Can High Activation Energy Explain the Rapid Discharge of Capacitors?

[volodymyr379]

Homework Statement

In all the sources, I've encountered, it is just stated that if you connect the plates of a capacitor by a wire, then the current from one plate will start to flow to another. However, if the plates of my capacitor are large, then, to a good degree of approximation, I can assume that they are infinite. In this case, the nonzero electric field will exist only between the capacitor plates, which means that no force will act on the electrons inside the wire. The electrons located at the surfaces of the capacitor plates will be "held" by the electric field between the plates. This brings me to the question of why charge will flow from one plate to the other when there is no force acting on the electrons?

Relevant Equations

E = sigma/epsilon 0 - the magnitude of the electric field between the infinite plates, sigma - surface density of the charge, epsilon 0 - vacuum permittivity.
E = 0 everywhere else.

I would be grateful if someone could explain where my reasoning or assumptions are wrong.

 

[jbriggs444]

However, if the plates of my capacitor are large, then, to a good degree of approximation, I can assume that they are infinite. In this case, the nonzero electric field will exist only between the capacitor plates, which means that no force will act on the electrons inside the wire.

So you are hand-waving away edge effects and then putting a wire at the edge? The potential difference does not go away that easily.

 

[alan123hk]

In all the sources, I've encountered, it is just stated that if you connect the plates of a capacitor by a wire, then the current from one plate will start to flow to another. However, if the plates of my capacitor are large, then, to a good degree of approximation, I can assume that they are infinite. In this case, the nonzero electric field will exist only between the capacitor plates, which means that no force will act on the electrons inside the wire. The electrons located at the surfaces of the capacitor plates will be "held" by the electric field between the plates. This brings me to the question of why charge will flow from one plate to the other when there is no force acting on the electrons?

I suspect that if the capacitor plates are infinitely large, how can you reach their edges to place a wire across that infinite.

In reality, no matter how large the two plates are and how narrow the gap between the two plates is, the fringe electric field extending from the edges of the two-plate capacitor always exists.

Also, when you place the wire, the electric field will be redistributed in space again.

 

[ehild]

Homework Statement:: In all the sources, I've encountered, it is just stated that if you connect the plates of a capacitor by a wire, then the current from one plate will start to flow to another. However, if the plates of my capacitor are large, then, to a good degree of approximation, I can assume that they are infinite. In this case, the nonzero electric field will exist only between the capacitor plates, which means that no force will act on the electrons inside the wire.

The ends of the wire, connected to the capacitor plates, are at different potentials. Potential difference inside the wire means electric field . and force on the electrons , so electric current flowing through the wire. This current removes electrons from the negative terminal of the wire which is connected to the negative plate of the capacitor. The place near the junction gets positive with respect to the the surrounding metal and attracts electrons. These electrons flow in the wire towards the positive terminal, enter the positive plate, and neutralize some positive ions there. The current is maintained till there is potential difference across the wire.

 

[vela]

The electric field is only approximately 0 everywhere else because the plates still have to be finite in size if you hope to connect the two with a wire outside the capacitor. The electrostatic force is conservative, so if you integrate the electric field along any closed loop, you have to get 0. Mathematically, if you integrate from the lower plate to the upper plate through the inside of the capacitor and then from the upper plate to the lower plate along the proposed path of the wire outside the capacitor, you have
$$\int_\text{lower}^\text{upper} \vec{E}_\text{inside}\cdot d\vec l + \int_\text{upper}^\text{lower} \vec{E}_\text{outside}\cdot d\vec l = 0,$$ which can't hold if $\vec{E}_\text{outside} = 0$ while $\vec{E}_\text{inside} \ne 0$.

It's good that you're thinking about this stuff so you can learn about the limitations of the approximations we often make in solving problems.

 

[alan123hk]

I think that another argument is that If the potential difference between the two plates is V, the line integral of electric field along the path outside the capacitor is still equal to V, which also indicates that the external electric field is nonzero.

 

[volodymyr379]

I would like to thank vela and alan123hk. Your replies helped a lot.

 

[hilbert2]

Some kind of chemical reaction could be too slow to happen in practice even if it's energetically favorable, if there's a high activation energy barrier for that reaction. The flow of electrons in a circuit can happen without having to cross that kind of barrier, so the capacitor will quickly end up in the lower-energy discharged state. When like charges are packed in a capacitor plate, they are more likely to move away from that plate than to stay there.