Almost everyone who works with electronic equipment has at some time verified that a capacitor can store energy. If the plates of a charged capacitor are connected by a conductor, such as a wire, charge moves between the plates and the connecting wire until the capacitor is uncharged. The discharge can often be observed as a visible spark. If you should accidentally touch the opposite plates of a charged capacitor, your fingers act as a pathway for discharge, and the result is an electric shock. The degree of shock you receive depends on the capacitance and on the voltage applied to the capacitor. Such a shock could be fatal if high voltages are present, such as in the power supply of a television set. Because the charges can be stored in a capacitor even when the set is turned off, unplugging the television does not make it safe to open the case and touch the components inside.
Consider a parallel-plate capacitor that is initially uncharged, such that the initial potential difference across the plates is zero. Now imagine that the capacitor is connected to a battery and develops a maximum charge Q. (We assume that the capacitor is charged slowly so that the problem can be considered as an electrostatic system.) When the capacitor is connected to the battery, electrons in the wire just outside the plate connected to the negative terminal move into the plate to give it a negative charge. Electrons in the plate connected to the positive terminal move out of the plate into the wire to give the plate a positive charge. Thus, charges move only a small distance in the wires.
To calculate the energy of the capacitor, we shall assume a different process—one that does not actually occur but gives the same final result. We can make this assumption because the energy in the final configuration does not depend on the actual charge-transfer process. We imagine that we reach in and grab a small amount of positive charge on the plate connected to the negative terminal and apply a force that causes this positive charge to move over to the plate connected to the positive terminal. Thus, we do work on the charge as we transfer it from one plate to the other. At first, no work is required to transfer a small amount of charge dq from one plate to the other.3 However, once this charge has been transferred, a small potential difference exists between the plates. Therefore, work must be done to move additional charge through this potential difference. As more and more charge is transferred from one plate to the other, the potential difference increases in proportion, and more work is required.
Suppose that q is the charge on the capacitor at some instant during the charging process. At the same instant, the potential difference across the capacitor is ΔV = q/C. We know that the work necessary to transfer an increment of charge dq from the plate carrying charge –q to the plate carrying charge q (which is at the higher electric potential) is
This is illustrated in Figure 26.11. The total work required to charge the capacitor from q = 0 to some final charge q = Q is
The work done in charging the capacitor appears as electric potential energy U stored in the capacitor. Therefore, we can express the potential energy stored in a charged capacitor in the following forms:
This result applies to any capacitor, regardless of its geometry. We see that for a given capacitance, the stored energy increases as the charge increases and as the potential difference increases. In practice, there is a limit to the maximum energy (or charge) that can be stored because, at a sufficiently great value of ΔV, discharge ultimately occurs between the plates. For this reason, capacitors are usually labeled with a maximum operating voltage.
We can consider the energy stored in a capacitor as being stored in the electric field created between the plates as the capacitor is charged. This description is reasonable in view of the fact that the electric field is proportional to the charge on the capacitor. For a parallel-plate capacitor, the potential difference is related to the electric field through the relationship ΔV = Ed. Furthermore, its capacitance is C = ε0A/d (Eq. 26.3). Substituting these expressions into Equation 26.11, we obtain
Because the volume V (volume, not voltage!) occupied by the electric field is Ad, the energy per unit volume uE = U/V = U/Ad, known as the energy density, is
Although Equation 26.13 was derived for a parallel-plate capacitor, the expression is generally valid. That is, the energy density in any electric field is proportional to the square of the magnitude of the electric field at a given point.
One device in which capacitors have an important role is the defibrillator (Fig. 26.13). Up to 360 J is stored in the electric field of a large capacitor in a defibrillator when it is fully charged. The defibrillator can deliver all this energy to a patient in about 2 ms. (This is roughly equivalent to 3 000 times the power output of a 60-W lightbulb!) The sudden electric shock stops the fibrillation (random contractions) of the heart that often accompanies heart attacks and helps to restore the correct rhythm.
A camera’s flash unit also uses a capacitor, although the total amount of energy stored is much less than that stored in a defibrillator. After the flash unit’s capacitor is charged, tripping the camera’s shutter causes the stored energy to be sent through a special lightbulb that briefly illuminates the subject being photographed.