Capacitance and Dielectrics

Capacitance¶

The capacitance of parallel plate capacitor is \(\(C = \frac{q}{\Delta V} = \frac{\sigma A}{\dfrac{\sigma}{\epsilon_0} d} = \frac{\epsilon_0 A}{d}\)\)

The capacitance of cylindrical capacitor (\(V = 0\) at \(r = b\)) is

\[ \begin{gathered} \Delta V = \int_a^b \frac{Q}{2\pi\epsilon_0 rL}dr = \frac{Q}{2\pi\epsilon_0 L} \ln{\frac{b}{a}} \\\\ C = \frac{Q}{\Delta V} = \frac{2\pi\epsilon_0 L}{\ln(b/a)} \end{gathered} \]

The capacitance of spherical capacitor is

\[ \begin{gathered} \Delta V = \int_a^b \frac{Q}{4\pi\epsilon_0 r^2}dr = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right) \\\\ C = \frac{Q}{\Delta V} = 4\pi\epsilon_0 \frac{ab}{b - a} \end{gathered} \]

When \(b \rightarrow \infty\), \(C = 4\pi\epsilon_0 a\).

When capacitors are in parallel, we have \(\(C = \frac{Q_1 + Q_2}{V} = \frac{C_1V + C_2V}{V} = C_1 + C_2\)\)

When capacitors are in series, we have \(\(\frac{Q}{C} = \frac{Q}{C_1} + \frac{Q}{C_2} \Rightarrow \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\)\)

The total work to charge a capacitor to \(Q\) is \(\(W = \int_0^Q \frac{q}{C} dq = \frac{Q^2}{2C} = \frac{1}{2} CV^2\)\)

Pull the plates apart from \(d\) to \(d'\). When the battery is disconnected, then \(\(C' = \frac{d}{d'}C, V' = \frac{d'}{d}V, U' = \frac{d'}{d}U\)\)

When the battery is connected, then \(\(C' = \frac{d}{d'}C, E' = \frac{d}{d'}E, U' = \frac{d}{d'}U\)\)

Energy is stored in the electric field. For a parallel plate capacitor, we have

\[ \begin{gathered} C = \frac{\epsilon_0 A}{d}, \quad E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A} \\\\ \Rightarrow U = \frac{Q^2}{2C} = \frac{1}{2} E^2\epsilon_0 Ad \end{gathered} \]

So the energy density is \(\(u = \frac{U}{Ad} = \frac{1}{2}\epsilon_0 E^2\)\)

It's also the force per unit area (the electrostatic stress) acting on the plates. Actually, this result is true, in general, for a conductor of any shape with an electric field \(E\) at its surface.

Dielectrics¶

The dielectric constant of a material is defined as \(\(\kappa_e = \frac{C}{C_0}\)\)

which is always greater than 1, e.g., glass = 5.6, water = 78.

Insert material with dielectric constant \(\kappa_e\) into parallel plates, when \(Q\) remains constant, we have \(\(V = \frac{Q}{\kappa_e C_0} = \frac{V_0}{\kappa_e}, \quad E = \frac{V_0}{\kappa_e d} = \frac{E_0}{\kappa_e}\)\)

when \(V\) remains constant, we have \(\(Q = \kappa_e C_0V = \kappa_e Q_0\)\)

The non-polar dielectrics in an electric field will generate induced electric dipole moment and undergo electric displacement polarization.

The polar dielectrics in an electric field will undergo alignment polarization.

The intensity of polarization is \(\(\vec{P} = \frac{\sum \vec{p}_m}{\Delta V}\)\)

Suppose \(n\) is the number of molecular per unit volume, then we have

\[ \begin{gathered} dq = qdN = q(ndV) = nqldA\cos\theta = PdA\cos\theta \\\\ \Rightarrow \iint \vec{P} \bullet d\vec{A} = \sum_{out} q = - \sum_{in} q \\\\ \Rightarrow \sigma = \frac{dq}{dA} = \vec{P} \bullet \vec{n} = P_n \end{gathered} \]

For a spherical dielectrics with uniform polarization \(\vec{P}\), the depolarization field at center is

\[ \begin{gathered} \begin{aligned} dE &= \frac{dq}{4\pi\epsilon_0 R^2} \cdot \cos\theta = \frac{\sigma dA}{4\pi\epsilon_0 R^2} \cdot \cos\theta \\\\ &= \frac{P\cos\theta \cdot 2\pi R\sin\theta \cdot Rd\theta}{4\pi\epsilon_0 R^2} \cdot \cos\theta \\\\ &= \frac{P\cos^2\theta\sin\theta}{2\epsilon_0} d\theta \\ \end{aligned} \\\\ E = - \frac{P}{2\epsilon_0} \int_0^{\pi} \cos^2\theta\sin\theta d\theta = - \frac{P}{3\epsilon_0} \end{gathered} \]

Define the electric displacement vector \(D = \epsilon_0E + P\), we have

\[ \begin{gathered} \epsilon_0 \iint E \bullet dA = \sum_{in}(q_0 + q') \\\\ \Rightarrow \iint D \bullet dA = \sum_{in}q_0 \end{gathered} \]

For general isotropic materials, there are \(P = \chi_e\epsilon_0 E\). Then we have

\[ \begin{gathered} D = \epsilon_0E + P = \epsilon_0E + \chi_e\epsilon_0E = (1 + \chi_e)\epsilon_0E = \kappa_e\epsilon_0E \\\\ \Rightarrow \kappa_e = 1 + \chi_e \end{gathered}\]

where \(\kappa_e\) is dielectric constant, \(\chi_e\) is polarization coefficient.

For a parallel plate capacitor, we have

\[ \begin{gathered} D_1\Delta A + D_2\Delta A = \sigma_{e0} \Delta A \\\\ E_1 = 0 \quad\Rightarrow\quad D_1 = \kappa_{e1}\epsilon_0E_1 = 0 \\\\ \Rightarrow D = D_2 = \sigma_{e0} = \epsilon_0E_0 \\\\ \Rightarrow E = \frac{D}{\kappa_e\epsilon_0} = \frac{E_0}{\kappa_e} \end{gathered} \]

For a dielectric sphere with a point charge \(q_0\) at center, we have \(D_{in} = D_{out} = \dfrac{q_0}{4\pi r^2}\), but \(E_{in} = \dfrac{D_{in}}{\kappa_e\epsilon_0} = \dfrac{E_{out}}{\kappa_e}\), and therefore \(\sigma_{ind} = \epsilon_0(E_{out} - E_{in}) = P\).

Shorthand¶

Capacitance of parallel plate capacitor
Capacitance of cylindrical capacitor
Capacitance of spherical capacitor
When the battery is disconnected and pull apart the plates, \(C\downarrow\), \(V\uparrow\), \(U\uparrow\)
When the battery is connected and pull apart the plates, \(C\downarrow\), \(Q\downarrow\), \(U\downarrow\)
Energy density: \(u = \dfrac{1}{2} \epsilon_0 E^2\)