Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_and_electronics_1:block11 [2025/10/31 11:58] – angelegt mexleadmin | electrical_engineering_and_electronics_1:block11 [2025/11/02 16:54] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== Block 11 — Influence and displacement field ====== | + | ===== Block 11 — Influence and Displacement Field ====== |
| ===== Learning objectives ===== | ===== Learning objectives ===== | ||
| < | < | ||
| After this 90-minute block, you can | After this 90-minute block, you can | ||
| - | * ... | + | * explain **electrostatic induction** on conductors and argue why the interior of a conductor is field-free (Faraday cage). |
| + | * distinguish the **electric field strength** $\vec{E}$ from the **electric displacement flux density** $\vec{D}$ and state $ \vec{D} = \varepsilon \vec{E} = \varepsilon_0 \varepsilon_{\rm r}\vec{E}$. | ||
| + | * apply **Gauss’s law** for the displacement field to simple closed surfaces to relate enclosed charge $Q$ and flux $\oint \vec{D}\cdot {\rm d}\vec{A}$. | ||
| + | * determine $E(r)$ for parallel-plate and coaxial geometries starting from $\vec{D}$, then using $\vec{E}=\vec{D}/ | ||
| + | * reason about **surface charge density** $\varrho_A = \Delta Q/\Delta A$ and the normal field at conductor surfaces. | ||
| + | * use typical **relative permittivities** $\varepsilon_{\rm r}$ to estimate field reduction in dielectrics. | ||
| + | * interpret **dielectric strength** $E_0$ (breakdown) and reason about its impact on design limits (safe $E$, spacing, material choice). | ||
| </ | </ | ||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| ===== Preparation at Home ===== | ===== Preparation at Home ===== | ||
| Zeile 16: | Zeile 23: | ||
| * ... | * ... | ||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| ===== 90-minute plan ===== | ===== 90-minute plan ===== | ||
| - | - Warm-up (x min): | + | - Warm-up (10 min): |
| - | - .... | + | - One-minute recap quiz (Block 10): equipotentials, |
| - | - Core concepts & derivations (x min): | + | - Demo: conductor in external field → Faraday cage effect (refer to the embedded sim in this block). |
| - | - ... | + | - Core concepts & derivations (45 min): |
| - | - Practice (x min): ... | + | - Induction on conductors: charge displacement, |
| - | - Wrap-up (x min): Summary box; common | + | - Polarization of dielectrics; |
| + | - Definitions: | ||
| + | - Worked derivations via $\vec{D}$: | ||
| + | * Parallel plates: $D=Q/A \; | ||
| + | * Coaxial cylinders: $D(r)=Q/ | ||
| + | - Material data: typical $\varepsilon_{\rm r}$; concept of **dielectric strength** $E_0$ and safe design margins. | ||
| + | - Practice (25 min): | ||
| + | - Short board tasks using pillbox surfaces to find $\varrho_A$ on a conductor. | ||
| + | - Mixed-dielectric capacitor slice: split voltages via constant $D$. | ||
| + | - Guided use of the embedded sims to observe field/ | ||
| + | - Wrap-up (10 min): | ||
| + | - Summary box (key formulas, when to start with $D$ vs. $E$). | ||
| + | - Common | ||
| + | - Preview to Block 12 (capacitors from field viewpoint). | ||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| ===== Conceptual overview ===== | ===== Conceptual overview ===== | ||
| <callout icon=" | <callout icon=" | ||
| - | - ... | + | - **Conductors in electrostatics: |
| + | - **Dielectrics (polarization): | ||
| + | - **Displacement field & Gauss’s law:** for any closed surface, the flux of $\vec{D}$ equals the enclosed charge: $Q=\oint \vec{D}\cdot{\rm d}\vec{A}$. \\ Choose the surface to exploit symmetry, get $\vec{D}$ first, then $\vec{E}$ via material law. | ||
| + | - **Permittivity: | ||
| + | - **Design limit:** when $|E|$ exceeds the **dielectric strength** $E_0$, breakdown occurs → current flows. \\ Safe design keeps $|E|\ll E_0$ by material choice and geometry (spacing, shaping to avoid high curvature). | ||
| </ | </ | ||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| ===== Core content ===== | ===== Core content ===== | ||
| - | ... | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| + | ==== Electric Field inside of a conductor ==== | ||
| + | As seen in [[https:// | ||
| + | |||
| + | Wo we want to have a look onto an uncharged object in an external field. Also here any hole inside does not show an electric field \\ | ||
| + | The reason for that is, that the outer field gets cancelled out by an opposing inner electric field. \\ | ||
| + | A charge displacement on the external surface (induced by the external field) is the reason for that opposing inner field. \\ | ||
| + | Please have a look onto the yellow and blue collor in <imgref ImgNr193> | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | <callout icon=" | ||
| + | |||
| + | Any external electric field causes a charge displacement on an conductor in such a way, that there in no internal field inside the material (neither in holes nor in the material itself) | ||
| + | </ | ||
| + | |||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| + | ==== Electrostatic Induction ==== | ||
| + | |||
| + | In a thought experiment, an uncharged conductor (e.g., a metal plate) is brought into an electrostatic field (<imgref ImgNr11> | ||
| + | The external field or the resulting Coulomb force causes the moving charge carriers to be displaced. \\ | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | < | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | </ | ||
| + | |||
| + | Results: | ||
| + | * The charge carriers are still distributed on the surface. | ||
| + | * Now, equilibrium is reached when just so many charges have moved, that the electric field inside the conductor disappears (again). | ||
| + | * The field lines leave the surface again at right angles. Again, a parallel component would cause a charge shift in the metal. | ||
| + | |||
| + | This effect of charge displacement in conductive objects by an electrostatic field is referred to as **electrostatic induction** (in German: // | ||
| + | Induced charges can be separated (<imgref ImgNr11> right). | ||
| + | If we look at the separated induced charges without the external field, their field is again just as strong in magnitude as the external field only in the opposite direction. | ||
| + | |||
| + | <callout icon=" | ||
| + | - The location of an induced charge is always on the conductor surface. This results in a surface charge density $\varrho_A = {{\Delta Q}\over{\Delta A}}$ | ||
| + | - The conductor surface in the electrostatic field is always an equipotential surface. Thus, the field lines always originate and terminate perpendicularly on conductor surfaces. | ||
| + | - The interior of the conductor is always field-free (Faraday effect: metallic enclosures shield electric fields). | ||
| + | </ | ||
| + | |||
| + | How can the conductor surface be an equipotential surface despite different charges on both sides? | ||
| + | Equipotential surfaces are defined only by the fact that the movement of a charge along such a surface does not require/ | ||
| + | Since the interior of the conductor is field-free, movement there can occur without a change in energy. | ||
| + | As the potential between two points is independent of the path between them, a path along the surface is also possible without energy expenditure. | ||
| + | |||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| + | ==== Electric Field inside of an Isolator ==== | ||
| + | |||
| + | But how is it like for an isolator in an external field? \\ | ||
| + | There are no free charges in an isolator - so, is there no compensation of the external field inside the isolator at all? | ||
| + | |||
| + | The simulation in <imgref ImgNr194> | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | So, let us look into an isolator: The <imgref ImgNr195> | ||
| + | Most of the isolators have a bipolar property (e.g. {{https:// | ||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | So the root-cause-path is: | ||
| + | - We have initially the external electric field. | ||
| + | - This induces the counteracting field by bending the dipoles. | ||
| + | - The internal measurable electric field is compensated | ||
| + | |||
| + | To have an uncompensated field in the following the **electric displacement flux density $\vec{D}$** is introduced. | ||
| + | The electric displacement flux density is only focusing on the __cause__ of the electric fields. | ||
| + | As we have seen, its effect can differ since the space can also " | ||
| + | |||
| + | The electric displacement flux density is only related to the causing charges $Q$. Thie relationship is shown in the following. | ||
| + | |||
| + | \begin{align} | ||
| + | \boxed{Q = \int {\rm d}Q = {\rlap{\rlap{\int_A} \int} \: \LARGE \circ} \vec{D} \cdot {\rm d} \vec{A} } | ||
| + | \end{align} | ||
| + | |||
| + | The symbol ${\rlap{\Large \rlap{\int} \int} \, \LARGE \circ}$ denotes that there is a closed surface used for the integration. | ||
| + | |||
| + | The meaning of the formula is: The " | ||
| + | |||
| + | This can be compared with a bordered swamp area with water sources and sinks: | ||
| + | * The sources in the marsh correspond to the positive charges, and 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. | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| + | ==== Dielectric Constant (Permittivity) ==== | ||
| + | |||
| + | Dielectric materials reduce the electric field inside them. How much die field is reduced is given by a material dependent constant the **dielectric constant** or **permittivity** $\varepsilon_r$. It is unitless and a ratio related to the unhindered field in vacuum. | ||
| + | |||
| + | \begin{align*} | ||
| + | {{D}\over{E}} = \varepsilon = \varepsilon_0 \cdot \varepsilon_{ \rm r} \\ | ||
| + | \boxed{D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E} | ||
| + | \end{align*} | ||
| + | |||
| + | Some values of the relative permittivity $\varepsilon_{ \rm r}$ for dielectrics are given in <tabref tab01> | ||
| + | |||
| + | |||
| + | <WRAP 30em> | ||
| + | |||
| + | < | ||
| + | ^ material | ||
| + | | air | $\rm 1.0006$ | ||
| + | | paper | $\rm 2$ | | ||
| + | | PE, PP | $\rm 2.3$ | | ||
| + | | PS | $\rm 2.5$ | | ||
| + | | hard paper | $\rm 5$ | | ||
| + | | glass | $\rm 6...8$ | ||
| + | | water ($20~°{ \rm C}$) | $\rm 80$ | | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| + | ==== Typical Geometries ==== | ||
| + | |||
| + | The " | ||
| + | This shall be shown with the two most common geometries (which are the only one necessary for this course). | ||
| + | |||
| + | <WRAP group> | ||
| + | <WRAP half column> | ||
| + | === Field of a parallel Plates === | ||
| + | \\ | ||
| + | * Nearly all of the field is between the plate (see <imgref ImgNr294> | ||
| + | * All of the $D$-field of the charges is between the plates, and therefore through the area $A$ of the plates (see <imgref ImgNr294> | ||
| + | * Given $D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E$, the eletric field $E$ is: $$ E = {{Q}\over{ \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot A}} $$\\ | ||
| + | |||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | | ||
| + | |||
| + | {{drawio> | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | === Field of a coaxial cylindrical Plates === | ||
| + | \\ | ||
| + | * All of the field is between the plate (see <imgref ImgNr295> | ||
| + | * All of the $D$-field of the charges on the inner plate penetrates through any cylintrical area $A(l,r) = 2 \pi \cdot l \cdot r$. \\ → The $D$-Field is given as: $$ D = {{Q}\over{A(l, | ||
| + | * Again given $D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E$, the eletric field $E$ is: $$ E = {{Q}\over{ \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot 2 \pi \cdot l \cdot r}} $$\\ | ||
| + | |||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | | ||
| + | |||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| + | ==== 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, | ||
| + | * The maximum electric field $E_0$ is referred to as ** dielectric strength** (in German: // | ||
| + | * $E_0$ depends on the material (see <tabref tab02>), but also on other factors (temperature, | ||
| + | |||
| + | <WRAP 30em> | ||
| + | < | ||
| + | ^ Material | ||
| + | | air | $\rm 0.1...0.3$ | ||
| + | | SF6 gas | $\rm 8$ | | ||
| + | | insulating oils | $\rm 5...30$ | ||
| + | | vacuum | ||
| + | | quartz | ||
| + | | PP, PE | $\rm 50$ | | ||
| + | | PS | $\rm 100$ | | ||
| + | | distilled water | $\rm 70$ | | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| ===== Common pitfalls ===== | ===== Common pitfalls ===== | ||
| - | * ... | + | * Mixing up **cause** and **effect**: using $\oint \vec{E}\cdot{\rm d}\vec{A}$ to count charge. Use **$\vec{D}$** for Gauss’s law with charge; convert to $\vec{E}$ only via $\vec{E}=\vec{D}/ |
| + | * Forgetting that the **interior of a conductor is field-free** in electrostatics and that $E$ is **normal** to an ideal conducting surface (no tangential $E$ on the surface). | ||
| + | * Assuming induced charges fill the **volume** of a conductor. They reside on the **surface**; | ||
| + | * Ignoring that **$D$ is continuous** in the normal direction across simple dielectric interfaces when no free surface charge is present; consequently, | ||
| + | * Treating $\varepsilon_{\rm r}$ as a constant in all contexts. Real materials can be frequency/ | ||
| + | * Checking breakdown with voltage only. The limit is on **field** $E$; always relate geometry (e.g., plate spacing, curvature) to $E$ and compare to **$E_0$** with units (e.g., $\,{\rm kV/mm}$). | ||
| ===== Exercises ===== | ===== Exercises ===== | ||
| - | ==== Worked | + | ==== Tasks ==== |
| + | |||
| + | <panel type=" | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | </ | ||
| + | |||
| + | In the simulation in <imgref ImgNr198>, | ||
| + | In the beginning, the situation of an infinitely long cylinder in a homogeneous electric field is shown. | ||
| + | The solid lines show the equipotential surfaces. The small arrows show the electric field. | ||
| + | |||
| + | - What is the angle between the field on the surface of the cylinder? | ||
| + | - Once the option '' | ||
| + | - What can be said about the potential distribution on the cylinder? | ||
| + | - On the left half of the field lines enter the body, on the right half, they leave the body. What can be said about the charge carrier distribution at the surface? Check also the representation '' | ||
| + | - Is there an electric field inside the body? | ||
| + | - Is this cylinder metallic, semiconducting, | ||
| + | |||
| + | </ | ||
| + | |||
| + | {{page> | ||
| + | {{page> | ||
| + | {{page> | ||
| + | |||
| + | <wrap anchor # | ||
| + | <panel type=" | ||
| + | |||
| + | Given is the two-dimensional component shown in <imgref ImgNr148> | ||
| + | In the picture, there are 4 positions marked with numbers. \\ \\ | ||
| + | |||
| + | Order the numbered positions by increasing charge density! | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | |||
| + | # | ||
| + | |||
| + | $\varrho_2 < \varrho_3 < \varrho_1 < \varrho_4$ | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | |||
| + | </ | ||
| + | |||
| + | # | ||
| + | |||
| + | </ | ||
| + | |||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | A plate capacitor with a distance of $d = 2 ~{ \rm cm}$ between the plates and with air as dielectric ($\varepsilon_{ \rm r}=1$) gets charged up to $U = 5~{ \rm kV}$. | ||
| + | In between the plates, a thin metal foil with the area $A = 45~{ \rm cm^2}$ is introduced parallel to the plates. | ||
| + | |||
| + | Calculate the amount of the displaced charges in the thin metal foil. | ||
| + | |||
| + | <button size=" | ||
| + | * What is the strength of the electric field $E$ in the capacitor? | ||
| + | * Calculate the displacement flux density $D$ | ||
| + | * How can the charge $Q$ be derived from $D$? | ||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | $Q = 10 ~{ \rm nC}$ | ||
| + | </ | ||
| + | |||
| + | </ | ||
| + | |||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | An ideal plate capacitor with a distance of $d_0 = 7 ~{ \rm mm}$ between the plates gets charged up to $U_0 = 190~{ \rm V}$ by an external source. | ||
| + | The source gets disconnected. After this, the distance between the plates gets enlarged to $d_1 = 7 ~{ \rm cm}$. | ||
| + | |||
| + | - What happens to the electric field and the voltage? | ||
| + | - How does the situation change (electric field/ | ||
| + | |||
| + | <button size=" | ||
| + | * Consider the displacement flux through a surface around a plate | ||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | - $U_1 = 1.9~{ \rm kV}$, $E_1 = 27~{ \rm kV/m}$ | ||
| + | - $U_1 = 190~{ \rm V}$, $E_1 = 2.7~{ \rm kV/m}$ | ||
| + | </ | ||
| + | |||
| + | </ | ||
| + | |||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | An ideal plate capacitor with a distance of $d_0 = 6 ~{ \rm mm}$ between the plates and with air as dielectric ($\varepsilon_0=1$) is charged to a voltage of $U_0 = 5~{ \rm kV}$. | ||
| + | The source remains connected to the capacitor. In the air gap between the plates, a glass plate with $d_{ \rm g} = 4 ~{ \rm mm}$ and $\varepsilon_{ \rm r} = 8$ is introduced parallel to the capacitor plates. | ||
| + | |||
| + | 1. Calculate the partial voltages on the glas $U_{ \rm g}$ and on the air gap $U_{ \rm a}$. | ||
| + | |||
| + | # | ||
| + | * Build a formula for the sum of the voltages first | ||
| + | * How is the voltage related to the electric field of a capacitor? | ||
| + | # | ||
| + | |||
| + | # | ||
| + | |||
| + | The sum of the voltages across the glass and the air gap gives the total voltage $U_0$, and each individual voltage is given by the $E$-field in the individual material by $E = {{U}\over{d}}$: | ||
| + | \begin{align*} | ||
| + | U_0 &= U_{\rm g} + U_{\rm a} \\ | ||
| + | &= E_{\rm g} \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a} | ||
| + | \end{align*} | ||
| + | |||
| + | The displacement field $D$ must be continuous across the different materials since it is only based on the charge $Q$ on the plates. | ||
| + | \begin{align*} | ||
| + | D_{\rm g} &= D_{\rm a} \\ | ||
| + | \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} &= \varepsilon_0 | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore, we can put $E_\rm a= \varepsilon_{\rm r, g} \cdot E_\rm g $ into the formula of the total voltage and rearrange to get $E_\rm g$: | ||
| + | \begin{align*} | ||
| + | U_0 &= E_{\rm g} \cdot d_{\rm g} + \varepsilon_{\rm r, g} \cdot E_{\rm g} \cdot d_{\rm a} \\ | ||
| + | &= E_{\rm g} \cdot ( d_{\rm g} + \varepsilon_{\rm r, g} \cdot d_{\rm a}) \\ | ||
| + | |||
| + | \rightarrow E_{\rm g} &= {{U_0}\over{d_{\rm g} + \varepsilon_{\rm r, g} \cdot d_{\rm a}}} | ||
| + | \end{align*} | ||
| + | |||
| + | Since we know that the distance of the air gap is $d_{\rm a} = d_0 - d_{\rm a}$ we can calculate: | ||
| + | \begin{align*} | ||
| + | E_{\rm g} &= {{5' | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | By this, the individual voltages can be calculated: | ||
| + | \begin{align*} | ||
| + | U_{ \rm g} &= E_{\rm g} \cdot d_\rm g &&= 250 ~\rm{{kV}\over{m}} \cdot 0.004~\rm m &= 1 ~{\rm kV}\\ | ||
| + | U_{ \rm a} &= U_0 - U_{ \rm g} &&= 5 ~{\rm kV} - 1 ~{\rm kV} & | ||
| + | |||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | |||
| + | # | ||
| + | $U_{ \rm a} = 4~{ \rm kV}$, $U_{ \rm g} = 1 ~{ \rm kV}$ | ||
| + | # | ||
| + | |||
| + | |||
| + | 2. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$? | ||
| + | |||
| + | # | ||
| + | Again, we can start with the sum of the voltages across the glass and the air gap, such as the formula we got from the displacement field: $D = \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} = \varepsilon_0 | ||
| + | Now we shall eliminate $E_\rm g$, since $E_\rm a$ is given in the question. | ||
| + | \begin{align*} | ||
| + | U_0 & | ||
| + | &= {{E_\rm a}\over{\varepsilon_{\rm r, | ||
| + | \end{align*} | ||
| + | |||
| + | The distance $d_\rm a$ for the air is given by the overall distance $d_0$ and the distance for glass $d_\rm g$: | ||
| + | \begin{align*} | ||
| + | d_{\rm a} = d_0 - d_{\rm g} | ||
| + | \end{align*} | ||
| + | |||
| + | This results in: | ||
| + | \begin{align*} | ||
| + | U_0 &= {{E_{\rm a}}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + E_{\rm a} \cdot (d_0 - d_{\rm g}) \\ | ||
| + | {{U_0}\over{E_{\rm a} }} &= {{1}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + d_0 - d_{\rm g} \\ | ||
| + | & | ||
| + | d_{\rm g} &= { { {{U_0}\over{E_{\rm a} }} - d_0 } \over { {{1}\over{\varepsilon_{\rm r,g}}} - 1 } } & | ||
| + | \end{align*} | ||
| + | |||
| + | With the given values: | ||
| + | \begin{align*} | ||
| + | d_{\rm g} &= { { 0.006 {~\rm m} - {{5 {~\rm kV} }\over{ 12 {~\rm kV/cm}}} } \over { 1 - {{1}\over{8}} } } &= { {{8}\over{7}} } \left( { 0.006 - {{5 }\over{ 1200}} } \right) | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | |||
| + | # | ||
| + | $d_{ \rm g} = 2.10~{ \rm mm}$ | ||
| + | # | ||
| + | |||
| + | </ | ||
| + | |||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | {{youtube> | ||
| + | |||
| + | </ | ||
| + | |||
| - | ... | ||
| ===== Embedded resources ===== | ===== Embedded resources ===== | ||
| <WRAP column half> | <WRAP column half> | ||
| - | Explanation (video): ... | + | Application of electrostatic induction: Protective bag against electrostatic charge/ |
| + | {{youtube> | ||
| </ | </ | ||