1
$\begingroup$

I have a question regarding the application of Gauss's law to find the E-fields produced by two parallel conducting plates. My textbook (Halliday and Resnick 9e) states that there is no E-field above and below two parallel conducting plates; an E-field only exists between the two plates.

I don't understand this. Wouldn't the charge on the two plates still have an effect on the outside of the plate?

Speaking in terms of Gauss's Law, aren't there two ways to draw the Gaussian pillbox (see options 1 and 2 on the image below). In Option 1, $q_{enc} = 0$, which doesn't take into account the effect of the positive charge on the upper plate. In Option 2, $q_{enc} = \sigma * A$, which does take the positive charge into account and gives a non-zero E-field above the parallel plates.

How do you choose between Option 1 and Option 2 when calculating the E-field?

Two ways to draw a Gaussian pillbox on parallel conducting plates

$\endgroup$
1
$\begingroup$

Conductors rearrange their charge so that the electric potential within the conductor is the same everywhere. This means the E-field within a conductor is always zero (at least in the absence of moving charges).

When you have two (wide) conductors with equal but opposite charge that are close together and parallel to one another as in this problem, the only configuration that yields the correct charge on each conductor and zero E-field inside the conductors is the one you have depicted, with all the charge gathered on surfaces facing the other conductor. You can try different configurations, but this is the only one that will work. The reason there is no E-field above and below the system is these charges shield each other's field perfectly. You can also see this explicitly by applying Gauss' law.

Pillbox #1 correctly indicates that the normal component of the E-field above the top conductor is zero.

Pillbox #2 does enclose a charge of $\sigma A$, however this is not because of a non-zero E-field on the top surface of the pillbox. This E-field is still zero, as indicated by pillbox #1. In this case, there is a downward E-field of $\sigma/\epsilon_0$ between the conductors (and hence on the bottom surface of the pillbox), which explains why there is a non-zero charge enclosed by the pillbox.

$\endgroup$

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.