What is a capacitor? Basically, it is any two conductors separated by an insulator. In practical applications, the conductors usually have charges of equal magnitude and opposite sign, so the net charge of the capacitor is zero. The electric field between the two conductors is proportional to the magnitude of this charge, and the potential difference Vab between the conductors is also proportional to the charge magnitude Q.

The capacitance (C) of a capacitor is defined as the ratio of the magnitude of the charge (Q) on either conductor to the magnitude of the potential difference (Vab):

C = Q/V[ab]
(Equation 6-37)

As you probably guessed capacitance is measured in coulombs per volt, which is called a farad (F) in honor of Michael Faraday. There is one convention you should probably keep in mind. When a capacitor is referred to having a charge of Q, it means that one conductor has charge +Q and the other Q. Overall the charge is still zero.

Parallel Plate Capacitors

A parallel plate capacitor is just two conducting plates parallel to each other, hence the name, separated by a small distance when comapared to the length of the plate. The field between the plates is basically uniform, except for some fringing at the edges, and the charges on the plates are uniformly distributed over the opposing surfaces.

In our example, we will assume that the plates are in vacuum, which they sometimes are. It already has been proved that the electric field magnitude between two plates is:

E = (sigma)/(epsilon 0) = Q/((epsilon 0)A)
(Equation 6-38)

Since the electric field between the plates is uniform, the potential difference between the plate is:

V[ab] = Ed = (Qd)/((epsilon 0)A)
(Equation 6-39)

d is the distance between the plates. So:

C = Q/V[ab] = (epsilon 0)A/d
(Equation 6-40)

Plates in Series

Capacitors in Series Illustration IllustrationIn the illustration at right, you can see that two capacitors are connected in series between points a and b, maintained at a constant potential difference Vab. They start initially uncharged and both capacitors always have the same charge Q. Remember that the capacitors need to have a net charge of zero. So the top and bottom of each capacitor need to be equal but oppositely charged of each other. Thus, in a series connection the magnitude of charge on all plates is the same. Thus:

Click to view Equation 6-41
(Equation 6-41)

Plates in Parallel

Capacitors in Parallel IllustrationIn the illustration at right, you can see that these capacitors are connected in parallel now between points a and b. Now, in this case the potential difference Vab is the same for both, but the charges on the capacitors aren’t necessarily equal since they may have different capacitances. The charges are:

Q1 = C1V and Q2 = C2V
And the total charge Q supplied by the potential difference is:

Q = Q1 + Q2; Q = V(C1 + C2); Q/V = C1 + C2; C = C1 + C2
(Equation 6-42)

Energy in a Charged Capacitor

You use capacitors to do many things, one of which is to store energy. But what good is storing energy if you don’t know how much energy you stored? In order to get energy, you need to store energy by moving charge from one plate to another. After that process the final charge Q and the final potential difference V is related by Q = CV.

Now between the point when the process starts and finishes, the net charge on either plate is q. The potential difference v between the plates is:

v = q/C
(Equation 6-43)

The work dW required to transfer the next charge and the total work required to charge from 0 to Q is:

Click to view Equation 6-44
(Equation 6-44)


Most capacitors have a nonconducting solid material, a dielectric, between their plates. The purpose of this is three fold. First, you don’t want the plates to touch so when you roll up a parallel plate to make it smaller, having something between the plates stops them from touching. Second, any dielectric material, when in a large electric field, experiences dielectric breakdown, a partial ionization which permits conduction through the material that is supposed to insulate. Most materials can tolerate stronger electric fields than air can. Thirdly, the capacitance of a capacitor of given dimensions is larger when there is a dielectric between them. With experimentation, it is shown that after inserting a dielectric, the potential difference decreases while the charge on each plate remains the same.

The original capacitance of the capacitor, C0, is:

C0 = Q/V0
(Equation 6-45)

When a dielectric is inserted between the plates, Q does not change and V is less than V0 (under experimentation). That means that C is greater than C0, and the ratio of the two is called the dielectric constant (k):

kappa = C/C0
(Equation 6-46)

Now, by definition a perfect vacuum is given a dielectric constant of 1, and a dielectric always increases capacitances so any material has a dielectric constant greater than that of a perfect vacuum. With a vacuum, the E field between the plates is:

E0 = V0/d = (sigma)/(epsilon 0)
(Equation 6-47)

A reduction in potential difference, with the dielectric, implies a reduction in the E field, which implies a reduction in the charge per unit area. Since Q has remained the same, it can only mean that the reduction was caused by induced charges on the surfaces of the dielectric. The negative plate induced a positive charge on one side while the positive plate induced a negative charge on the other side of the dielectric. The electric field in the dielectric must be:

E = V/d = (sigma - sigma[i])/(epsilon 0)
(Equation 6-48)

And the capacitance with the dielectric is:

C = (kappa)(epsilon 0)A/d
(Equation 6-49)

R-C Series Circuit

In all the circuits we did before, we always used a constant EMF and a constant resistance. Now we will play with something where the potentials and currents are changing and dependent on time.

R-C Series Circuit IllustrationRefering the the circuit diagram at right, when the double-pole, double-throw (dpdt) switch is in the upper position, it is connected to the battery which charges the capacitor through resistor R. This type of circuit is called an R-C series circuit (aptly named because a resistor and capacitor are in series). When the capacitor is uncharged, initially, the initial potential difference across the capacitor is zero, the entire battery voltage appears across the resistor, causing an initial current of I = V/R. As the capacitor charges, its voltage increases, so the potential difference across the resistor decreases, which corresponds to a current decrease. After the capacitor is fully charged, the entire battery voltage is across the capacitor, no potential difference is across the resistor, and the current becomes zero.

Before we begin, we will use q as the charge on the capacitor. It is lowercase because it is not the total charge but a changing charge. i (same reason why it is not capitalized) is the charging current after the switch is thrown up. Thus:

click to view Equation 6-50
(Equation 6-50)

At the moment when the switch is flipped, q = 0 and the current is V/R, as if the capacitor wasn’t there. However, as q increases, the current decreases until it becomes zero. When i = 0, then:

V/R = q/RC; q = CV; q = Q
(Equation 6-51)

Of course that really isn’t useful to us if we wanted to know how much charge was located on the capacitor after three seconds or something like that. So to get a quantitative description of the variation with time, we replace i in the equation with dq/dt, which gives us:

Click to view Equation 6-52
(Equation 6-52)

Now we have that pesky constant of integration (K in this example; we usually use C but C is capacitance) that always haunts us when we don’t have limits of integration. So to replace that with something more useful, we set t = 0 (and at t = 0q = 0):

Click to view Equation 6-53
(Equation 6-53)

Now if we take the derivative of this with respect to time, we get the current as a function of time:

i = dq/dt = (V/R)e^(-t/RC) = (I0)e^(-t/RC)
(Equation 6-54)

We see that the charge and current are both exponential functions of time. You will notice that RC appears quite a lot, that is because RC is called the time constant, or relaxation time, of the circuit, and is denoted by t.

That’s just charging. Now comes when you flip the switch down and discharge. The way to solve this is similar to discharging, so we will skip the steps in between. When the switch is thrown, the current will begin at its highest and then begin to decrease. Thus the current, or rate of charge is:

dq/dt = -q/RC
(Equation 6-55)

Do the calculus and you will eventually find:

q = (Q0)e^(-t/RC); i = (I0)e^(-t/RC)
(Equation 6-56)