law of conservation of energy.

Fallen Seraph

Condsider the situation:

You have a small alley made entirely out of an insulator, and in it you place a ball with an induced positive charge. At the other end of the alley you have pair of metal plates connected to a high volatage circuit, and the one with the positive charge on it is closest to the alley, the negative one far enough away not to cancel out any effect the charge will have on the ball. You push the ball towards the charged plate (with an insulator) investing your effort in the potential energy the ball is gaining due to its position with relation to the electric charge on the plate, and as you get the ball very close to the plate so that it has a quite reasonable amount of potential energy you turn the circuit off. Now the ball has no potential energy.

Where did it go? Or rather what's wrong with my reasoning?

ApeXaviour

I don't think what you mention could really be described as a circuit. Well from what I'm visualising in my head anyway. The two plates wouldn't hold much charge really. I mean if you think of them as a disjointed parallel plate capacitor. Where charge Q is proportional to capacitance C which in turn is inversely proportional to the distance between the plates d.

But anyway say it did have some charge. As you push the ball towards the positive plate, you'd also be increasing the potential of the plate. This potential energy of the ball you mentioned is actually the potential energy between the ball and the plate. So it could theoretically be used by either, ie if you let go of the ball it would be repelled away. Or in the case you describe; if you turn off the circuit. I think it's important to distinguish how you turn it off. Are you disconnecting it entirely such that it's no longer a circuit? The plates would then retain their charge since they'd be no more than independent charged plates.
If you disconnect the power while retaining a circuit of sorts, well the potential (including the extra potential added from the proximity of the ball) has to go somewhere. And it's going to be dissipated through the circuit either by short circuiting or through resistors.


Well that's my very tired and hungover look on it anyhow. Found myself saying that positive charge repels negative (since edited out), so I've probably missed a detail or two. If nobody else replies then I'll come back when I've a clearer head and have a look over what I said.

Man I should learn latex, equations make things so much easier to explain and less ambiguous

Civilian_Target

Think about it. You're increasing the potential energy of the positively charged ball as you push it towards the plate. On the other side of the energy equation, there's the electrical energy on the plate's circuit, providing this potential energy (this is the potential energy of the battery or mains). You disconnect the mains, removing the potential energy (voltage is a form of potential energy) from the plate, and on the other side of the equation, you lose the potential energy on the charged ball.

Fallen Seraph

Ah, yes. Thank you.