Capacitors
Two conductors separated by an insulator are not particularly interesting. When, however, two large, thin, flat
conductors are separated a small distance by a good insulator, they form a device called a capacitor.
Capacitors can be very small or very large. Most are made of metal foils separated by thin plastic or paper
sheets. They are then rolled up and put into little metal cans (for large ones) or covered with a glob of plastic
or ceramic if they are small.
Here is a good schematic model of a capacitor.
The red part is charged positive, and the blue part is negative. The material between the metal foil is called the
dielectric. It can be almost any insulator (e.g. paper, plastic, rubber, wax, air). Capacitors can store and
release charge and are very useful in electronics. They are rated on their capacitance, which is the ratio of
stored charge to the potential difference.
C=Q/V
C is the capacitance
Q is the stored charge
V is the potential difference across the two plates
The unit for capacitance is the farad (F), named for physicist Michael Farady.
One farad is a coulomb per volt, which is a very large unit.
Capacitors are typically rated in µF and pF.
Exercise: A 3pF capacitor is connected to a 12V battery.
What is the charge on the capacitor?
C = Q /V
Q = VC
Q = 3•10-12F • 12V
Q = 3.6•10-11C
Finding Capacitance
To calculate capacitance:
C = K ∈0 A / d
C : Capacitance
k : Dielectric constant (unique to a material)
∈ 0 : Permittivity of free space : 8.85•10-12 C2 / Nm2
A : Area of the plates
d : distance between the plates
Exercise: An air capacitor has a 0.1mm plate spacing
between 2cm2 plates. K for air is 1. What is the capacitance?
2cm2 = 0.0002m2
0.1mm = 0.001m
C = K ∈0 A / d
C = 8.85•10-12 C2 / Nm2 • 0.0002m2 / 0.001m
C = 1.77pF
The way this concept is commonly assed is by way of a multiple-choice question. Typically, the question asks
what happens to the capacitance when changes are made to the makeup of the capacitor.
What happens to the capacitance when the area increases?
If A goes up, the capacitance goes up. The capacitor can hold more charge per volt.
What happens to the capacitance when the plates are moved closer together?
If d goes up, the capacitance goes down. The capacitor can hold less charge per volt.
What happens to the capacitance when the dielectric is replaced with a material with higher k?
If k goes up, the capacitance goes up. The capacitor can hold more charge per volt.
Energy Stored in Capacitor
A capacitor with uncharged plates is boring; it’s just metal plates. When a battery applies a voltage difference,
electrons begin to flow. One plate of the capacitor loses a few electrons, and the other plate gains a few. It
takes time for this to happen. Eventually the capacitor will have all the charge that it can hold (when Q = CV),
and electricity will stop flowing. The capacitor will have maximum potential energy at this point and store its
maximum amount of charge.
1. Both plates are uncharged, there are as many negative charges as positive charges.
2. There is a growing charge imbalance as negative charges leave the left plate.
It’s getting harder and harder to move charge to the negative plate.
3. The battery has separated as much charge as it can, there is maximum charge separation.
The battery is not strong enough to move any more charge to the negative plate.
Once a capacitor is charged, charges cease to flow. The excess free electrons on the negative plate really want
to get away (negative repels negative), but the battery maintains the charge separation. If the circuit is broken,
then the electrons have no path to follow. In fact, if the charged capacitor is completely removed from the
circuit, it is still charged. If the two plates are connected by a conducting path (such as through a light bulb or
other electrical device), all the imbalanced charges will immediately rush out. Capacitors can be used to
deliver large amounts of sudden charge, such as in a camera flash.
A charged capacitor has the ability to do work, so it must have some potential energy. To calculate the
amount of energy the capacitor can release, use:
PEcapacitor = ½ QV
Of course, because Q = CV, we can use two different forms of that equation.
PEcapacitor = ½ QV = ½ CV2
Exercise: How much energy can a 12.5 pF capacitor store
when connected to a 9.0 V battery.
PE = ½ CV2
PE = ½ 12.5•10-12F (9V)2
PE = 5.1•10-10J