2
$\begingroup$

Let's say you have a parallel plate capacitor and you connect one plate to the positive terminal of a battery and the other plate to the negative end. So this is like a static situation, you have a large positive charge (the cathode in the battery, which keeps recharging)) connected to a metal plate, and a big negative charge (the anode) connected to another metal plate. From what I've learned about static situations, the charges should equalize so that the plate connected to the positive charge should have half the charge of the cathode, and the same for the negative side. However, my physics teacher said this would not happen and he even did an experiment to show me. However, I don't understand why my reasoning doesn't work. Could someone explain to me why I am wrong? Thanks!

$\endgroup$

2 Answers 2

2
$\begingroup$

The charges move so that the voltages on the connected items equalize. (If they're not equal, the resultant electric field would cause charges to move.) If you were connecting two identical capacitors, the charges would balance as you hypothesized. But the battery is effectively a huge capacitor, so from $Q=CV \, \, (\text{or} \,\, V=Q/C)$, equalizing the voltages requires only a little charge on the capacitor (compared with the stored charge in the battery), because its capacitance is so small compared to the battery's. (Note it's a bit of a cheat to talk about the capacitance of a battery, since it's a non-constant quantity, but it's still a useful concept for situations like this one.)

$\endgroup$
0
0
$\begingroup$

Well, suppose the charges equalize so half the charge of the cathode goes to the plate connected to it. What happens next? The redox reaction inside the battery charges the cathode back up to full value (the value needed for electric field inside the battery to balance the chemical gradient). But it's connected to that metal plate, so they should equalize: 1 + 1/2 = 3/2 = 3/4 + 3/4, so now the metal plate has 3/4 the charge of the cathode.

And so on, and so on...

$\endgroup$

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.