0
$\begingroup$

When you attach a capacitor to a battery via two wires, charge transfer occurs from one of the plates to the other. However, in an open circuit, there is infinite resistance in the dead-ended wire, and no current flows through. I’m sure I’ve got a misconception here somewhere, because isn’t the situation with the capacitor like an open circuit? If there’s no current, why does the capacitor become charged?

$\endgroup$
2
1
$\begingroup$

(1) The voltage across a capacitor is proportional to the charge $Q$ (where it is understood that there is $+Q$ charge on one plate and $-Q$ charge on the other plate).

(2) The voltage across the capacitor is initially zero so the $Q$ is initially zero.

(3) The voltage across the battery is not zero so, if the battery and capacitor are connected together, there must be a flow of charge (a current) until the capacitor and battery voltage are equal, i.e., until there is equal and opposite charge $Q$ on the plates such that

$$Q = CV_{BAT}$$

Your misconception is that the capacitor is an open circuit but it isn't. The capacitor has finite impedance (except at zero frequency) and thus, there can be a time varying current through associated with a time varying voltage across. Only for steady voltage across is there zero current through.

$\endgroup$
1
  • $\begingroup$ That's the reward:) $\endgroup$
    – user36790
    Nov 3 '15 at 11:27
0
$\begingroup$

Electrons repel each other, which means they stay distributed around a circuit (unless pushed hard, e.g. by a large voltage). However, capacitors are a counterexample, because they have, well, capacitance. In capacitors, the negative charge of the electrons one one plate is partially cancelled by the positive charge of the electron vacancies on the adjacent plate. This means it takes much less voltage to accumulate charge in a capacitor; the larger the capacitance, the less voltage it takes per unit charge.

Your open circuit also has capacitance between the two dead-end wires, but only a very tiny amount. When you connected your battery, charge did flow, but only a very tiny amount accumulated on the ends of the wires before the resulting voltage equalled the battery voltage. Closing the circuit with a real capacitor greatly increases the charge that flows before the capacitor's voltage equals the battery.

$\endgroup$
0
$\begingroup$

Yes, there will be current because charge will flow towards the capacitor plates. But only for a while, and then it stops.

Think of it like this.

  • When the battery is initially attached, the charges do not "know" that there is a hole in the circuit. They just want to move, because they are repelled from the battery pole and now they have a free uncharged (unrepelling) conducting path to move along. So they do. They move away from the battery pole.

  • When the charges reach the capacitor plate they stop. After some time, a lot of charge has reached this plate and accumulated here. Since like charges repell each other, this bunch of charge accumulated at the plate starts to repel further charge that comes towards it. So the current is resisted and thus reduced.

  • Sooner or later there will be exactly the amount of charge gathered at the plate that corresponds to an electric potential equal to the battery pole's potential. This means that the battery's "push" on the charge in the wire equals the "push" back from the plate. Nothing will move now, and the current stops.

Bottomline is that there will be current, but not a steady current. A current will start as if the circuit was closed and then it will decrease until it is zero, which corresponds to the open circuit that you correctly talk about.

$\endgroup$

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.