capacitor question

[CityBird]

I would like to know if my understanding of capacitors is correct.

 

a) If you move the plate of a charged hypothetical capacitor further apart from each other the voltage must increase since V = Q/C and C = εA/d. The stored energy E = 1/2CV² increases because V² increases faster then the capacitance C drops. My understanding is that this energy rise strictly comes from the mechanical energy from the lateral pull.

 

b) If one would insert a dielectric with an εr of 2 into an unconnected and charged air plate capacitor εr of 1, capacitance would obviously double. There's however no conversion of energy (At least that's my line of thought) since the work done is at a 90 degrees angle with respect to the capacitor. Also when you would remove the dielectric again everything is as before. So one can't speak about energy loss and gain. The drop in voltage must therefore come from inside. My understanding is that the dielectric is charged due to the electric field inside the capacitor. This produces a voltage that counters the already present voltage on the plates which results in an overall voltage drop.

 

Is a) and b) correct?

 

[Danijel Gorupec]

Yes. As far as I know, you are correct.

 

a) Note: C = εA/d might not work for very big 'd' (relative to 'A'). Thus if you spread the plates to infinite distance you should not obtain infinite V and infinite E.

 

b) What you say is true in first approximation. I however suspect that in a more detailed picture, the dielectric might heat up for just a little bit when inserted/removed. So if you keep inserting/removing it, a thousand times in a second, maybe you should count for some additional energy loss.

 

 

BTW, your thinking is clear and I like it. I hope you will stay on the forum.

[studiot]

 

My understanding is that this energy rise strictly comes from the mechanical energy from the lateral pull.

 

Ultimately, yes.

 

But in pulling the plates apart your pull is separating the charges against the electric field, thereby doing electrical work.

[Enthalpy]

a) Yes

 

b) The capacitor supplies mechanical work to the inserted insulator, that is, it attracts the insulator. This compensates the decrease of electric energy.

 

This attraction and work is possible because the electric field is not perpendicular to the plates, but has a component parallel to the insulator's movement. Near the edge of the insulator film, the field lines go to the side to plunge in to insulator. They take this longer way through the air because the distance through the dielectric is less costly.

 

b2) The same happens with ferromagnetic materials and coils. In an electric motor, the magnetic field lines flow approximately perpendicular through the gap between the rotor and the stator, except near the slits where pass the wires carrying the current. At the slits, the magnetic field gets tilted to reach the more favourable slit's side. This is where force and work are made, and the magnetic lines have at these locations a component parallel to the movement.

[Delbert]

I would like to know if my understanding of capacitors is correct.

 

a) If you move the plate of a charged hypothetical capacitor further apart from each other the voltage must increase since V = Q/C and C = εA/d.

Correct. Or expressed another way: if you want to generate a voltage higher than you can generate, charge a capacitor to the voltage you can generate and pull it apart.