Capacitor connected to voltage source, vary plate separation

[greypilgrim]

Hi.

A plate capacitor is connected to a constant voltage source. The stored energy is
$$W=\frac{1}{2}\cdot C\cdot U^2=\frac{1}{2}\cdot \varepsilon_0\frac{A}{d}\cdot U^2\propto\frac{1}{d}$$
if the voltage source remains connected when varying $d$.

So the energy decreases with increasing $d$. This is a bit counterintuitive to me. If the plates are allowed to move freely, they should attract because they are oppositely charged. On the other hand, systems strive for lower energy states, which means they should repel.

My guess is that I have to somehow include the currents to or from the voltage source while $d$ changes (since $Q$ is not constant), but I'm not sure how to do so.

 

[Delta2]

What is happening in real world scenario where the source and the conducting paths have ohmic resistance is that the energy is lost as heat in these ohmic resistance as the charge flows from one plate to the other through the source.

When we increase the distance d the energy stored in the capacitor becomes that ohmic heat.

When we decrease the distance d the source provides the energy both for the increase in energy in capacitor and for the energy lost as heat.

I don't know how one can explain it in case of ideal source and ideal conducting paths. But I know that there are many little paradox hidden in physics when we take ideal cases.

 

[vanhees71]

In this scenario you keep the voltage constant. The energy you describe is the energy of the electric field between the plates. When making $d$ larger you have to do work against the attracting force between the plates, and keeping $U$ constant means that energy is stored in the electric field between the plates.

 

[Delta2]

In this scenario you keep the voltage constant. The energy you describe is the energy of the electric field between the plates. When making $d$ larger you have to do work against the attracting force between the plates, and keeping $U$ constant means that energy is stored in the electric field between the plates.

I thought of that too but since we provide work, the energy in the capacitor should increase not decrease.

 

[Delta2]

Oh wait, don't tell me the energy in the capacitor is in fact negative number, since the plates are charged opposite, so when it decreases (in absolute value as d increases) in fact it is increasing...

 

[Steve4Physics]

So the energy decreases with increasing $d$. This is a bit counterintuitive to me. If the plates are allowed to move freely, they should attract because they are oppositely charged. On the other hand, systems strive for lower energy states, which means they should repel.

For objects which repel, a 'lower energy state' requires them to be further apart.

For objects which attract, a 'lower energy state' requires them to be closer.

And don't forget the capacitor is not an isolated system - energy (and charge) flow between the capacitor and the supply when $d$ changes.

My guess is that I have to somehow include the currents to or from the voltage source while $d$ changes (since $Q$ is not constant), but I'm not sure how to do so.

The plates are attracted. To increase the separation (I'll use '$x$' rather than '$d$') the plates must be pulled apart.

Say one plate is fixed and a net outwards force, $F$, is applied to the other plate to maintain equilibrium.

Now suppose $F$ is (very slowly) increased so that $x$increases. Note that $F$ will decrease as $x$ increases. The work done by $F$ is $\int_{x_1}^{x_2} F(x)dx$.

$C, Q$ (charge on capacitor plates) and $E$ (electrical potential energy stored in capacitor) will decrease.

Charge is being ‘pushed back’ into the supply (like charging a battery). So the supply’s energy increases.

 

[vanhees71]

I thought of that too but since we provide work, the energy in the capacitor should increase not decrease.

The field energy changes by
$$\mathrm{d} W=\frac{U^2}{2} \mathrm{d} C=-\frac{\epsilon_0 A U^2 }{2d^2 }\mathrm{d} d.$$
For the charge transport you need the energy (to be brought up from the battery)
$$\mathrm{d} W_Q=U \mathrm{d} Q=U^2 \mathrm{d} C=2 \mathrm{d} W<0,$$
i.e., in fact the battery takes up energy, which has to be put in as work energy when increasing the distance of the capacitor plates, i.e., the work is used to charge the battery.

 

[Delta2]

Charge is being ‘pushed back’ into the supply (like charging a battery). So the supply’s energy increases.

This sounds like a very good explanation to me.

BUT

The increase in the energy of the supply comes from the energy stored in capacitor (which decreases), the work done by the external force that separates the plates or both?

 

[Steve4Physics]

The increase in the energy of the supply comes from the energy stored in capacitor (which decreases), the work done by the external force that separates the plates or both?

That's a good question. I think the answer is 'both'. I'm assuming ohmic heating and EM radiation losses (due to accelerating charges) can be neglected here - but am happy to be corrected!

 

[jeffinbath]

Hi.

A plate capacitor is connected to a constant voltage source. The stored energy is
$$W=\frac{1}{2}\cdot C\cdot U^2=\frac{1}{2}\cdot \varepsilon_0\frac{A}{d}\cdot U^2\propto\frac{1}{d}$$
if the voltage source remains connected when varying $d$.

So the energy decreases with increasing $d$. This is a bit counterintuitive to me. If the plates are allowed to move freely, they should attract because they are oppositely charged. On the other hand, systems strive for lower energy states, which means they should repel.

My guess is that I have to somehow include the currents to or from the voltage source while $d$ changes (since $Q$ is not constant), but I'm not sure how to do so.

I think you are not giving enough thought about the energy or work done to produce the voltage that you are applying to the plates in the first place. It is that energy that is being used to attract the capacitor plates to each other. It is best just to go back to first principles by considering the 18th century electrophorus device. Here is a section taken from Wikipedia after entering Electrophorus in the search panel.

It fascinated people at the time because it seemed as though one could pick up a charge onto the upper plate and then make that plate do work by discharging it elsewhere and then replace it onto the resin (i.e. the dielectric) to recharge it and keep repeating the trick “endlessly” in the short term at least. What is actually going on is that the polarized resin is acting as an electret which is the electrostatic equivalent of a permanent magnet. Each time the upper electrode is placed on it there is an attractive force acting on it which requires energy from somebody’s arm muscles to lift it off . Also the device had to have arm energy used to transfer electrons from a piece of fur onto the device in the first place.