanks! $\endgroup$
    – Axodarap
    Commented Jun 6, 2023 at 15:08
1
$\begingroup$

For symmetric arrangements like hollow sphere or infinitely long hollow cylinder opposite fields tends to cancel out. Gauss's law tells us more about the flux through a surface. For e.g. if we take a spherical concentric gaussian surface inside a hollow sphere(with no thickness) with charge distributed on surface of hollow sphere,then by gauss's law:- $${\subset\!\supset}\llap{\iint}\left(\vec{E}\cdot \hat{n}\right)da=\frac{Q_{enc}}{\epsilon_0}=0$$ It tells us that flux through the inner surface is zero. And the symmetry of the given shape also validates this, because due to symmetry, field inside the hollow cylinder becomes zero.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.