A capacitor is simply a circuit element used for storing charge. It is generally constructed of two closely spaced metal plates. When we apply a voltage across the plates, a positive charge will develop on one plate, a negative charge on the other. When that voltage has magnitude V, and a charge of magnitude Q develops on one of the plates, then we define the capacitance C of the capacitor to be
C = Q/V .
Capacitance is a purely geometrical quantity. This means it depends only on, for example, the arrangement of the two plates in relation to each other. It does not depend on the voltage placed across the capacitor – if we double the voltage, the charge will also double, leaving C the same.
The SI unit of capacitance is called the farad. As can be seen from the equation above, if we use the SI units of charge (coulombs) and voltage (volts), then 1 farad = 1 coulomb/volt.
1. The Parallel Plate Capacitor
The most basic kind of capacitor consists of two flat plates placed parallel to each other. Suppose each plate has area A, and that the distance between the plates is s. We can then calculate what the capacitance should be for this arrangement.
Let us connect the plates to a battery that produces voltage V across them. Charge will quickly build up on the capacitor plates until an electric field E is developed in the space between them such that the voltage drop between the capacitor plates, V=Es, is the same as the battery voltage.
If we can now write E in terms of Q, we can get C = Q/V. To do this we can use Gauss’s law. Please see another source if you need to review Gauss’s law, as it is not the point of this article. Suffice it to say for us that the use of Gauss’s law gives an electric field between the plates of
E = Q/(epsilon_0*A)
where epsilon_0 is the permittivity of free space. Plugging this in,
V = (Q*s)/(epsilon_0*A)
C = Q/V = (epsilon_0*A)/s .
Voila! Note that C depends only on A, the area of the plates, and s, their spacing. If we make the plates bigger, increasing A, we can store a greater charge for a given voltage, increasing the capacitance. If we bring the plates closer together, decreasing s, then the charges on the plates will “tug” at each other more, and we can store more charges on the plates for a given voltage, also increasing the capacitance.
2. Calculating Equivalent Capacitance
Just as you may have seen in the case of resistors, when multiple capacitors are used in a circuit, it can be helpful when analyzing the circuit to instead treat them as one single capacitor, whose equivalent capacitance depends on the arrangement and rating of the capacitors we are considering. That may sound a little confusing, so let’s consider a couple of examples.
Suppose we have two identical capacitors in parallel. (If you don’t know what that means, you should review circuit basics elsewhere.) Then each capacitor is connected to the same voltage V, and each will store the same charge Q. The total charge stored is then 2Q, and so the equivalent capacitance is just the sum of the two capacitances. Generally, if capacitances C_1, C_2, C_3, … are connected in parallel, then can be replaced with an equivalent capacitance
C_equivalent = C_1 + C_2 + C_3 + …
The series case is just slightly more difficult to think about. Suppose we have two capacitor in series, connected to a battery of voltage V. Now, instead of each capacitor having the same voltage across it, each capacitor must instead store the same charge. Let’s think about why this is so.
Each capacitor is made of two parallel plates. Let’s say the first capacitor is storing a charge Q. This means one plate, say the “top” plate, has charge +Q, while the “bottom” plate has charge -Q. Say this bottom plate is connected to the top plate of the next capacitor. Then this top plate of the second capacitor must have charge +Q. This is essentially because the charges that left the bottom plate of capacitor one have all ended up in the top plate of capacitor two. If this did not happen then there would be charges building up somewhere in the connecting wire between them, which does not happen. So, the bottom plate of the second capacitor has charge -Q, and each capacitor is storing the same charge, Q.
OK. Now, say capacitor one has capacitance C_1, and capacitor two C_2. Since we know the charge, we know the voltage across each capacitor.
V_1 = Q/C_1
V_2 = Q/C_2
But, V = V_1+V_1, so
V = (Q/C_1)+(Q/C_2) = Q*[(1/C_1)+(1/C_2)]
(V/Q) = (1/C_equivalent) = (1/C_1)+(1/C_2)
In other words, capacitances in series add reciprocally. This rule holds true no matter how many capacitors we are considering.
3. Capacitors and Dielectrics
Earlier, when we were discussing the parallel plate capacitor, we said that one way to increase capacitance was to bring the plates closer together. Another common technique is to insert an electrical insulator between the plates known as a dieletric. Let’s see how this works.
First suppose we have a parallel plate capacitor with no dielectric. We connect is to a voltage source V, and the plates become charged until an electric field exists between them such that V=Es. Now let’s drop in the dielectric. What happens? The dielectric sees the electric field between the plates and reacts by polarizing. In other words, the molecules in the dielectric will rotate to place their negative ends toward the positive plate, and vice versa. As a result, the dielectric causes a new electric field, E_new, that points in the opposite direction of the capacitor field E. The net electric field magnitude would then seem to be E-E_new, and the new voltage V=(E-E_new)s.
But this situation cannot persist. The capacitor must maintain a voltage between its plates equal to the voltage source it is connected to. As a result, still more charge flows onto the capacitor plates to generate a stronger electric field to compensate for the E_new field generated by the dielectric, until the net electric field is again E, with V=Es.
Therefore, adding a dielectric to a capacitor allows us to store more charge for a given voltage. Specifically, if C_new is the capacitance with the dielectric, C_old the capacitance without, we say,
C_new = K*C_old,
where K is called the dielectric constant.
4. Energy Stored by a Capacitor
Let us conclude our discussion by considering the energy stored by a capacitor. Remember that voltage, also known as electrical potential, is nothing but electrical potential energy per unit charge. To determine the energy stored we need to calculate the work W done to move the charges from one plate to the other.
To do this calculation, we’ll have to perform an integral. This is because the amount of work needed to move a charge changes as the capacitor is charged. Suppose we start with an uncharged capacitor. Then the first charge we move is basically free, but as charge builds up on the capacitor, it becomes harder and harder to move more charges, requiring more and more work per charge. (It becomes harder and harder to stuff more positive charges on the positive plate, for example.) At any time, moving an infinitesimal charge dq requires work,
dW = v*dq,
where v is the current voltage across the plates. (v will start at 0, and end up at V.) Well,
v = q/C,
from the definition of capacitance, where q is the amount of charge currently on the plates. This will range from 0 to Q. So,
dw = (1/C)*q*dq
Integrating both sides, we get,
W = (Q^2)/(2C) .
This is the potential energy stored in the capacitor.