# Capacitor charging and Energy storage

I am currently on the concept of energy density and storing electric potential energy on the field itself (which is quite a new and cool concept to me). However, I still don’t have a solid grasp on how potential ENERGY is being stored in the first place. I would like to ask for some advice/corrections for my chain of reasoning.

Consider the following setup.

I place two sufficiently large conducting sheets reasonably close together (but not touching). This forms a parallel-plate capacitor. I have also learned that a sufficiently large flat plane of charge produces a uniform and perpendicular field. Placing two parallel plates with opposite charge density would form an electric field but ONLY within the space between them as outside fields will totally cancel out. Let us consider a case when there are no fringing fields and this is approximated by assuming a sufficiently large plate.

Now, if I want to charge the capacitor, this means pumping charges on one of the plates which, by induction, produces an equal but opposite charge on the opposite plate. Electrical potential energy is supposedly stored because it takes work to move charge against the electric field (and in fact equal to the work if we set 0 potential energy to an uncharged state). As an analogy, I imagine a capacitor as a spring where “compressing the spring” means adding charge.

But in the scenario I have presented, there is NO electric field outside the space between the plates. There is no repulsive force. No “spring force” can kick out charges.

How can we store electric potential energy if there is no force (due to above reason) repelling our efforts to accumulate charges on a conducting plate? Is there anything wrong with my reasoning?

The force is obviously there. Any two electric charges exert force on each other, repulsive or attractive, depending on the type of charge. Even if there is no external force, there is force within a plate itself. To begin with, you have a neutral plate. But as you start to charge it, the plate accumulates electrons. These accumulated electrons are extra for the plate and makes the plates, as a whole, charged. Thus when you try to put more charge on the plate, the existing charges of the plate repel any extra charges. Thus you have to do work to overcome this repelling force. As a result, potential energy is stored to the capacitor.

Note that, although I mentioned only 'charge' here, I mean negative charges only as, they are the ones that do the moving. Only electrons are light enough to move around and not the positively charged nucleus. Thus I hope there is no confusion when, I say charges of the repel each other. I mean the electrons for the main plate that is being charged directly and static ions for the induced plate. The static ions are positive as they lack electrons and their positivity restricts further depletion of electrons.

If there is any confusion, feel free to comment.

• But the other plate will also have accumulated equal and opposite charge, right? And repulsive forces on the plate in which you are trying to add the electrons on are completely “countered” by attractive forces by the electron deficient plate.
– Lyle
Oct 4, 2020 at 5:37
• No, that is not true. If that was the case, then one could infinite amount of charge within a finite amount of area. That certainly cannot be true right? But I probably understand your confusion. You are probably thinking that since both the plates have equal amounts of charge on each other, and they are of opposite type to each other as well, any new charge added, within a plate will feel net zero force. But this is not true. The force not only depend on the amount of charge, it depends on distance as well in a inverse square manner. Thus any nearby charge exerts more force than one at far. Oct 4, 2020 at 5:43
• That sounds fair. I would want to clarify one more thing. Let’s look at from an electric field perspective. For electric plates, electric field seems only to reside in between the plates (i.e no field outside the area between the plates). But I was told that the potential energy increases because you have to do work AGAINST the field. Yet, in charging the capacitor, you didn’t even have to pass through the field in the first place. What gives?
– Lyle
Oct 4, 2020 at 5:49
• Let me clarify first that the electric doesn't vanish abruptly outside the plates. The field dies out quickly but not like it is 10 N/C at the edge of the plates but 0 just outside them. But the problem you are wondering is not related to do this. The electrons need to do the work against the field within the plate not between the plates. The plates are made of atoms right! By charging the plate, we actually provide excess electrons to this array of atoms. Although atoms are neutral, they contain electrons and when the excess electrons pile up they are affected by the electrons of the atom. Oct 4, 2020 at 11:59
• When you provide further electrons, they get repelled by already existing extra electrons. This makes it more and more difficult for more electrons to gather. Due to this, we have the phenomenon of capacitance. It is literally the capacity of the material to hold charge. If you continue to provide any extra charge, it leaks out. Note that this is all happening within the plates only. It is not due to the electric field between the plates or outside them. Oct 4, 2020 at 12:08

Now, if I want to charge the capacitor, this means pumping charges on one of the plates which, by induction, produces an equal but opposite charge on the opposite plate.

The charge on the opposite plate is not produced by induction. An external voltage source, such as a battery, does work to remove electrons from one plate and deposit them on the other, resulting in a net negative charge on one plate and net positive charge on the other. The total electrical charge (protons and electrons) on the plates remain unchanged. The charges are just redistributed establishing a uniform electric field between the plates and a potential difference equal to the product of the electric field and separation of the plates.

Electrical potential energy is supposedly stored because it takes work to move charge against the electric field (and in fact equal to the work if we set 0 potential energy to an uncharged state).

Essentially correct.

As an analogy, I imagine a capacitor as a spring where “compressing the spring” means adding charge.

The ideal spring is the mechanical analogue of a capacitor, as discussed further below.

But in the scenario I have presented, there is NO electric field outside the space between the plates. There is no repulsive force. No “spring force” can kick out charges.

Neglecting fringe effects, there is no electric field in the space outside the capacitor due to the field between the plates. But the charges are moved from one plate to the other by an externally applied field of an external voltage source, such as a battery.

The positive terminal of the battery pulls electrons off the plate connected to it. The negative terminal of the battery deposits and equal number of electrons on the plate attached to it. It requires a force to perform work to pull and push charges from and to the plates. That force is supplied by electric field of the battery.

How can we store electric potential energy if there is no force (due to above reason) repelling our efforts to accumulate charges on a conducting plate? Is there anything wrong with my reasoning?

What's wrong is you are not considering the external electrical force supplied by the battery to move charges against the attraction or repulsive forces. In this regard, perhaps the ideal spring analogy will help.

SPRING ANALOGY:

An ideal spring is the mechanical analogue of an ideal capacitor (and vice versa). The analogous parameters are:

1. Voltage $$V$$ applied to the capacitor is analogous to the force $$F$$ applied to the spring.
2. The displacement $$x$$ of the spring is analogous to the charge $$Q$$ delivered to the capacitor.
3. The capacitance $$C$$ of the capacitor is analogous to the inverse of the spring constant $$k$$ of the spring.

I need to emphasize that charge does not equal displacement, voltage does not equal force, and capacitance does not equal the inverse of the spring constant. These are analogies, not identities.

The electrical potential energy stored in the electric field of the charged capacitor is commonly shown as

$$E_{C}=\frac{CV^2}{2}$$

The relationship between voltage, capacitance, and charge for a capacitor is

$$V=\frac{Q}{C}$$

Substituting this in the previous equation we obtain

$$E_{C}=\frac{Q^2}{2C}$$

The elastic potential energy stored in a spring that is compressed (or extended) a displacement of $$x$$ is given by

$$E_{S}=\frac{kx^2}{2}$$

Demonstrating that $$C$$ is analogous to the inverse of $$k$$ and $$Q$$ is analogous to $$x$$.

Since for a spring, $$F=kx$$, or $$x=F/k$$, we can express the elastic potential energy stored in the spring as

$$E_{S}=\frac{F^2}{2k}$$

Comparing that to the first equation for the energy stored in a capacitor shows that the spring force $$F$$ is analogous to the capacitor voltage, $$V$$.

If two springs with different spring constants are subjected to identical forces, more elastic potential energy will be stored in the spring with the lower spring constant. This is because the stored elastic potential energy varies as the square of the displacement and linearly with the spring constant. So if the spring constant of spring 1 is half the spring constant of spring 2, for the same force the displacement $$x$$ of spring 1 will be twice that of spring 2. The elastic potential energy stored in spring 1 will be twice that of spring 2.

One final point. As charge is moved from one plate to the other it becomes more and more difficult to move additional charge. That's because when the initial charge is moved the plates are essentially neutral so that little work is required. This explains why during the initial phase of charging a capacitor the current (rate of charge delivery) is maximum.

However as net charge builds up, the attraction and repulsion forces increase resisting the transfer of additional charge. So now the current (rate of charge delivery) is decreasing as the voltage across the capacitor builds.

Once the voltage across the capacitor equals the source (e.g. battery) voltage, the net driving voltage becomes zero and current cease. The capacitor is fully charged.

The spring analogy is the force required to cause displacement increases as the amount of displacement increases.

Hope this helps.

• This eliminated the misconceptions I seem to have. Thank you so much.
– Lyle
Oct 4, 2020 at 18:06
• @Lyle I've edited to include a final point that you may find additionally helpful. Oct 4, 2020 at 18:30
• Thank you so much for these analogies and explanation. This is truly a lifesaver for me.
– Lyle
Oct 5, 2020 at 4:24