# In a capacitor, is there energy in the electric field, is there potential energy, or both?

The electric field between two capacitor plates is very simple.

$$\vec{E} = \frac{Q}{\epsilon_0 A} \vec{e}_z$$

I can get the energy stored in the field by integrating the energy density, $$u_e$$, over the volume (between the plates).

$$U = \int_V u_e \; \text{d}^3\!x = \int_V \frac{\epsilon_0}{2} E^2 \; \text{d}^3\!x$$

Since the field is constant, if I pull the plates appart—say that I double the distance—the integration volume is now twice what is was, and the energy stored in the field doubles. Fine!

Simultaneously, we can make an argument from potential energy of the charges in the plates. The charges in each plate are attracted to the other, so when I pull them appart there is a force, and I'm doing work which gives the charges additional potential energy; in virtue of their increased separation.

My question is: Are these two separate processes, where energy stored in the field AND in the potential energy of the charges. Or, are these two different ways of describing the same physical fact that the energy of the system is increasing?

Cheers!

• Awesome! I will look it up in Griffiths. You should post your comment as a proper post so I can flag the question as answered. Thanks! Mar 15, 2022 at 10:31

Energy stored in a capacitor is electrical potential energy, and it is thus related to the charge $$Q$$ and voltage $$V$$ on the capacitor. We must be careful when applying the equation for electrical potential energy $$\Delta PE = q \Delta V$$ to a capacitor. Remember that $$\Delta PE$$ is the potential energy of a charge q going through a voltage $$\Delta V$$. But the capacitor starts with zero voltage and gradually comes up to its full voltage as it is charged. The first charge placed on a capacitor experiences a change in voltage $$\Delta V = 0$$ since the capacitor has zero voltage when uncharged. The final charge placed on a capacitor experiences $$\Delta V = V$$, since the capacitor now has its full voltage $$V$$ on it. The average voltage on the capacitor during the charging process is $$\frac{V}{2}$$ , and so the average voltage experienced by the full charge $$q$$ is $$\frac{V}{2}$$ . Thus the energy stored in a capacitor is $$Q \frac{V}{2}$$ , where $$Q$$ is the charge on a capacitor with a voltage $$V$$ applied. (Note that the energy is not $$Q V$$, but $$Q \frac{V}{2}$$ ). Charge and voltage are related to the capacitance $$C$$ of a capacitor by $$Q = C V$$, and so the expression for $$E_{cap}$$ can be algebraically manipulated into three equivalent expressions:
$$E_{cap} = Q \frac{V}{2} = C \frac{V^{2}}{2} =\frac{Q^{2}}{2C}$$
where $$Q$$ is the charge and $$V$$ is the voltage on a capacitor $$C$$. The energy is in joules for a charge in coulombs, a voltage in volts, and capacitance in farads.