1. Home
2. Lessons
3. #275

# Plate Capacitor: Voltage, Capacitance and Eletric Force

Plate capacitor - two electrically charged plates lying close together.
Level 2
Level 2 requires school mathematics. Suitable for pupils.

## Basic setup

A plate capacitor usually consists of two round or rectangular conductive plates (also called Electrodes). These have an area $$A$$ and are located at a distance $$d$$ from each other. Both the area and the distance between the plates are two important parameters that geometrically characterize a plate capacitor.

So far there are only two plates. Only when you put positive and negative electric charges on the two plates, the whole setup becomes a plate capacitor. Charge one plate with positive charge and the opposite plate with the same amount of negative charge. So the total charge on one plate is $$+Q$$ and on the other plate $$-Q$$ . The amount $$Q$$ is the same on both plates.

The positive and negative electric charges on the separated plates now attract each other. If they were free, they would simply move towards each other. But since the plates are spaced at a fixed distance $$d$$ from each other, they cannot do that.

From this you probably already recognize the first possible application of a plate capacitor. If you connect the two charged plates with a conducting wire and a small lamp, an electric current flows from one plate to the other and causes the lamp to light up until there is no more difference in charge on the plates. So with a capacitor you can store electric energy. But you can do much more with it, for example create a frequency filter, which is built into the charging cable of your smartphone and is there to protect the microelectronics from external electromagnetic interference. But that is by far not all. So it's worth getting to know them a little better!

## Voltage between capacitor plates

If you place a small charge $$q$$, that is a freely moving positive charge (lets call it a test charge), directly on the positive plate, the positive plate repels the positive charge and the negative plate attracts it. The free charge experiences an electric force $$F$$ inside the plate capacitor, which accelerates the test charge straight ahead. The charge accelerates until it reaches the opposite negative plate.

Before it hits the plate, it has received a velocity $$v$$ from the acceleration and thus also a kinetic energy $$W$$. This energy, which the charge has gained by moving from one plate to the other, is characterized by the electric voltage $$U$$ between the plates. Electric voltage $$U$$ between two plates is the energy $$W$$ that a small test charge gains when it moves from one plate to the other. Divided by the charge $$q$$. So voltage $$U$$ is energy per charge.

You can influence the voltage $$U$$ between the plates and thus also the energy $$W$$ gained from the test charge by charging the plates even more, that is increasing the charge $$Q$$ on both plates. Then the electric force $$F$$ on the test charge will increase. The test charge would then accelerate even more and thus achieve a greater velocity $$v$$ at the end, so, gain a greater kinetic energy $$W$$. If you double the electric charge $$Q$$, then the electric voltage $$U$$ doubles as well. A test charge $$q$$ would then gain twice as much energy $$W$$ when it moves from one plate to another.

## Capacitance of a plate capacitor

Charge $$Q$$ and voltage $$U$$ are proportional to each other, where the constant of proportionality $$C$$ is the so-called capacitance:1$Q = C \, U$

Unit of the capacitance is $$\left[\frac{\text{As}}{\text V}\right] = [\text{F}]$$ ("Farad").

The capacitance indicates by how much the voltage U changes when the charge Q on the plates is changed.

Capacitance is an important characteristic quantity of a capacitor, which depends mainly on its geometry, that is on the distance $$d$$ and on the plate area $$A$$. The capacitance also depends on the material with which the space between the plates is filled. Here we assume that between the plates there is a vacuum or at least only air ($$\varepsilon_{\text r} = 1$$). You can calculate the capacitance of a plate capacitor as follows:

2$C ~=~ \varepsilon_0 \, \frac{A}{d}$

Capacitance $$C$$ is the electric field constant $$\varepsilon_0 = 8.854 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}}$$ multiplied by the plate area $$A$$ and divided by the distance $$d$$. So to find out the capacitance of a plate capacitor you only have to determine the plate area and measure the distance between the plates.

To get the largest possible capacitance of the plate capacitor, you have to make the plate area $$A$$ as large as possible and the plate distance $$d$$ as small as possible.
Example: CapacitanceA plate capacitor with the plate area $$A = 1 \, \text{cm}^2$$ and the plate distance $$d = 1 \, \text{mm}$$ has the capacitance:$C ~=~ 8.854 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ \frac{ 0.0001 \, \text{m}^2 }{0.001 \, \text{m}} ~=~ 8.854 \cdot 10^{-13} \, \text{F}$

The capacitance of $$0.8854 \, \text{pF}$$ (picofarad) is a very small number. A large capacitance is not so easy to realize experimentally. Even if the plate capacitor is additionally immersed in non-conducting water, its capacitance will only increase by a factor of 80. It will still remain a very small capacitance.

## Electric force on a charge

No matter where you place the test charge $$q$$ inside the plate capacitor, it will always move straight ahead to the other plate everywhere and experience the same force $$F$$. A force field, that is the entirety of all force vectors in space, is homogeneous here. Homogeneous means that it does not matter where you place the test charge. The test charge experiences the same electric force everywhere in the plate capacitor. You can calculate the force on a test charge. The force $$F$$ is - without deriving the formula here:

3$F ~=~ q\,\frac{U}{d}$
The force on a test charge is greater if the test charge and the voltage are greater and the plate distance is smaller.
If you divide the force 3 by the test charge $$q$$, you get the quantity electric field $$E$$:
4$E ~=~ \frac{U}{d}$

So the E-field is nothing else than force per charge. The electric field $$E$$ in a plate capacitor is determined by the voltage $$U$$ and the plate distance $$d$$.

The greater the voltage and the smaller the distance, the greater the electric field.

Since the force field is homogeneous, the electric field in the plate capacitor is also homogeneous. Instead of drawing the vector arrows, the electric field is often illustrated with field lines.