0
$\begingroup$

I've learned that current will flow to bring the potential of both ends of a wire to the same value if there is a difference in potential between them.

Consider a cell of emf, enter image description here, and negligible internal resistence.Then, suppose I connect one end conducting wire of negligible resistence to the positive terminal of the battery while leaving the other end free.This would create a potential difference accross the ends of the wire(initial potential at the ends is equal to zero).

IS THE CURRENT (flow of charge) GOING TO FLOW (with charge entering cell)?


WHAT I THOUGHT WOULD HAPPEN:

My understanding is that the positive terminal will attract electrons, and as electrons move from the wire to the terminal, the potential of that terminal will decrease ( this flow is happening since the terminal is in contact with the wire  and so electron can get on the termminal).

Next, due to the Electromotive force mechanism (which isn't actually a force) in the cell , the electron will then be moved (kind of ) to the negative terminal of the cell, lowering the potential of the negative terminal in order maintain the initial potential at positive terminal.

I'm unsure of what happen next.I believe that until a static condition is reached, this process will continue.

Was my thinking correct?

Would something like this actually occur? If it does, won't disconnecting the wire after sometime change the emf of the cell?


Note--- I believe I must be mistaken after giving it some additional thought, but what is correct in that case?

I simply want to know what is actually going on and what function the cell's EMF and chemical reaction are performing.

I would value a thoughtful response that could help me figure things out.

$\endgroup$
12
  • $\begingroup$ We call electrical networks "circuits" because current flows in circles. Where's the circle here? $\endgroup$
    – John Doty
    Jul 16 at 18:46
  • $\begingroup$ @JohnDoty I mean just flow of electrons , which require only potential difference. $\endgroup$ Jul 16 at 18:51
  • $\begingroup$ @JohnDoty Kind of an irrelevant comment. You can analyse an open loop just aswell as a closed one $\endgroup$ Jul 16 at 18:51
  • 1
    $\begingroup$ In all honestly, I don't really know what happens in batteries when one side is disconnected.I think Normally negative charge on the positive plate is brought to a higher potential energy and then leaves through the negativeplate. In the case one side is connected but charge is still conected on one side, I would guess that because batteries are designed to maintain a constant potential difference, the charge buildup doesn't actually go across into the battery to reach the negative plate. I could be wrong. However what I do know is that the conductor is now in equipotential. $\endgroup$ Jul 16 at 19:24
  • 1
    $\begingroup$ But cannot tell you what happens to the pd across the battery as I don't really know about how batteries create and maintain this potential difference. In eithercase , steady state is that no current flows in the conductor $\endgroup$ Jul 16 at 19:25

2 Answers 2

4
$\begingroup$

Consider a cell of emf $\epsilon$, and negligible internal resistence.Then, suppose I connect one end conducting wire of negligible resistence to the positive terminal of the battery while leaving the other end free.

This sort of circuit may seem difficult to analyze, but it is actually rather easy. What you describe is the following circuit:

Original circuit

This circuit doesn’t look correct because it isn’t a closed loop. The key is to remember that every conductor has a stray (aka parasitic) capacitance. This is the conductor’s capacitance with ground. It is unavoidable and based on the geometry of the conductor.

In many circuits this stray capacitance is so small that we neglect it. However, for “free” conductors it is non-negligible. Adding them in transforms your circuit to this one:

Transformed circuit

Where $C_1$ is the stray capacitance of the negative terminal and $C_2$ is the stray capacitance of the conductor attached to the positive terminal.

Now, the circuit is easy to analyze. It is just a pair of capacitors in series and a battery. The stray capacitances are already very small, and capacitors in series are smaller than the smallest capacitance. So the capacitance is very small, and therefore the charge on the conductor is correspondingly small. But it is not zero.

By the way, the current stops as soon as the capacitors finish charging. Since there is no resistance this happens instantly. Realistically there is some internal resistance and the capacitance with this resistance will form a time constant that will determine how quickly it charges.

$\endgroup$
2
  • $\begingroup$ In an actual lab bench situation, the charge in the capacitors will wind up being of order a trillionth of the charge the battery can deliver. Zero for practical purposes. $\endgroup$
    – John Doty
    Jul 17 at 14:42
  • $\begingroup$ @JohnDoty I agree $\endgroup$
    – Dale
    Jul 17 at 15:08
1
$\begingroup$

Usually by "current", we're talking about large-scale, steady-state bulk flows of charge. These are easy to analyze with concepts like Ohm's law and others.

But transitory charge movement may exist as well. But this will sometimes not be characterized as "current".

You're correct, the non-zero capacitance of the wire means that some amount of charge may need to move to bring the wire and the cell terminal to the same potential. But this will be tiny and is usually ignored.

Remember it's not like either side of the cell has a defined potential. The cell just makes sure they have a difference. If we ground one of the terminals, then the cell will move some (tiny) charge to make the other terminal be the correct offset potential.

Once the other terminal is at the correct offset, the cell won't be able to move any more charge.

$\endgroup$

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.