# Does a capacitor experience a Newtonian reaction when being charged?

Say I have a capacitor that is being charged with a battery. There are two wires, one on the positive and one on negative terminal of the battery which go to a capacitor. During the charging, a positive charge accumulates on one plate and a negative charge forms on the other plate. But electrons can not pass through the capacitor physically, so I assume that the electrons accumulate on one plate (positive charge), but the other plate (negative charge) has less or no electrons because they are drawn towards the battery via the wire.

Is this assumption correct? Are both wires carrying electrons, one towards and one away from the battery while the capacitor is being charged, thus making the net force after some time on the capacitor and battery system zero? Was there an initial force on the capacitor and battery system when the first wire had moving electrons away from the battery but the other wire had no electrons yet moving towards the battery? Could I say that the force that the moving current exerts on a closed circuit is asymmetrical at first for a very short time (pushing the circuit to the left) until the current reaches the other battery terminal and then the force of the current becomes symmetrical through the circuit (circuit pushed to the right and thus ends up in the same starting spot)?

## 1 Answer

The net effect of the charging process is the movement of electron from one plate which then has a net positive charge to the other plate which then has a net negative charge. The battery facilitates this by creating an electric field in the wires and it is this electric field which applies forces on the electrons which makes them move. The movement of electrons continues until the charges on the two plates set up an electric field in opposition to that produced by the battery.

In terms of momentum if the circuit has no external forces acting its centre of mass will not move so as the electrons move towards one plate the rest of the circuit will move the other by an amount which is not detectable because the mass of all the electrons which moved is very small compared with the mass of the rest of the circuit.

• Thank you. When you say "as the electrons move towards one plate the rest of the circuit will move the other by an amount" that is the reaction the circuit experiences from the electron flow. But, in one terminal wire the electrons move in one direction, and in the other terminal wire the electrons move in the opposite direction, thus making the net force on the system zero during the whole charging cycle? – Arundel Feb 9 '16 at 17:05
• I agree with you for the whole system but II was comparing the motion of one plate with the motion of the rest of the circuit. – Farcher Feb 9 '16 at 20:56