# I Discharge only one plate of a capacitor?

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1. Aug 30, 2016

### greypilgrim

Hi.

Assume we connect both plates of a capacitor to ground and disconnect again, leaving them at ground potential. Then we bring electrons from one plate to the other, until there is a voltage difference $\Delta U$ between the plates. Compared to ground, one plate has the potential $\frac{1}{2}\Delta U$ and the other $-\frac{1}{2}\Delta U$.

What happens now if we connect one plate to ground with a wire? Common knowledge says nothing happens since there is no closed circuit. But on the other hand, there's a voltage difference of $\pm\frac{1}{2}\Delta U$ across the wire that should cause a current, emptying the grounded plate.

So what's correct?

2. Aug 30, 2016

I think the answer is that it only takes a very infinitesimal amount of charge to change the potential from $+ (1/2) \Delta U$ to zero. Unlike the voltage in a capacitor plate that is developed across the short distance between the plates, (and the electric field is assumed zero everywhere else from the charges $+Q$ and $-Q$), an unbalanced charge will have a longer range (distance) effect for the voltage it creates. The charge that flows between the plate and the ground to balance the voltages (and bring the plate to ground) when it is connected may be measurable, but it is likely to be very small. The charge on the capacitor plates, for all practical purposes, will still be $+Q$ and $-Q$. As an example, a bird can sit on a high voltage wire without getting zapped. The bird lands there and a very small charge will redistribute itself throughout the bird to bring it to nearly the same potential as the high voltage wire. The charge that redistributes itself is quite minute. The bird will only get zapped if it makes contact to ground.

Last edited: Aug 30, 2016
3. Aug 30, 2016

### greypilgrim

But imagine we now connect the second plate to ground as well. In this second step, no current flows between ground and the first plate since they're already at the same potential. This means all charge must have left the first plate already when we connected it to ground in the first step. This charge is just the charge of the electrons we brought from one plate to the other, with enough energy it could be anything, far from infinitesimal.

4. Aug 30, 2016

When the first plate is at ground, the second plate is at $-\Delta U$. When the second plate is grounded,a charge $+Q$ will flow into it, and a charge $-Q$ will flow into the first plate to neutralize everything. Even when the first plate was at ground, it still held a charge of $+Q$. If the second plate is neutralized to zero charge, the first plate, if it doesn't empty itself, will be an unbalanced positive charge. In order for this charge to not create a voltage, it needs to flow into the very large ground.(i.e. electrons (charge of $-Q$) will flow from the ground to neutralize it.)

Last edited: Aug 30, 2016
5. Aug 31, 2016

### greypilgrim

Interesting. I have a question about the fields: When the capacitor is charged and both plates disconnected from the ground, we could find the potential of the second plate by computing the line integral of the electric field from the second plate to the ground, where we should find $-\frac{1}{2}\Delta U$.
If this is true, only an infinitesimal amount of charge is shifted in the whole system, so the electric field should remain practically the same, and also line integrals. But you say
How can this be?

6. Aug 31, 2016

### CWatters

It's the way capacitors work. Capacitors "like" to maintain voltage (Inductors "like" to maintain current). It's almost as if they give voltage (current) inertia.

Equations for a capacitor..
Q=CV
dQ/dt = I = CdV/dt
dV/dt = I/C ...............................................(1)

In Mechanics Newton said
f = ma
a = dv/dt
so
dv/dt = f/m ............................................... (2)

Compare (1) and (2).

So capacitance acts a bit like mass does in mechanics. It slows down the rate of change of voltage which means the plates "like" to stay at the same voltage with respect to each other. The bigger the capacitance the more current you have to pump in to make the voltage change.

The plates start off with potential +0.5ΔU and -0.5ΔU wrt to ground. If you change the voltage of one plate by 0.5ΔU (for example by connecting it to ground) then the other plate will also change by 0.5ΔU to maintain the same dV.

So if you ground the +0.5ΔU plate then it falls by 0.5ΔU. That means the -0.5ΔU will also fall by 0.5ΔU and end up at -ΔU.

7. Sep 1, 2016

### greypilgrim

But what can (physically) change on the other plate? It's disconnected from anything, so its total charge remains the same. It could redistribute, but I can't see how this could affect its potential.

8. Sep 1, 2016

The unconnected plate has a wire attached to it. The charge can very easily redistribute itself to raise the voltage. For a wire, it only tales a very small amount of charge, a very infinitesimal amount, to raise the voltage appreciably. The redistribution can easily affect the voltage. By changing the voltage of the other plate, the electric field from it changes (and reaches the unconnected plate). Thereby the charge on the unconnected plate will necessarily redistribute.

9. Sep 1, 2016

### CWatters

+1
Perhaps visualise your circuit like this...

The main capacitor is shown in bold and the left hand side is the unconnected plate. Charles mentions a wire and I have shown a small stray capacitance between the wire and ground. The voltage on that small capacitor is given by..

ΔV=ΔQ/C

If C is very small or approaches zero then any change in Q produces a large voltage change so the charge doesn't have to redistribute much.

You can argue that there need not be a wire but that just makes the stray capacitor even smaller.