Well, again
For checking your understanding please do the following exercises:
$\rightarrow$ 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:
The capacitance $C$ of an idealized plate capacitor is defined as
\begin{align*} \boxed{C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{A}\over{l}} = {{Q}\over{U}}} \end{align*}
Some of the main results here are:
This relationship can be examined in more detail in the following simulation:
The Abbildung 1 shows the topology of the electric field inside a plate capacitor.
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 2).
| Shape of the Capacitor | Parameter | Equation for the Capacity |
|---|---|---|
| plate capacitor | area $A$ of plate distance $l$ between plates | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{A}\over{l}} \end{align*} |
| cylinder capacitor | radius of outer conductor $R_{ \rm o}$ radius of inner conductor $R_{ \rm i}$ length $l$ | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot 2\pi {{l}\over{{\rm ln} \left({{R_{ \rm o}}\over{R_{ \rm i}}}\right)}} \end{align*} |
| spherical capacitor | radius of outer spherical conductor $R_{ \rm o}$ radius of inner spherical conductor $R_{ \rm i}$ | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot 4 \pi {{R_{ \rm i} \cdot R_{ \rm o}}\over{R_{ \rm o} - R_{ \rm i}}} \end{align*} |
In Abbildung 3 different designs of capacitors can be seen:
In Abbildung 2 are shown different capacitors:
Various conventions have been established for designating the capacitance value of a capacitor various conventions.
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.
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 $d_1 = 7 ~{ \rm cm}$.
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 $U_{ \rm a}$.
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 \cdot E_{\rm a} \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} &= {{5'000 ~\rm V}\over{0.004 ~{\rm m} + 8 \cdot 0.002 ~{\rm m}}} \\ &= 250 ~\rm{{kV}\over{m}} \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} &= 4 ~{\rm kV}\\ \end{align*}
2. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$?
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} \cdot ({{1}\over{\varepsilon_{\rm r,g}}} - 1) + d_0 \\ d_{\rm g} &= { { {{U_0}\over{E_{\rm a} }} - d_0 } \over { {{1}\over{\varepsilon_{\rm r,g}}} - 1 } } &= { { d_0 - {{U_0}\over{E_{\rm a} }} } \over { 1 - {{1}\over{\varepsilon_{\rm r,g}}} } } \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) {~\rm m} \end{align*}
Two concentric spherical conducting plates set up a spherical capacitor. The radius of the inner sphere is $r_{ \rm i} = 3~{ \rm mm}$, and the inner radius from the outer sphere is $r_{ \rm o} = 9~{ \rm mm}$.
The background behind the dielectric constant $\varepsilon_{ \rm r} $ and the field is explained in the following video
Electrolytic capacitors can explode!