0
$\begingroup$

I am having trouble understanding Gauss's Law. Suppose we wish to find the electric field strength between two parallel plates with charge density $\sigma$. I know it should be $E=\sigma/(2\epsilon_0)$.

But if we place a cylinder between the plates with base parallel to the plates such that the cylinder does not intersect the plates, then the charge enclosed is $0$ and the flux is $2EA$ where $A$ is the area of the base of the cylinder. Thus by Gauss's Law, $2EA=0$, which gives $E=0$. What is the error in my understanding?

$\endgroup$
3
$\begingroup$

Your equation is wrong. The flux is not $2EA$. Rather, is $-EA + EA = 0$. The reason is because Gauss's law involves the dot product of $E$ and $dA$. The direction of $dA$ is the way out of the cylinder. When the electric field goes through the bottom of the cylinder, the electric field is in the opposite direction of the area vector, so the electric flux is $-EA$. When the electric field vector exits the cylinder, it is in the same direction as the area vector, so the electric flux is $EA$. Hence, the net electric flux is $-EA + EA = 0$, not $2EA$. Hence, $0 = 0$, not $2EA = 0$. To solve for the electric field, take a cylinder which goes through one plate. Then, the electric field goes outside the cylinder through the inner base, so the electric field is $EA$. The charge enclosed is $\sigma A$, so $EA = \sigma A/\epsilon_0 $, and thus $E = \sigma /\epsilon_0$.

$\endgroup$
0
$\begingroup$

For a Gaussian surface between the two plates, the total flux through the surface is zero. For the particular surface you give, all of the electric field lines crossing one of the circular faces cross the other face but in an opposite sense.

This is because the outward normal vector for one face of the cylinder is opposite in direction to the outward normal for the other face.

In other words, there is a sign difference; there is $EA$ flux through one face and $-EA$ flux through the other so the total flux is zero.


As an aside, if the (infinite) parallel plates have the same charge density $\sigma$, the electric field between the plates is actually zero.

If one plate has charge density $\sigma$ and the other has charge density $-\sigma$, then the electric field between the plates is $E = \frac{\sigma}{\epsilon_0}$, not $\frac{\sigma}{2\epsilon_0}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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