# Capacitors: How do they store energy ?

by hms.tech
Tags: capacitors, energy, store
 P: 247 A Capacitor stores energy when it is charging up but what is the intuition behind such a process ? I, in fact, think that as electrons are being stored on one of the plate, positive charge is being build up on the other plate, an electric field is set up as there is a separation of charges, and this separation of charges would bring about a change in Potential energy of the system (the charges on the plate). According to Coulomb's law, this energy change must be negative. We can infer that the potential energy would really decrease ie become more negative. If that is true, how can a charged capacitor(with negative energy) do positive work while it discharges to power a bulb? (its absurd)
 PF Gold P: 10,790 Separating charges would not cause them to develop a negative potential energy. It takes energy to separate them, and they give up energy whenever they return.
P: 247
 Quote by Drakkith Separating charges would not cause them to develop a negative potential energy. It takes energy to separate them, and they give up energy whenever they return.
Would you like to do the math to prove your point ?
or maybe refer to some website which has done the math ?

PF Gold
P: 10,790

## Capacitors: How do they store energy ?

 Quote by hms.tech Would you like to do the math to prove your point ? or maybe refer to some website which has done the math ?
I'd be glad to if I had any idea how to do the math or where to find an example. Have you tried google?

Also, could you explain how coloumbs law leads to negative potential energy in the case? Perhaps we can work this out step by step.
 Sci Advisor Thanks P: 2,105 The math can help to understand what's going on here a lot (math is always good for understanding). Let's first consider the most simple case of a two-plate capacitor with vacuum between its plates. Then you can think of charging it that you transport charges (say electrons) from one plate to the other. Due to charge conservation both plates carry the same but opposite charges. The charge transport needs energy because the more you charge the plates with opposite charges the larger the electric field between the plates becomes, leading to a force against further charge transport. Due to energy conservation, valid for the motion of charges in static electric fields, it doesn't matter, how you move the charge from one plate to the other, you always need the same energy to reach a certain charge state of the capacitor. Suppose one plate (sitting at $x=0$) is already charged by an amount $+Q$. Then necessarily the other plate (sitting at $x=d$) parallel to the first plate has to carry the charge $-Q$. In the stationary state both charges are located on the surface of the plates inside the capacitor. Using Gauß's Law to a box parallel to the plates with one side within the conducting capacitor plate and one side inside the vacuum of the capacitor and the symmetry assumption (neglecting the edge effects of the finite plates, assuming that the distance $d$ between the plates is much smaller than their size) leads to an electric field $$\vec{E}=\frac{Q}{4 \pi \epsilon_0 A} \vec{e}_x,$$ where $A$ is the area of the plates. If you want to transport another infinitesimal amount of (negative!) charge $-\mathrm{d} Q$ from the left plate to the right plate you have to do work against the Force $\vec{F}=-\mathrm{d} Q \vec{E}$. Since the path along which you carry the charge doesn't matter, you can take a straight line along $\vec{E}$ to get the work needed to do that: $$\mathrm{d} W=\mathrm{d} Q d E_x =\mathrm{d} Q d \frac{Q}{\epsilon_0 A}.$$ Since you start from an uncharged capacitor, to reach a total charge $Q$ you have to integrate this expression from $0$ to $Q$ wrt. $Q$: $$W=\int_0^Q \mathrm{d} Q' \frac{Q'd}{\epsilon_0 A} = \frac{Q^2 d}{2 \epsilon_0 A}.$$ This we can easily rewrite in terms of the finally reached electric field within the capacitor: $$E_x=\frac{Q}{4 \pi \epsilon_0 A}.$$ Plugging this in our formula for the total work done when carrying the charges from one to the other plate in favor of $Q$, we find $$W=\frac{\epsilon_0}{2} E_x^2 A d.$$ Now $A d$ is the volume between the plates and $\frac{\epsilon_0}{2} \vec{E}^2$ is the energy density of the electric field! Thus the total work done to carry the charges from one plate to the other, is now stored as field energy in the electric field between the plates. If you put a dielectricum between the plates for not too high fields the response of this medium is that a polarization by slightly moving the bound charges inside the medium a little bit from their equilibrium place, which needs further work against the binding forces of these charges, which then is stored in this additional electric field, i.e., the polarization of the medium. This leads to an additional factor $\epsilon_r$ in the formula for the work: $$W=\frac{\epsilon_r \epsilon_0}{2} E_x^2 A d.$$
 Sci Advisor P: 3,017 Nice job Van !!! Maybe it'll help also to think about what goes on in a good dielectric vs in free space: Dielectric materials contain polar molecules, ie they have a + and a - end. Water is a good example. Pure water has a dielectic constant around 80, meaning that a capacitor with pure water between its plates would have 80X the capacitance of one with nothing but free space between them. (Water Molecules image courtesy of these guys: http://users.humboldt.edu/rpaselk/C1...C109_lec10.htm and it's an interesting page. In presence of an increasing electric field those polar molecules will begin to align with it, abandoning their preferred random orientations, and that takes mechanical work . Discharging the capacitor removes the field so the dielectric relaxes. That's why oil is used for severe duty AC capacitors - its slippery molecules don't heat up so much as they oscillate with the field. Plastic capacitors will melt in some applications where oil thrive, like commutating or snubbing SCR's. It's analogous to a mechanical spring. You doubtless noticed the similarity - Van's W = K E^2, for a spring it's K X^2 Doubtless this is oversimplified but it helped me in my early days. Now - if someone can explain why it is that empty space has a dielectric constant - i'd be much obliged. thanks, old jim
P: 247
 Quote by vanhees71 Let's first consider the most simple case of a two-plate capacitor with vacuum between its plates. Then you can think of charging it that you transport charges (say electrons) from one plate to the other.
My notes say the exact same thing , I specifically don't understand how can one even imagine this phenomenon ? It is simply false because there is an insulator (air/dielectric)
between the two plates and no charge flows between this space !

Not during charging , not during discharging, Never !

Can you justify your claim ?

Here is a portion of my notes :
Attached Thumbnails

 Sci Advisor PF Gold P: 2,176 You obviously have to connect the two plates first, but once the capacitor is charged you can disconnect it and it will (in an ideal world) maintain its charge indefinitly.
 P: 5,634 The plates are 'connected' [prior post] via the electric circuit.... Think of a battery included, the battery provides chemical energy to do the work of moving the charges,,,,,this energy is then present in the field energy of the charged capacitor less any losses due to battery heating, resistance losses in the circuit,etc..
P: 247
 Quote by f95toli You obviously have to connect the two plates first, but once the capacitor is charged you can disconnect it and it will (in an ideal world) maintain its charge indefinitly.
Do we ? From what I recall the plates are connected to the oppositely charged terminals of a battery and that is it .

We don't have to complete the circuit ...
 P: 5,634 that is what the two posts prior to yours are saying.
PF Gold
P: 2,176
 Quote by hms.tech Do we ? From what I recall the plates are connected to the oppositely charged terminals of a battery and that is it . We don't have to complete the circuit ...
The battery is part of the circuit, so if a capacitor is connected in parallel to a battery electrons can redistribute so that you get "extra" electrons on one plate, and a "shortage" of electrons on the other. The amount of energy stored this way will depend on the voltage (squared) of the battery and the capacitance.
P: 247
 Quote by f95toli The battery is part of the circuit, so if a capacitor is connected in parallel to a battery electrons can redistribute so that you get "extra" electrons on one plate, and a "shortage" of electrons on the other. The amount of energy stored this way will depend on the voltage (squared) of the battery and the capacitance.

How does that prove the point that electrons can travel through air towards the other plate ?
P: 5,634
 How does that prove the point that electrons can travel through air towards the other plate ?
they do not....unless there is a spark due to dielectric breakdown.
P: 247
 Quote by Naty1 they do not....unless there is a spark due to dielectric breakdown.
Then why is this "false concept" used to derive the equation for energy on a capacitor ?

see the attachments, where it is clearly said that an electron moves one by one ....

 Sci Advisor PF Gold P: 2,176 It's not. Where does it say that the electron moves through the air? All it says is that you can calculate the stored energy by considering the amount of energy required to move electrons from one plate to the other. HOW they get from one plate to the other is irrelevant.
Thanks
P: 2,105
 Quote by hms.tech My notes say the exact same thing , I specifically don't understand how can one even imagine this phenomenon ? It is simply false because there is an insulator (air/dielectric) between the two plates and no charge flows between this space ! Not during charging , not during discharging, Never ! Can you justify your claim ? Here is a portion of my notes :
That's the magic of the fact that the electrostatic field has a potential. Of course, in reality there are no charges transported through the non-conducting vacuum or dielectric between the plates, but that doesn't matter, as I've emphasized in my posting! No matter, how you imagine to transport the charges such that you finally have a charged capacitor, the work to be done for this is the same. This work is stored as energy in the electric field between the plates and in the polarization of the dielectric (if there is one between the plate). So I can just use this simple picture of moving the electrons through the space between the plates to get the capacitor in this charged state. Of course, in practice of usual capacitors nobody does this, but thanks to the curl-free nature of the electrostatic field the energy stored in the charged capacitor (in precisely this field) is the same, no matter, how the charges were moved. A discription of the real processes happening when connecting a battery with the capacitor is way more complicated und fortunately not needed to answer the question.

To answer Jim Hardy's question. Of course vacuum has no "dielectric" constant. The $\epsilon_0$ is in the equations from the choice of the units in the Systeme International (SI). From a physical point of view it's a very unnatural choice of units and sometimes leads to misunderstandings, because the constants $\epsilon_0$ and $\mu_0$ seem pretty mysterious to the beginner, because they are quite artificial to make the numbers easier to handle for every-day electrical engineering.

The only fundamental constant in classical electromagnetism is the speed of light, indicating that Maxwell's electromagnetic theory is a relativistic theory.
Mentor
P: 16,298
 Quote by hms.tech My notes say the exact same thing , I specifically don't understand how can one even imagine this phenomenon ? It is simply false because there is an insulator (air/dielectric) between the two plates and no charge flows between this space ! Not during charging , not during discharging, Never ! Can you justify your claim ? Here is a portion of my notes :
The electrostatic force is conservative. That means that it does not matter what path it takes, the work is the same. It doesn't matter if the charge goes through the air (which is possible but not common) or if it goes through wires. Either way the work is the same.

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