How is capacitor energy the same as electrostatic potential energy?

[versine]

The energy stored in a capacitor is derived by integrated the work needed to move charge dQ from one plate to another. I'm confused on how this energy is the same as electrostatic potential energy, the energy needed to assemble this configuration from infinity. In the case of capacitor energy, doesn't it neglect the energy needed to construct the two neutral charge plates?

 

[anuttarasammyak]

For a simple model of N + ions and N - electrons in a plate, could you explain what kind of procedure do you have in you mind about " electrostatic potential energy, the energy needed to assemble this configuration from infinity " , e.g. all ions and electrons are at infinity ?

 

[versine]

My thought is to alternate bringing ions and electrons.
Bringing the first ion would take zero work. Bringing the electron would take work. Then bringing another ion would take zero work since the the ion and electron are effectively neutral. And so on. So doesn't bringing each electron take work? I don't see how it ends up as zero.

 

[Ibix]

Why would it take zero work to bring in the first ion? You have to separate it from its electrons, which would take work. Then you could recover that work by bringing in the electrons - so the end result is zero work.

 

[versine]

What do you mean separate from electrons? Aren't we bringing in ions/electrons from infinity?

 

[Ibix]

What do you mean separate from electrons? Aren't we bringing in ions/electrons from infinity?

Where in the universe do you have separate piles of electrons and ions? Matter is neutral, so you need to separate the ion from its electrons in order to bring it in on its own.

 

[DaveE]

Suppose you move the capacitor plates in from infinity as electrically neutral atoms?

EM forces are conservative, which means that you can evaluate the electrical potential energy difference between two configurations without regard to the path or method that those configurations were created by. The work involved in moving an electron away from a proton is exactly equal and cancels the work to bring them back together, however you do it.

 

[nasu]

@versine Why would you pick on the capacitor, specifically? You can ask the same thing about a block sliding down an incline. Do you consider the energy involved in assembling the block from proton and electrons when you use conservation of energy to find the velocity, for example? What about asembling the nucleus from nucleons?
The energy is there but is not relevant, same as it is not relevant to a discharging capacitor. Everywhere in electrostatics we are considering only the extra charges added or removed from the neutral, equilibrium configuration of the system. And this works OK because you can consider the energy of a neutral object as a zero level and measure from here. When you discharge the object you get back to this "zero level". If you can produce an electric field so strong that the object's ion lattice will blow into elementary particles then this energy may become relevant. But this is not the case in electrostatics.

 

[anuttarasammyak]

My thought is to alternate bringing ions and electrons.
Bringing the first ion would take zero work. Bringing the electron would take work. Then bringing another ion would take zero work since the the ion and electron are effectively neutral. And so on. So doesn't bringing each electron take work? I don't see how it ends up as zero.

Plates made of metal have negative bound energy below zero which infinitely separated dilute + ions and - electrons at still have. That is why metals are stable and do not dissemble to + ions and - electrons.
In that sense as for the title of the thread

How is capacitor energy the same as electrostatic potential energy?​

is not accurate. We may regard that electrostatic potential energy of the system would consist of minus binding energy and capacity energy. It should be still minus unless too much charged plates could make the metal states unstable.

 

[vanhees71]

You can derive the expressions for the energy-momentum-stress tensor of the em. field, mudulo a socalled "pseudo-gauge freedom", from Noether's theorem (see Sect. 4.7 in https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf ).

For electrostatics you have
$$u=\frac{\epsilon_0}{2} \vec{E}^2.$$
For a charge distribution within a volume $V$ you get
$$\mathcal{E}=\frac{\epsilon_0}{2} \int_V \mathrm{d}^3 x \vec{E} \cdot (-\vec{\nabla} \phi)=\frac{\epsilon_0}{2} \int_V \mathrm{d}^3 x (\vec{\nabla} \cdot \vec{E}) \phi = \frac{1}{2} \int_V \mathrm{d}^3 x \rho \phi.$$

 

[sophiecentaur]

The energy stored in a capacitor is derived by integrated the work needed to move charge dQ from one plate to another. I'm confused on how this energy is the same as electrostatic potential energy, the energy needed to assemble this configuration from infinity. In the case of capacitor energy, doesn't it neglect the energy needed to construct the two neutral charge plates?

I think this (basic) answer may help, without getting in too deep.
It is the Potential Difference between the capacitor plates that accounts for the stored energy. You are effectively bringing all charges from infinity and arranging them on the capacitor. You can ignore the absolute potential -it's only the difference that counts.

 

[vanhees71]

This you can of course also calculate for an arbitrary capacitor by considering the work needed for charging the capacitor from 0 charge to charge $Q=CU$. Having charge $q$ (i.e. $+q$ on one plate $-q$ on the other) you need the energy of $\mathrm{d} q q/C$ to transport the next infinitesimal charge $\mathrm{d} q$ from one to the other plate, i.e., for the total energy you get
$$\mathcal{E}=\int_0^Q \mathrm{d} q \frac{q}{C}=\frac{Q^2}{2C}=\frac{C}{2} U^2.$$

 

[DaveE]

You can derive the expressions for the energy-momentum-stress tensor of the em. field, mudulo a socalled "pseudo-gauge freedom", from Noether's theorem (see Sect. 4.7 in https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf ).

For electrostatics you have
$$u=\frac{\epsilon_0}{2} \vec{E}^2.$$
For a charge distribution within a volume $V$ you get
$$\mathcal{E}=\frac{\epsilon_0}{2} \int_V \mathrm{d}^3 x \vec{E} \cdot (-\vec{\nabla} \phi)=\frac{\epsilon_0}{2} \int_V \mathrm{d}^3 x (\vec{\nabla} \cdot \vec{E}) \phi = \frac{1}{2} \int_V \mathrm{d}^3 x \rho \phi.$$

I thought I was pushing the limits of "B level" with my comment about conservative fields. But I guess I'm out of touch with what's taught in high schools these days.

 

[vanhees71]

Sorry, I always forget to consider the level. Of course, this you can't use in high school. You have simply to admit that it cannot be explained with the language available at this level.

 

[anuttarasammyak]

We may regard that electrostatic potential energy of the system would consist of minus binding energy and capacity energy.

Feynmann lectures on physics https://www.feynmanlectures.caltech.edu/II_08.html
8–3The electrostatic energy of an ionic crystal
shows good example of calculation of electrostatic energy of materials though it is about ionic crystal not about metal that condenser plates consist of.

​

 

[anuttarasammyak]

You can derive the expressions for the energy-momentum-stress tensor of the em. field, mudulo a socalled "pseudo-gauge freedom", from Noether's theorem (see Sect. 4.7 in https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf ).

For electrostatics you have
$$u=\frac{\epsilon_0}{2} \vec{E}^2.$$
For a charge distribution within a volume $V$ you get
$$\mathcal{E}=\frac{\epsilon_0}{2} \int_V \mathrm{d}^3 x \vec{E} \cdot (-\vec{\nabla} \phi)=\frac{\epsilon_0}{2} \int_V \mathrm{d}^3 x (\vec{\nabla} \cdot \vec{E}) \phi = \frac{1}{2} \int_V \mathrm{d}^3 x \rho \phi.$$

I am puzzled that u > 0 leads to $\mathcal{E} >0$ but not so e.g.
$$\mathcal{E}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{d}<0$$
for two different sign charges separated d. How should I find an arbitration ?

 

[vanhees71]

I'm puzzled bout what you mean. In the above equation $\rho$ is the charge distribution which leads to the electrostatic potential $\phi$, and the expression is indeed manifestly positive (semi-)definite. So how can you get something negative? It's clear that you cannot use a point particle as an example, because it's self-energy diverges.

 

[anuttarasammyak]

So how can you get something negative?

Thanks. Now I know that
$$\frac{1}{2}\int \rho \phi dV$$
includes self energy of charges and it cannot be negative even in the case I referred. It may diverge to infinity but we can take out finite part from it for our use, e.g. electrostatic energy difference between same and different sign charges of same distance d apart the value of which is
$$2\frac{1}{4\pi\epsilon_0}\frac{q^2}{d}$$

 

[vanhees71]

What you discribe is an interaction energy, and this can of course take both signs.

For a set of point charges you can define only such a quantity, which is
$$\mathcal{E}_{\text{int}}=\frac{1}{2} \sum_{i \neq j} \frac{q_i q_j}{4 \pi \epsilon_0 r_{ij}}.$$
You have to take out the self-energies.

For continuous distributions you can include the self-energy, and then the electrostatic energy is always positive, as is clear from
$$\mathcal{E}=\frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{E}^2(\vec{x}).$$

 

[anuttarasammyak]

Re #19

I'm confused on how this energy is the same as electrostatic potential energy, the energy needed to assemble this configuration from infinity.

OP sees $\mathcal{E}_{int}$ and regard $\mathcal{E}_{self}$ as ground energy to be zero. We should share in the discussion what is ground level of energy in the system.

 

[DaveE]

Re #19

OP sees $\mathcal{E}_{int}$ and regard $\mathcal{E}_{self}$ as ground energy to be zero. We should share in the discussion what is ground level of energy in the system.

I'm guessing the OP left a long time ago, when you guys started arguing about vector calculus.

 

[hutchphd]

I'm guessing the OP left a long time ago, when you guys started arguing about vector calculus.

But the OP had a good answer by post #7 so I don't see a serious issue