As written in Abbildung 26 a), the electrostatic induction in a single plate is not considered. Rather, we are now interested in what happens based on the electrostatic induction when the plates are brought together. The electrostatic induction will again move charges inside the conductors. Near to the negative outer plate (1) positive charges get induced on (X). Equally, near to positive outer plate (2) negative charges get induced on (Y). Graphically speaking, for each field line ending on the pair of plates, a single charge must move from one plate to the other. The direction of the movement is in similar to the direction of $\vec{E}$. This ability to separate charges (i.e. to generate electrostatic induction) is another property of space. This property is independent of any matter inside of the space.

This movement is represented with the displacement flux $\Psi$. The displacement flux is given by the amount of moved charge $\Psi = n \cdot e = Q$, with the unit $[\Psi]= [Q] = 1 C$. When looking on Abbildung 26 b) and c), it is evident, that for larger plates (X) and (Y) more charges gets displaced. So, in order to get a constant value by dividing displacement flux by the corresponding area. This leads to the electric displacement field $D$ (sometimes also displacement flux density), which is defined as:

\begin{align*} \boxed{ D = {{\Psi}\over{A}} } \end{align*}

ON the other hand one could also only focus onto the induced charges on the surfeces: In the shown arrangement (homogeneous field, all surfaces parallel to each other), the surface charge density $\varrho_A = {{\Delta Q}\over{\Delta A}}$ thus electrostatic induction is proportional to the external field $E$. It holds:

\begin{align*} \varrho_A = {{\Delta Q}\over{\Delta A}} \sim E \\ \varrho_A = {{\Delta Q}\over{\Delta A}} = \varepsilon \cdot E \end{align*}

Since the induced charges $\Delta Q$ are equal to the flux $\Psi$ the electric displacement field is also given by:

\begin{align} \boxed{\vec{D} = \varepsilon \cdot \vec{E}} \end{align}

• Similar to the electric field $\vec{E}$ also the flux density is a field.
• It can be interpreted as an vector field. pointing in the same direction as the electic field $\vec{E}$.
• The electric displacement field has the unit „charge per area“, i.e. $As/m^2$.

Why is now a second field introduced? This shall become clearer in the following, but first it shall be considered again how the electric field $\vec{E}$ was defined. This resulted from the Coulomb force, i.e. the action on a sample charge. The electric displacement field, on the other hand, is not described by an action, but caused by charges. The two are related by the above equation. It will be shown in later sub-chapters that the different influences from the same cause of the field can produce different effects on other charges.

The permittivity (or dielectric conductivity) $\varepsilon$ thus results as a constant of proportionality between $D$-field and $E$-field. The inverse ${{1}\over{\varepsilon}}$ is a measure of how much effect ($E$-field) is available from the cause ($D$-field) at a point. In vacuum, $\varepsilon$ is $\varepsilon_0$, the electric field constant.

General relationship between charge Q and electric displacement field D

Up to now, only a homogeneous field was considered and only a surface perpendicular to the field lines. Thus only equipotential surfaces (e.g. a metal foil) were investigated. In that case it was found that the charge is equal to the electric displacement field on the surface: $\Delta Q = D\cdot \Delta A$.

This formula is now to be extended to arbitrary surfaces and inhomogeneous fields. As with the potential and other physical problems, the problem is to be broken down into smaller subproblems, solved and then summed up. For this purpose a small area element $\Delta A = \Delta x \cdot \Delta y$ is needed. In addition, the position of the area in space should be taken into account. This is possible if the cross product is chosen: $\Delta \vec{A} = \Delta \vec{x} \times \Delta \vec{y}$, since so is the surface normal. In what follows, the cross product will be relevant to the calculation, but the consequences of the cross product will be:

• The magnitude of $\Delta \vec{A}$ is equal to the area $\Delta A$.
• The direction of $\Delta \vec{A}$ is perpendicular to the area.

In addition, let $\Delta A$ now become infinitesimally small, that is, $dA = dx \cdot dy$.

1. problem: inhomogeneity → solution: infinidesimal area

First, we shall still assume an observation surface perpendicular to the field lines, but an inhomogeneous field. In the inhomogeneous field, the magnitude of $D$ is no longer constant. In order to correct this, $dA$ is chosen so small that just „only one field line“ passes through the surface. In this case D is homogeneous again. Thus holds:

$Q = D\cdot A$

\begin{align*} Q = D\cdot A \quad \rightarrow \quad dQ = D\cdot dA \end{align*}

2nd problem: arbitrary surface → solution: vectors

Now assume an arbitrary surface. Thus the $\vec{D}$-field no longer penetrates through the surface at right angles. But for the electrostatic induction only the rectangular part was relevant. So only this part has to be considered. This results from consideration of the cosine of the angle between (right-angled) area vector and $\vec{D}$-field:

\begin{align*} dQ = D\cdot dA \quad \rightarrow \quad dQ = D\cdot dA \cdot cos(\alpha) = \vec{D} \cdot d \vec{A} \end{align*}

The area vector and the surface normal can be seen in Abbildung 29.

3. Summing up

Since so far only infinitesimally small surface pieces were considered must now be integrated again to a total surface. If a closed enveloping surface around a body is chosen, the result is:

\begin{align} \boxed{\int dQ = \iint_{\text{closed surf.}} \vec{D} \cdot d \vec{A} = \iiint_V \varrho_V d\vec{V} = Q} \end{align}

The „sum“ of the $D$-field emanating over the surface is thus just as large as the sum of the charges contained therein, since the charges are just the sources of this field. This can be compared with a bordered swamp area with water sources and sinks:

• The sources in the marsh correspond to the positive charges, the sinks to the negative charges. The formed water corresponds to the $D$-field.
• The sum of all sources and sinks equals in this case just the water stepping over the edge.

Applications

Are calculated in the course.

Spherical Capacitor

Spherical capacitors are now rarely found in practical applications. In the Van-de-Graaff generator, spherical capacitors are used to store the high DC voltages. The earth also represents a spherical capacitor. In this context, the electric field of $100...300 V/m$ in the atmosphere is remarkable, since several hundred volts would have to be present between head and foot (for resolution, see the article Electricity from the air in Bild der Wissenschaft).

Plate Capacitor

The relation between the $E$-field and the voltage $U$ on the ideal plate capacitor is to be derived from the integral of displacement flux density $\vec{D}$: \begin{align*} Q = \iint_{\text{closed surf.}} \vec{D} \cdot d \vec{A} \end{align*}

tasks

1.6 Non-Conductors in electrostatic Field

The material law of electrostatics

First of all, a thought experiment is to be carried out again (see Abbildung 31):

1. First a charged plate capacitor in vacuum is assumed, which is separated from the voltage source after charging.
2. Next, the intermediate region is to be filled with a material.

Think about how $E$ and $D$ would change before you unfold the subsection.
Why might which of the two quantities change?

You may have considered what happens to the charge $Q$ on the plates. This charge cannot leave the plates. So $Q = \iint_{\text{closed surf.}} \vec{D} \cdot d \vec{A}$ cannot change.
Since the fictitious surface around an electrode does not change either, $\vec{D}$ cannot change either.

On the other hand, polarizable materials in the capacitor can align themselves. This dampens the effective field. Maybe you remember what the „acting field“ was: the $E$-field. So the $E$-field becomes smaller (see Abbildung 32).

Previously:

\begin{align*} D = \varepsilon_0 \cdot E \end{align*}

The determined change is packed into the material constant $\varepsilon_r$. This gives the material law of electrostatics:

\begin{align*} \boxed{D = \varepsilon_r \cdot \varepsilon_0 \cdot E} \end{align*}

Since the charge $Q$ cannot vanish from the capacitor in this experimental setup and thus $D$ remains constant, the $E$ field must become smaller for $\varepsilon_r>1$.

Abbildung 32 is drawn here in a simplified way: the alignable molecules are evenly distributed over the material and are thus also evenly aligned. Accordingly, the E-field is uniformly attenuated.

Some values of the relative permittivity $\varepsilon_r$ for dielectrics are given in Tabelle 1.

Dielectric strength of dielectrics

• The dielectrics act as insulators. The flow of current is therefore prevented
• The ability to insulate is dependent on the material.
• If a maximum electric field $E_0$ is exceeded, the insulating ability is eliminated
  • One says: The insulator breaks down. This means that above this electric field a current can flow through the insulator.
  • Examples are: Lightning in a thunderstorm, ignition spark, glow lamp in a phase tester
  • The maximum electric field $E_0$ is called dielectric strength (in German: Durchschlagfestigkeit or Durchbruchfeldstärke).
  • $E_0$ depends on the material (see Tabelle 2), but also on other factors (temperature, humidity, …).

tasks

1.7 Capacitors

Capacitor and Capacitance

• A capacitor is defined by the fact that there are two electrodes (= conductive areas), which are separated by a dielectric (= non-conductor).
• This makes it possible to build up an electric field in the capacitor without charge carriers moving through the dielectric.
• The characteristic of the capacitor is the capacitance $C$.
• In addition to the capacitance, every capacitor also has a resistance and an inductance. However, both of these are usually very small.
• Examples are
  • the electrical component „capacitor“,
  • an open switch,
  • a wire to ground,
  • a human being

$\rightarrow$ Thus, for any arrangement of two conductors separated by an insulating material, a capacitance can be specified.

The capacitance $C$ can be derived as follows:

1. It is known that $U = \int \vec{E} d \vec{s} = E \cdot l$ and hence $E= {{U}\over{l}}$ or $D= \varepsilon_0 \cdot \varepsilon_r \cdot {{U}\over{l}}$.
2. Furthermore, $\iint_{\text{closed surf.}} \vec{D} \cdot d \vec{A} = Q$ by the idealized form of the plate capacitor: $Q=D \cdot A$.
3. Thus, the charge $Q$ is given by: \begin{align*} Q = \varepsilon_0 \cdot \varepsilon_r \cdot {{U}\over{l}} \cdot A \end{align*}
4. This means that $Q \sim U$, given the geometry (i.e., $A$ and $d$) and the dielectric ($\varepsilon_r $).
5. So it is reasonable to determine a proportionality factor ${{Q}\over{U}}$.

The capacitance $C$ of an idealized plate capacitor is defined as

\begin{align*} \boxed{C = \varepsilon_0 \cdot \varepsilon_r \cdot {{A}\over{l}} = {{Q}\over{U}}} \end{align*}

This relationship can be examined in more detail in the following simulation:

capacitor lab

If the simulation is not displayed optimally, this link can be used.

The Abbildung 33 shows the topology of the electic field inside of an plate capacitor.

Designs and types of capacitors

To calculate the capacitance of different designs, the definition equations of $\vec{D}$ and $\vec{E}$ are used. This can be viewed in detail e.g. in this video.
Based on the geometry, different equations result (see also Abbildung 34).

In Abbildung 35 different designs of capacitors can be seen:

1. rotary variable capacitor (also variable capacitor or trim capacitor).
   1. A variable capacitor consists of two sets of plates: a fixed set and a movable set (stator and rotor). These represent the two electrodes.
   2. The movable set can be rotated radially into the fixed set. This covers a certain area $A$.
   3. The size of the area is increased by the number of plates. Nevertheless, only small capacities are possible because of the necessary distance.
   4. Air is usually used as the dielectric, occasionally small plastic or ceramic plates are used to increase the dielectric constant.
2. multilayer capacitor
   1. In the multilayer capacitor, there are again two electrodes. Here, too, the area $A$ (and thus the capacitance $C$) is multiplied by the finger-shaped interlocking.
   2. Ceramic is used here as the dielectric.
   3. The multilayer ceramic capacitor is also called KerKo or MLCC.
   4. The variant shown in (2) is an SMD variant (surface mound device).
3. Disk capacitor
   1. A ceramic is also used as dielectric for the disk capacitor. This is positioned as a round disc between two electrodes.
   2. Disc capacitors are designed for higher voltages, but have a low capacitance (in the microfarad range).
4. Electrolytic capacitor, in German also called Elko for Elektrolytkondensator
   1. In electrolytic capacitors, the dielectric is an oxide layer formed on the metallic electrode. the second electrode is the liquid or solid electrolyte.
   2. Different metals can be used as the oxidized electrode, e.g. aluminium, tantalum or niobium.
   3. Because the oxide layer is very thin, a very high capacitance results (depending on the size: up to a few millifarads).
   4. Important for the application is that it is a polarized capacitor. I.e. it may only be operated in one direction with DC voltage. Otherwise, a current can flow through the capacitor, which destroys it and is usually accompanied by an explosive expansion of the electrolyte. To avoid reverse polarity, the negative pole is marked with a dash.
   5. The electrolytic capacitor is built up wound and often has a cross-shaped predetermined breaking point at the top for gas leakage.
5. film capacitor, in German also called Folko, for Folienkondensator.
   1. A material similar to a „chip bag“ is used as an insulator: a plastic film with a thin, metallized layer.
   2. The construction shows a high pulse load capacitance and low internal ohmic losses.
   3. In the event of electrical breakdown, the foil enables „self-healing“: the metal coating evaporates locally around the breakdown. Thus the short-circuit is cancelled again
   4. With some manufacturers this type is called MKS (Mmetallized foilccapacitor, Polyester).
6. Supercapacitor (engl. Super-Caps)
   1. As a dielectric is - similar to the electrolytic capacitor - very thin. In the actual sense, there is no dielectric at all.
   2. The charges are not only stored in the electrode, but - similar to a battery - the charges are transferred into the electrolyte. Due to the polarization of the charges, they surround themselves with a thin (atomic) electrolyte layer. The charges then accumulate at the other electrode.
   3. Supercapacitors can achieve very large capacitance values (up to the kilofarad range), but only have a low maximum voltage

In Abbildung 34 are shown different capacitors:

1. Above two SMD capacitors
   1. On the left a $100\mu F$ electrolytic capacitor
   2. On the right a $100nF$ MLCC in the commonly used Surface-mount_technology 0603 (1.6mm x 0.8mm)
2. below different THT capacitors (Through Hole Technology)
   1. a big electrolytic capacitor with $10mF$ in blue, the positive terminal is marked with $+$
   2. in the second row is a Kerko with $33pF$ and two Folkos with $1,5\mu F$ each
   3. in the bottom row you can see a trim capacitor with about $30pF$ and a tantalum electrolytic capacitor and another electrolytic capacitor

Various conventions have been established for designating the capacitance value of a capacitor various conventions.

1.8 Circuits with Capacitors

Series Circuit of Capacitor

If capacitors are connected in series, the charging current $I$ into the individual capacitors $C_1 ... C_n$ is equal. Thus, the charges absorbed $\Delta Q$ are also equal: \begin{align*} \Delta Q = \Delta Q_1 = \Delta Q_2 = ... = \Delta Q_n \end{align*}

Furthermore, after charging, a voltage is formed across the series circuit which corresponds to the source voltage $U_q$. This results from the addition of the partial voltages across the individual capacitors. \begin{align*} U_q = U_1 + U_2 + ... + U_n = \sum_{k=1}^n U_k \end{align*}

It holds for the voltage $U_k = \Large{{Q_k}\over{C_k}}$.
If all capacitors are initially discharged, then $U_k = \Large{{\Delta Q}\over{C_k}}$ holds. Thus \begin{align*} U_q &= &U_1 &+ &U_2 &+ &... &+ &U_n &= \sum_{k=1}^n U_k \\ U_q &= &{{\Delta Q}\over{C_1}} &+ &{{\Delta Q}\over{C_2}} &+ &... &+ &{{\Delta Q}\over{C_3}} &= \sum_{k=1}^n {{1}\over{C_k}}\cdot \Delta Q \\ {{1}\over{C_{ges}}}\cdot \Delta Q &= &&&&&&&&\sum_{k=1}^n {{1}\over{C_k}}\cdot \Delta Q \end{align*}

Thus, for the series connection of capacitors $C_1 ... C_n$ :

\begin{align*} \boxed{ {{1}\over{C_{ges}}} = \sum_{k=1}^n {{1}\over{C_k}} } \end{align*} \begin{align*} \boxed{ \Delta Q_k = const.} \end{align*}

For initially uncharged capacitors, (voltage divider for capacitors) holds: \begin{align*} \boxed{Q = Q_k} \end{align*} \begin{align*} \boxed{U_{ges} \cdot C_{ges} = U_{k} \cdot C_{k} } \end{align*}

In the simulation below, besides the parallel connected capacitors $C_1$, $C_2$,$C_3$, an ideal voltage source $U_q$, a resistor $R$, a switch $S$ and a lamp are installed.

• The switch $S$ allows the voltage source to charge the capacitors.
• The resistor $R$ is necessary because the simulation cannot represent instantaneous charging. The resistor limits the charging current to a maximum value.
  This leads to the DC circuit transients, explained in the last semester.
• The capacitors can be discharged again via the lamp.

This derivation is also well explained, for example, in this video.

Parallel Circuit of Capacitors

If capacitors are connected in parallel, the voltage $U$ across the individual capacitors $C_1 ... C_n$ is equal. It is therefore valid:

\begin{align*} U_q = U_1 = U_2 = ... = U_n \end{align*}

Furthermore, during charging, the total charge $\Delta Q$ from the source is distributed to the individual capacitors. This gives the following for the individual charges absorbed: \begin{align*} \Delta Q = \Delta Q_1 + \Delta Q_2 + ... + \Delta Q_n = \sum_{k=1}^n \Delta Q_k \end{align*}

If all capacitors are initially discharged, then $Q_k = \Delta Q_k = C_k \cdot U$
Thus \begin{align*} \Delta Q &= & Q_1 &+ & Q_2 &+ &... &+ & Q_n &= \sum_{k=1}^n Q_k \\ \Delta Q &= &C_1 \cdot U &+ &C_2 \cdot U &+ &... &+ &C_n \cdot U &= \sum_{k=1}^n C_k \cdot U \\ C_{ges} \cdot U &= &&&&&&&& \sum_{k=1}^n C_k \cdot U \\ \end{align*}

Thus, for the parallel connection of capacitors $C_1 ... C_n$ :

For initially uncharged capacitors, (charge divider for capacitors) holds: \begin{align*} \boxed{\Delta Q = \sum_{k=1}^n Q_k} \end{align*}

\begin{align*} \boxed{ {{Q_k}\over{C_k}} = {{\Delta Q}\over{C_{ges}}} } \end{align*}

In the simulation below, again besides the parallel connected capacitors $C_1$, $C_2$,$C_3$, an ideal voltage source $U_q$, a resistor $R$, a switch $S$ and a lamp are installed.

This derivation is also well explained, for example, in this video.

Tasks

1.9 Configurations of multiple Dielectrics

Up to now was assumed only one dielectricum resp. only vacuum within capacitor. Now is looked at more detailed, how multi-layered construction between sheets affects capacity. Thereby several dielectrics build boundary layers between each other. Different variants can be distinguished (Abbildung 37):

1. perpendicular layering: There are different dielectrics perpendicular to the field lines.
   Thus, the boundary layers are parallel to the capacitor plates.
2. lateral layering: There are different dielectrics parallel to the field lines.
   So the boundary layers are perpendicular to the capacitor plates.
3. arbitrary configuration: The boundary layers are neither parallel nor perpendicular to the capacitor plates.

Lateral Configuration

First, the situation is considered that the boundary layers are parallel to the electrode surfaces. A voltage $U$ is applied to the structure from the outside.

The layering is now parallel to equipotential surfaces. In particular, the boundary layers are then also equipotential surfaces.
The boundary layers can be replaced by an infinitesimally thin conductor layer (metal foil). The voltage $U$ can then be divided into several partial areas:

\begin{align*} U = \int \limits_{total \, inner \\ volume} \! \! \vec{E} \cdot d \vec{s} = E_1 \cdot d_1 + E_2 \cdot d_2 + E_3 \cdot d_3 \tag{1.9.1} \end{align*}

Since there are only polarized charges in the dielectrics and no free charges, the $\vec{D}$ field is constant between the electrodes.

\begin{align*} Q = \iint_{A} \vec{D} \cdot d \vec{A} = const. \end{align*}

Now, in the setup, the area $A$ of the boundary layers is also constant. Thus:

\begin{align*} \vec{D_1} \cdot \vec{A} & = & \vec{D_2} \cdot \vec{A} & = & \vec{D_3} \cdot \vec{A} & \quad \quad \quad & | \:\: \vec{D_k} & \parallel \vec{A} \\ D_1 \cdot A & = & D_2 \cdot A & = & D_3 \cdot A & \quad \quad \quad & | \:\: A & = const. \\ D_1 & = & D_2 & = & D_3 & \quad \quad \quad & | D_k & = \varepsilon_{rk} \varepsilon_0 \cdot E_k \\ \varepsilon_{r1} \varepsilon_0 \cdot E_1 &= &\varepsilon_{r2} \varepsilon_0 \cdot E_2 &= &\varepsilon_{r3} \varepsilon_0 \cdot E_3 \\ \end{align*} \begin{align*} \boxed{ \varepsilon_{r1} \cdot E_1 = \varepsilon_{r2} \cdot E_2 = \varepsilon_{r3} \cdot E_3 } \tag{1.9.2} \end{align*}

Using $(1.9.1)$ and $(1.9.2)$ we can also derive the following relationship: \begin{align*} E_2 = & {{\varepsilon_{r1}}\over{\varepsilon_{r2}}}\cdot E_1 , \quad E_3 = {{\varepsilon_{r1}}\over{\varepsilon_{r3}}}\cdot E_1 \\ \end{align*} \begin{align*} U = & E_1 \cdot d_1 + & E_2 & \cdot d_2 + & E_3 & \cdot d_3 \\ U = & E_1 \cdot d_1 + & {{\varepsilon_{r1}}\over{\varepsilon_{r2}}}\cdot E_1 & \cdot d_2 + & {{\varepsilon_{r1}}\over{\varepsilon_{r3}}}\cdot E_1 & \cdot d_3 \\ \end{align*} \begin{align*} U = & E_1 \cdot (d_1 + {{\varepsilon_{r1}}\over{\varepsilon_{r2}}} \cdot d_2 + {{\varepsilon_{r1}}\over{\varepsilon_{r3}}}\cdot d_3 ) \\ E_1 = & {{U}\over{ d_1 + \large{{\varepsilon_{r1}}\over{\varepsilon_{r2}}} \cdot d_2 + \large{{\varepsilon_{r1}}\over{\varepsilon_{r3}}}\cdot d_3 }} \end{align*} \begin{align*} \boxed{ E_1 = {{U}\over{ \sum_{k=1}^n \large{{\varepsilon_{r1}}\over{\varepsilon_{rk}}} \cdot d_k}} } \quad \text{and} \; E_k = {{\varepsilon_{r1}}\over{\varepsilon_{rk}}}\cdot E_1 \end{align*}

The situation can also be transferred to a coaxial structure of a cylindrical capacitor or concentric structure of spherical capacitors.

Perpendicular Configuartion

Now the boundary layers should be perpendicular to the electrode surfaces. Again a voltage $U$ is applied to the structure from the outside.

The layering is now perpendicular to equipotential surfaces. However, the same voltage is applied to each dielectric. Thus it is valid:

\begin{align*} U = \int \limits_{total \, inner \\ volume} \! \! \vec{E} \cdot d \vec{s} = E_1 \cdot d = E_2 \cdot d = E_3 \cdot d \end{align*}

Since $d$ is the same for all dielectrics, $\large{ E_1 = E_2 = E_3 = {{U}\over{d}} }$

with the electric flux density $D_k = \varepsilon_{rk} \varepsilon_{0} \cdot E_k$ results:

\begin{align*} { { D_1 } \over { \varepsilon_{r1} } } = { { D_2 } \over { \varepsilon_{r2} } } = { { D_3 } \over { \varepsilon_{r3} } } = { { D_k } \over { \varepsilon_{rk} } } \end{align*}

Since the electric flux density is just equal to the local surface charge density, the charge will no longer be uniformly distributed over the electrodes.
Where a stronger polarization is possible, the $E$-field is damped in the dielectric. For a constant $E$-field, more charges must accumulate there.
Concretely, more charges accumulate just around the dielectric constant $\varepsilon_{rk}$.

This situation can also be transferred to a coaxial structure of a cylindrical capacitor or concentric structure of spherical capacitors.

Arbitrary Configuration

With arbitrary configuration, simple observations are no longer possible.
However, some hints can be derived from the previous types of layering:

• Electric field $\vec{E}$:
  • The normal component $E_{n}$ is discontinuous at the interface: $\varepsilon_{r1} \cdot E_{n1} = \varepsilon_{r2} \cdot E_{n2}$
  • The parallel component $E_{p}$ is continuous at the interface: $ E_{p1} = E_{p2}$
• Electric displacement flux density $\vec{D}$:
  • The normal component $D_{n}$ is continuous at the interface: $ D_{n1} = D_{n2}$
  • The parallel component $D_{p}$ is discontinuous at the interface: $ {{1}\over \Large{\varepsilon_{r1}}}\cdot D_{p1} = {{1}\over \Large{\varepsilon_{r2}}} \cdot D_{p1} $

Since $\vec{D} = \varepsilon_{0} \varepsilon_{r} \cdot \vec{E}$ the direction of the fields must be the same.
Using the fields, we can now derive the change in the angle:

\begin{align*} \boxed { { { tan \alpha_1 } \over { tan \alpha_2 } } = { { \varepsilon_{r1} } \over { \varepsilon_{r2} } } } \end{align*}

The formula obtained represents the law of refraction of the field line at interfaces. There is also a hint that for electromagnetic waves (like visible light) the refractive index might depend on the dielectric constant. In fact, this is the case. However, in the calculation presented here, electrostatic fields were assumed. In the case of electromagnetic waves, the distribution of energy between the two fields must be taken into account. This is not considered in detail in this course.

Different dielectrics in the capacitor

Tasks

1.10 Summary

Further links

• Online Bridge Course Physics KIT: This semi-interactive course contains some of the information from my course. Furthermore, videos, exercises and more can be found there

additional Links

Illustrative and interactive examples

• What is an Electric Field?
• Conducting and Insulating Sphere
• Work and Equipotential Surfaces
• Capacitors, Charge, and Electric Potential

A really great introduction in electric and magnetic fields (but a bit too deep for this course) can be found in the physics lecture of Walter Lewin

examples: