TheCelt said:
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 = C
V^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.