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Plate Capacitor: Voltage, Capacitance and Eletric Force

Important Formula

Formula: Plate Capacitor

$$\class{purple}{E} ~=~ \frac{U}{d}$$ $$\class{purple}{E} ~=~ \frac{U}{d}$$ $$U ~=~ \class{purple}{E}\,d$$ $$d ~=~ \frac{U}{\class{purple}{E}}$$

Rearrange formula

What do the formula symbols mean?

Electric field (E field)

$$ \class{purple}{\boldsymbol E} $$ Unit $$ \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{N}}{\mathrm{C}} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{A} \, \mathrm{s}^3} $$

Electric field tells how large the electric force on a test charge would be if it were placed between the capacitor plates (electrodes). In a plate capacitor whose electrodes are larger compared to the distance between the electrodes, the electric field between the electrodes is homogeneous. This means that it does not matter where the test charge is placed between the electrodes, the charge will always experience the same force.

Voltage

$$ U $$ Unit $$ \mathrm{V} = \frac{ \mathrm J }{ \mathrm C } = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{A} \, \mathrm{s}^3 } $$

Voltage tells how large the potential difference between the two electrodes is, that is how large the difference in potential energies per charge is. Thus, the voltage indicates how much energy a test charge $q$ gains or loses as it travels from one electrode to the opposite electrode: $q \, U $.

Distance

$$ d $$ Unit $$ \mathrm{m} $$

Distance between the two electrodes. The greater the distance, the smaller the electric field $E$ (with constant voltage $U$).

Table of contents

Basic setup

Plate capacitor with a voltage $U$ between the two electrodes. The distance between the electrodes is $d$ and the area of an electrode is $A$. In between is a dielectric material $\varepsilon_{\text r}$.

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.

Dielectric medium:
A plate capacitor can be filled with a non-conductive material (called dielectric). For example, the dielectric could be the air, vacuum, water, wood, ceramic, or other non-conductor. This dielectric is characterized by the relative permittivity $\varepsilon_{\text r}$. The dielectric should not be conductive at all, because otherwise the charges would pass through the dielectric to the other plate, thus equalizing the charge difference (this is not the purpose of a capacitor). The dielectric is useful for manipulating the physical properties of the capacitor, such as its capacitance.

Table : Examples of relative permittivity of some materials
Dielectric	Relative Permittivity $\varepsilon_{\text r}$
Vacuum	1
Air	1.0006
Water	80
Glas	6 bis 8

Voltage between the plates

Hover the image!

Electric force $F$ on a charge $q$ in the plate capacitor.

When a small charge $q$, for example a free-moving positive charge (let's call it test charge) is placed next to the positive plate, the positive plate will repel the positive test charge and the negative plate will attract it. The test charge experiences an electric force $F$ inside the plate capacitor, which accelerates the small charge straight toward the negative plate. The charge continues to accelerate until it arrives at the opposite negative plate.

Before the test charge hits the negative plate, it has gained velocity $v$ due to acceleration and thus has also gained kinetic energy $W$. This kinetic energy gained by the test charge moving from one charged plate to the other is characterized by the voltage $U$ between the plates. Voltage $U$ between two plates is the energy $W$ gained by a small test charge as it moves from one plate to the other, divided by the charge $q$. Voltage $U$ is therefore energy per charge.

The voltage $U$ between the plates and thus also the gained energy $W$ of the test charge can be manipulated by charging the plates even more. This is done by increasing the charge $Q$ on both plates. This increases the electric force $F$ on the test charge. The test charge would then accelerate even more and thus reach a greater velocity $v$ at the end, so gain a larger kinetic energy $W$. If you double the electric charge $Q$, then the voltage $U$ also doubles. Thus, a test charge $q$ would then gain twice as much energy $W$ after traversing the voltage $U$.

Electric potential in plate capacitor

Electric potential $\varphi(x) $ is the the current potential energy of a charge at position $ x $, per charge. You get the potential $\varphi$ between the electrodes by solving the one-dimensional Laplace equation. The result is a potential $\varphi$ that depends linearly on the spatial coordinate $x$:

$$ \begin{align} \varphi(x) ~=~ - \frac{U}{d} \, x ~+~ \varphi_1 \end{align} $$

Electric potential $\varphi$ of a plate capacitor. One electrode was placed on the coordinate $x=0$ and the other electrode on the coordinate $x=d$.

The potential difference corresponds here to the voltage between the electrodes:

$$ \begin{align} U ~=~ \varphi_1 - \varphi_2 \end{align} $$

If you then plot the electric potential $\varphi$ behind and between the electrodes in a diagram ($\varphi$, $x$), you get a constant potential $\varphi_1$ in the range $x \le 0$, that is up to the first electrode. Also behind the second electrode, that is for $x \ge d$, the potential is constant $\varphi_2$. Between the electrodes, that is in the region between $x=0$ and $x=d$, the potential increases linearly from one electrode to the other.

Electric field and force inside a plate capacitor

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:

$$ \begin{align} F ~=~ \class{red}{q} \, \frac{U}{d} \end{align} $$

When is the force on a test charge the greatest?

The force on a test charge is greater if the test charge and the voltage are greater and the plate distance is smaller.

Electric field $E$ of a plate capacitor. One electrode was placed on the coordinate $x=0$ and the other electrode on the coordinate $x=d$.

If you divide the force 3 by the test charge $q$, then you obtain electric field $E := \frac{F}{q} $:

$$ \begin{align} \class{purple}{E} ~=~ \frac{U}{d} \end{align} $$

The typical unit of electric field $ E $ is $ \frac{\mathrm V}{\mathrm m} $ (volts per meter) or alternatively $ \frac{\mathrm N}{\mathrm C}$ (newtons per coulomb).

The electric field is therefore nothing else than force per charge. The electric field in a plate capacitor depends only on the voltage $U$ and on the plate distance $d$. The larger the voltage and the smaller the distance, the larger 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. In the case of a plate capacitor, the field lines are straight parallel lines running from one plate to the other. Such straight lines characterize a homogeneous E-field. A test charge then moves on such a straight line.

Electric field lines of the plate capacitor run parallel to each other. Behind the electrodes, the E-field cancels out. Between the electrodes the E-field is amplified.

By definition, the electric field lines run away from the positive plate and towards the negative plate. Consequently, the field lines exit the positively charged plate from both sides. And the field lines enter the negatively charged plate on both sides.

The field lines of the negative and positive plates point in opposite directions behind the plates and thus cancel each other out.

Where is the plate capacitor field free?

The plate capacitor is field-free outside the electrodes!

The field lines between the electrodes, on the other hand, point in the same direction, which is why the electric field between the electrodes is amplified.

If you plot the electric field behind and between the electrodes in a coordinate system ($E$, $x$), then the E-field up to the first plate at $x=0$ is zero. The E-field behind the second plate, which is at $x=d$, is also zero. Between the electrodes, that is in the region between $x=0$ and $x=d$, the E-field has a constant value $U/d$.

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:

$$ \begin{align} Q ~=~ C \, U \end{align} $$

Unit of the capacitance $ C $ is $ \frac{\mathrm{As}}{\mathrm V} $ (Ampere-second per volt) or abbreviated $ \mathrm{F} $ (Farad).

Capacitance is an important characteristic quantity of a capacitor, which depends mainly on its geometry, that is on the distance $d$ between the plates and on the plate area $A$. The capacitance also depends on the material, called dielectric, 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. You can calculate the capacitance of a plate capacitor as follows:

$$ \begin{align} C ~=~ \varepsilon_0 \, \frac{A}{d} \end{align} $$

Here, $\varepsilon_0$ is the electric field constant, which provides the correct unit of capacitance. This is a natural constant with the value: $\varepsilon_0 = 8.854 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} $.

If another dielectric is used between the electrodes instead of vacuum ($\varepsilon_{\text r} = 1 $), such as ceramic, then the capacitance of the capacitor changed by this can be taken into account by the relative permittivity $\varepsilon_{\text r}$. The capacitance changes by the factor $\varepsilon_{\text r}$:

$$ \begin{align} C ~=~ \varepsilon_0 \, \varepsilon_{\text r} \, \frac{A}{d} \end{align} $$

Which plate capacitor has the largest capacitance?

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: Calculate the capacitance of the plate capacitor

A plate capacitor with the plate area $A = 1 \, \text{cm}^2$ and the plate distance $d = 1 \, \text{mm} $ has the capacitance:

$$ \begin{align} C &~=~ 8.854 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ \frac{ 0.0001 \, \text{m}^2 }{0.001 \, \text{m}} \\\\
&~=~ 8.854 \cdot 10^{-13} \, \text{F} \end{align} $$

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.

Electrical energy in the field of the plate capacitor

If you integrate the voltage over the charge, you get the energy $W_{\text{e}}$, which is necessary to bring the charge $Q$ to the electrodes:

$$ \begin{align} W_{\text e} ~=~ \frac{1}{2} \, C \, U^2 \end{align} $$

Gespeicherte Energie im Plattenkondensator.

Example: Calculate the electrical energy of the plate capacitor

A capacitor has the capacitance $ C = 10 \, \text{nF} $ and is charged to $1.5 \, \text{V} $ with the help of a battery. Thus, the electrical energy of the capacitor is:

$$ \begin{align} W_{\text{e}} &~=~ \frac{1}{2} ~\cdot~ 10 \cdot 10^{-9} \, \text{F} ~\cdot~ (1.5 \, \text{V})^2 \\\\
&~=~ 11.25 \, \text{nJ} \end{align} $$

Equation 9 can also be expressed using the volume $V$ enclosed by the E-field between the electrodes:

$$ \begin{align} W_{\text{e}} ~=~ \frac{1}{2} \, \varepsilon_0 \, \varepsilon_{\text r} \, V \, \class{purple}{E}^2 \end{align} $$

The equation 11 can be interpreted in such a way that the electrical energy of the plate capacitor is not somehow in the electrodes, but stored in the electric field between the electrodes (because for the energy only the volume enclosed by the field is relevant).

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: Create a Plate Capacitor with a Certain Capacitance

Capacitors are characterized by the electrical capacitance. It tells you how good a capacitor can "store" electric charge. The goal of modern technology is usually to produce the smallest possible capacitors with the largest possible capacitance so that such a component also fits into your smartphone.

You want to have a plate capacitor which has a capacitance of $ C = 0.5 \, \text{nF} $ (nano farad). The area of a capacitor plate is given as $ A = 12 \, \text{cm}^2 $.

How large do you have to choose the distance $ d $ of the plates to reach this capacitance?
What else can you do to achieve the specified capacitance when the plate distance is set to $ d = 1.5 \, \text{mm} $?

Solution to exercise #1.1

We use the following formula: $$C ~=~ \varepsilon_{\text r} \, \varepsilon_{\text 0} \, \frac{A}{d}$$

Assuming a vacuum (or air for an approximation), the relative permittivity $ \varepsilon_{\text r} $ is approximately $ 1 $. Substitute this and rearrange equation for the unknown plate distance $ d $: $$d = \varepsilon_{\text 0} \, \frac{A}{C}$$

Substitute the given values, namely $ C = 0.5 \, \text{nF} = 0.5 \cdot 10^{-9} \, \text{F} $, $ A = 12 \, \text{cm}^2 = 12 \cdot 10^{-4} \, \text{m}^2 $, and $ \varepsilon_{\text 0} = 8.8 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} $: $$\begin{align}d ~&=~ 8.8 \cdot 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} ~\cdot~ \frac{12 \cdot 10^{-4} \, \mathrm{m}^2}{0.5 \cdot 10^{-9} \, \mathrm{F}} \\\\ ~&=~ 2.1 \cdot 10^{-5} \, \text{m} ~=~ 0.021 \, \mathrm{mm}\end{align}$$

As you can see from the result, the plates must be very close together to achieve such a small capacitance of $ C = 0.5 \, \text{nF} $.

Solution to Exercise #1.2

Since the plate separation determined in Exercise 1.1 $ d = 0.021 \, \text{mm} $ is incredibly small, we aim to achieve the desired capacitance using an alternative approach without reducing the distance to such a small value. In this exercise, we want to have a separation of at least $ d = 1.5 \, \text{mm} $.

To achieve this, when looking at the first formula in Exercise 1.1, you need to select an appropriate dielectric (i.e., a specific material between the capacitor plates), characterized by the relative permittivity $ \varepsilon_{\text r} $. Rearrange the first formula in Exercise 1.1 to solve for $ \varepsilon_{\text r} $: $$\varepsilon_{\text r} = \frac{C \, d}{\varepsilon_{\text 0} \, A}$$

Substitute the given values (including the desired distance $ d = 1.5 \, \text{mm} = 1.5 \cdot 10^{-3} \, \text{m} $): $$\begin{align}\varepsilon_{\text r} ~&=~ \frac{0.5 \cdot 10^{-9} \, \text{F} ~\cdot~ 1.5 \cdot 10^{-3} \, \text{m}}{8.8 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ 12 \cdot 10^{-4} \, \text{m}^2} \\\\ ~&\approx~ 89\end{align}$$

This value is approximately that of pure water. So, you would need to immerse the plate capacitor in pure water to achieve the desired capacitance of $ 0.5 \, \text{nF} $ with a plate separation of $ 1.5 \, \text{mm} $.

Exercise #2: Charge inside a Thundercloud

How much charge $Q$ is contained in a thundercloud with an area $A = 1 \, \text{km}^2 $ and an electric field $E$ - of magnitude $ 3 \times 10^6 \, \text{V}/\text{m} $ - built up between the upper and lower parts of the cloud? Thundercloud and earth as plate capacitor

Solution to the Exercise #2

Assume that the upper and lower parts of the cloud behave like two capacitor plates. Then the following equation is valid for the voltage: 1 \[ U ~=~ E \, d \]

The capacitance $ C $ of a plate capacitor is given by the following formula: 2 \[ C ~=~ \varepsilon_0 \, \frac{A}{d} \]

In general, the charge is proportional to the voltage, where the constant of proportionality is the capacitance: 3 \[ Q ~=~ C \, U \]

Insert Eq. 1 and 2 into Eq. 3 and eliminate the distance $ d $. Then you get: 4 \[ Q ~=~ \varepsilon_0 \, A \, E \]

Inserting the concrete values from the exercise results in the approximate amount of charge in a thundercloud: 5 \begin{align} Q &~=~ 8.854 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ 10^6 \, \text{m}^2 ~\cdot~ 3 \cdot 10^6 \, \frac{\text{V}}{\text{m}} \\\\ &~=~ 26.6 \, \text{C} \end{align}