Determining field between oppositely charged conducting plates

[Natha]

Homework Statement

Two large plane parallel conducting plates are given charges of equal magnitude and opposite sign; the surface charge densities are +s and -s. Find the electric field in the region between the plates using Gauss' Law.

Relevant Equations

Gauss' Law ##\phi=\frac{q_{in}}{\epsilon_0}##

I'm using a cylindrical gaussian surface that is right inside the positively charged conducting plate and has the other end in-between both plates. I'm having trouble discerning whether the charge density(##q_{in}##) should be ##\frac{\sigma}{2 \epsilon_0}## since the cylinder is only encompassing one charged surface of the positive plate. However, my textbook (Young physics: Example 22.9 Chapter 22.4 13th edition: pg. 760) has no mention of the 1/2 factor, instead mentioning that ##q_{in}=\sigma A##.

Thank you so much for the help!

 

[haruspex]

There are two ways to look at this.

You can consider the fields the two plates generate independently then add them.
Each generates a field of magnitude ##\frac {\sigma}{2\epsilon_0}## on each side. Between the plates, they add, leading to a magnitude ##\frac {\sigma}{\epsilon_0}##.

The other approach is to recognise that the fields cancel outside the pair of plates, so all the flux is from one plate to the other. A Gaussian box that is half way into one plate and, on its other side, is between the plates, captures all these flux lines. Gauss' law then gives ##\frac {\sigma}{\epsilon_0}##, as before.

 

[SammyS]

There are two ways to look at this.

You can consider the fields the two plates generate independently then add them.
Each generates a field of magnitude ##\frac {\sigma}{2\epsilon_0}## on each side. Between the plates, they add, leading to a magnitude ##\frac {\sigma}{\epsilon_0}##.

The other approach is to recognise that the fields cancel outside the pair of plates, so all the flux is from one plate to the other. A Gaussian box that is half way into one plate and, on its other side, is between the plates, captures all these flux lines. Gauss' law then gives ##\frac {\sigma}{\epsilon_0}##, as before.

Expect push back on this. I was about to post a reply challenging "Each generates a field of magnitude ##\frac {\sigma}{2\epsilon_0}## on each side." but thinking it through, you are correct. You won't get this result using Gauss's Law, but you will get it applying Coulomb's Law and integration.

 

[Natha]

Hi! Thank you all for your reply,

I'm still having trouble seeing how using the below setup (focusing on the Gaussian surface S1) can reveal the total magnitude of the electric force between the plates, ##\frac{\sigma}{\epsilon_0}##. Wouldn't you get the same result ##\frac{\sigma}{\epsilon_0}##, regardless of whether the second plate exists?

Edit: ohh, I just realized something. Is the reason why this works because if you remove the second surface, the positive charges will spread out on both sides of the conductor, thus resulting in charge density inside the Gaussian surface S1 to be ##\frac{\sigma A}{2}##?

 

[TSny]

Edit: ohh, I just realized something. Is the reason why this works because if you remove the second surface, the positive charges will spread out on both sides of the conductor, thus resulting in charge density inside the Gaussian surface S1 to be ##\frac{\sigma A}{2}##?

Yes.

 

[hutchphd]

Bingo! Nice to see a good aha moment.

 

[Natha]

Wonderful! Thank you all so much for your help :)