Confused by capacitors and potential energy

[TheCelt]

HiI was reading about capacitors and potential energy. But the equation seems counter to how i thought.

For potential energy you have:
U = Q^2 / 2C
or
U = CV^2 / 2

But doesn't this suggest you lose potential energy the more capacitance you have? Since in the first equation as C increases U decreases.

In the second equation, a dielectric surely reduces V between two plates as it becomes neutral in between them ? So you also lose U for the second equation as V decreases...

This is confusing me, isn't this contradicting the idea of potential energy here, if you have more charge storage, surely the potential should increase not decrease?

Hope you can clarify where i am confused here so i can understand better. Thanks.

 

[sophiecentaur]

It takes less energy to charge a large capacitor to a low voltage than to push the same charge into a smaller capacitor. You can't equate Charge with Energy,

 

[FactChecker]

But doesn't this suggest you lose potential energy the more capacitance you have? Since in the first equation as C increases U decreases.

For a fixed $Q$.

In the second equation, a dielectric surely reduces V between two plates as it becomes neutral in between them ? So you also lose U for the second equation as V decreases...

In this equation, it is hard to see how what it means to hold $Q$ fixed. Since $Q = C*V$, holding $Q$ fixed means that $V$ must decrease as $C$ increases. That would make $V^2$ decrease more than $C$ increases. The net result is that $U$ decreases here also.

 

[sophiecentaur]

In this equation, it is hard to see how to hold QQQ fixed. Since Q=C∗VQ=C∗VQ = C*V, holding QQQ fixed means that VVV must decrease as CCC increases. That would make V2V2V^2 decrease more than CCC increases. The net result is that UUU decreases here also.

You are just pointing out that some experiments are harder to carry out than others. It is not hard to put a measured amount of charge into a capacitor.

 

[FactChecker]

You are just pointing out that some experiments are harder to carry out than others. It is not hard to put a measured amount of charge into a capacitor.

I need to clarify my post. I only meant that the second equation hides what it means to hold $Q$ fixed, not that there is any experiment that is more difficult. I will edit it.

 

[PeroK]

This is confusing me, isn't this contradicting the idea of potential energy here, if you have more charge storage, surely the potential should increase not decrease?

Hope you can clarify where i am confused here so i can understand better. Thanks.

If you take an example of a parallel plate capacitor, with plates of area $A$ and a separation distance $d$. One plate is positively charged and the other negatively charged.

Capacitance is defined as charge/voltage. High capacitance means a lot of charge for a given voltage. In the example, we get:

$C = \frac{A \epsilon_0}{d}$

The only variable is $d$. If we keep the plates close together, the system has a low potential energy. You could imagine creating the system by separating positive and negative changes - and having to do work to achieve that. In this case you have stored a certain charge with little work and that equates to high capacitance.

If you move the plates further apart, then you have to do more work and the potential energy of the system increases. And, the capacitance reduces.

You can see from this example that capacitance (as defined in fact) is a measure of how much charge you can store in a system and not how much potential energy a system has.

 

[particlezoo]

HiI was reading about capacitors and potential energy. But the equation seems counter to how i thought.

For potential energy you have:
U = Q^2 / 2C
or
U = CV^2 / 2

But doesn't this suggest you lose potential energy the more capacitance you have? Since in the first equation as C increases U decreases.

In the second equation, a dielectric surely reduces V between two plates as it becomes neutral in between them ? So you also lose U for the second equation as V decreases...

This is confusing me, isn't this contradicting the idea of potential energy here, if you have more charge storage, surely the potential should increase not decrease?

Hope you can clarify where i am confused here so i can understand better. Thanks.

It may be more instructive to think about multiple capacitors in series vs. parallel before modifying an individual capacitor by itself.

When you have capacitors in parallel, they have the same voltage V. So using U = CV^2 / 2 makes sense for capacitors in parallel. Also, capacitors in parallel have their capacitance added linearly. You could do a sigma summation of these if you wanted to know the total energy stored in them. In this case, the parallel capacitor with the highest capacitance value would have the highest amount of stored energy where C_1+C_2+C_3+... = C = total parallel capacitance

When you have capacitors in series, the charge added to one capacitor is the same as the charge Q added to another capacitor in series with it. Warning: That does not mean they started off with the same charge nor does it mean that they will have the same charge. Nevertheless, we can imagine what happens when, starting with no charge on either, we decide to energize these capacitors. So using U = Q^2 / 2C, we can see that the series capacitor with the least capacitance will actually store the most energy. You could do a sigma summation of these if you wanted to know the total energy stored in them where 1 / (1/C_1+1/C_2+1/C_3+...) = C = total series capacitance

If you try to think of adding a dielectric between the plates of a capacitor, remember:
1) The case of adding a capacitor in parallel is analogous to adding a dielectric between the plates while applying an external voltage. If you have a constant applied voltage when adding a dielectric, this would increase the charge stored on the plates. The additional energy stored in this case is provided from the power source responsible for the applied voltage.
2) There is no corresponding analogy if the capacitor is disconnected. Then Q on the plates will be fixed, and some of the energy stored on the plates would be used in doing work on the dielectric. In that case, the charges of the dielectric cancel all but Q/ε_r of the charge on the plates, where ε_r is the dielectric constant (relative permittivity); this way it is as if one were to "partially short" the plates of the capacitor. That is why its energy value drops.

Also, if you try to think of changing the distance between a capacitor's plates, remember:
1) If you apply a fixed voltage externally while you change the plate distance, Q will change, which in this case makes it as if you added or removed capacitors in series.
2) If the capacitor is disconnected while you change the plate distance, Q will be fixed, then its V will change while positive or negative work will be done by the charges on the plates.

 

[sophiecentaur]

It needs to be pointed out that any experiment in which anyone of V,Q OR C is changed will involve putting energy into or taking it out of the system. This added or subtracted energy accounts for any difference that people may be attributing to some suspected paradoxical behaviour. Discussions about more and more complicated models tend to confuse the issue, rather than resolving problems. There have been millions of experiments which confirm the basic equations and, luckily, the errors in many real experiments are small enough for confidence.
Re-arranging the components and values, countless times in discussions is really not necessary, imo. You can BELIEVE Q=CV and all the rest.