Capacitors

Capacitors are devices that store electrical energy. The charge is proportional to the voltage between them:

\begin{align*} Q &= C|U|, \\ C &= \frac{Q}{|U|}. \end{align*}

The unit of capacitance is farad ( $F$ ).

Capacitance of Two Parallel Planes

Consider two infinite parallel planes at a distance $d$ with opposite charges of equal magnitude $Q$ :

The electric field between the planes as derived in Gauss's law chapter is equal to:

\begin{align*} \boldsymbol{E} &= -\frac{\sigma}{\epsilon_0}\boldsymbol{\hat{z}}, \\ E &= \frac{\sigma}{\epsilon_0}. \end{align*}

The voltage is equal to:

\begin{align*} U &= - \int_C \boldsymbol{E} \cdot d\boldsymbol{s} \\ &= -Ed, \\ |U| &= Ed. \end{align*}

The capacitance is equal to:

\begin{align*} C &= \frac{Q}{|U|} \\ &= \frac{Q}{Ed} \\ &= \frac{Q \epsilon_0}{\sigma d} \\ &= \frac{A \epsilon_0}{d}, \end{align*}

where $A$ is the area of each of the planes.

Cylindrical Capacitor

Cylindrical capacitor consists of a cylinder of radius $a$ and a cylindrical shell of radius $b$ of equal length $L$ . The inner cylinder has a positive charge $Q$ and the outer shell has a negative charge $-Q$ :

We choose cylinder with radius $r$ and equal length $L$ as the gaussian surface:

The electric field is constant on the cylinder. Using Gauss's law to get the electric field:

\begin{align*} \Phi_E &= \oiint_S \boldsymbol{E} \cdot d\boldsymbol{A} \\ &= E A \\ &= 2 E \pi r L = \frac{q}{\epsilon_0}, \\ E &= \frac{q}{2 \epsilon_0 \pi r L}, \end{align*}

where $q$ is the charge enclosed by the gaussian surface (black cylinder). If $r < a$ then $q = 0$ . If $r > b$ then $q~=~Q~+~(-Q) = 0$ . This implies $a < r < b$ and thus $q = Q$ . The voltage is equal to:

\begin{align*} U &= \int_C \boldsymbol{E} \cdot d\boldsymbol{s} \\ &= \frac{Q}{2 \epsilon_0 \pi L} \int_a^b \frac{1}{r} dr \\ &= \frac{Q}{2 \epsilon_0 \pi L} \left[\ln r\right]_a^b \\ &= \frac{Q}{2 \epsilon_0 \pi L} \ln \frac{b}{a} = |U|. \end{align*}

The capacitance is equal to:

\begin{align*} C &= \frac{Q}{|U|} \\ &= \frac{2 \epsilon_0 \pi L}{\ln \frac{b}{a}} \end{align*}

Spherical Capacitor

Spherical capacitor consists of two cocentric spheres. The inner sphere has radius $a$ and a positive charge $Q$ . The outer sphere has a negative charge $-Q$ and a radius $b$ . We choose a cocentric sphere of radius $r$ as our gaussian surface:

The electric field is constant on the surface of the gaussian surface. Using Gauss's law, the electric field is equal to:

\begin{align*} \phi_E &= \oiint_S \boldsymbol{E} \cdot d \boldsymbol{A} \\ &= EA \\ &= 4 E \pi r^2 \\ &= \frac{q}{\epsilon_0}, \\ E &= \frac{q}{4 \epsilon_0 \pi r^2}, \end{align*}

where $q$ is the charged enclosed by the gaussian surface. Similarly to the cylindrical capacitor, the charge is nonzero only when $a < r < b$ , thus $q = Q$ . We can now calculate the voltage:

\begin{align*} U &= \int_C \boldsymbol{E} \cdot d\boldsymbol{s} \\ &= \frac{Q}{4 \epsilon_0 \pi} \int_a^b \frac{1}{r^2} dr \\ &= \frac{Q}{4 \epsilon_0 \pi} \left[\frac{1}{r}\right]_b^a \\ &= \frac{Q}{4 \epsilon_0 \pi} \left(\frac{1}{a} - \frac{1}{b}\right) \\ &= \frac{Q (b - a)}{4 \epsilon_0 \pi ab} = |U|. \end{align*}

The capacitance is equal to:

\begin{align*} C &= \frac{Q}{|U|} \\ &= \frac{4 \epsilon_0 \pi ab}{b - a}. \end{align*}

An isolated capacitor is similar to a setup where the negative capacitor is placed at infinity:

\begin{align*} C &= \lim_{b \to \infty} \frac{4 \epsilon_0 \pi ab}{b - a} \\ &= 4 \epsilon_0 \pi \lim_{b \to \infty} \frac{a}{1 - \frac{a}{b}} \\ &= 4 \epsilon_0 \pi a. \end{align*}

Potential Energy of a Capacitor

Capacitors can be used to store energy. Consider a capacitor with initial charge equal to zero. The potential energy is the external work done to charge the capacitor is given by:

\begin{align*} E_p &= W_{ext} = \int_0^Q |U| dq \\ &= \int_0^Q \frac{q}{C} dq \\ &= \frac{Q^2}{2 C} \\ &= \frac{1}{2} C |U|^2, \end{align*}

where $Q$ charge after the charging process and $C$ is the capacitance.

Consider a parallel plate capacitor, where $C = A \epsilon_0/d$ and $|U| = Ed$ then the potential energy is equal to:

\begin{align*} E_p &= \frac{1}{2} \frac{A \epsilon_0}{d} E^2 d^2 \\ &= \frac{1}{2} \epsilon_0 E^2 Ad \\ &= \frac{1}{2} \epsilon_0 E^2 V, \end{align*}

where $V$ is the volume.

The energy density is the energy per unit volume:

e_p = \frac{1}{2} \epsilon_0 E^2

For example, consider a spherical conductor of radius $a$ . As derived with Gauss's law, the electric field outside is equal to:

\begin{align*} E &= \frac{a^2 \sigma}{r^2 \epsilon_0} \\ &= \frac{4 \pi r_c^2 \sigma}{4 \pi r^2 \epsilon_0} \\ &= \frac{Q}{4 \pi r^2 \epsilon_0}. \end{align*}

Then the energy density is equal to:

e_p = \frac{Q^2}{32 \pi^2 r^4 \epsilon_0}.

The potential energy is equal to:

\begin{align*} E_p &= \iiint_V e_p\ dV \\ &= \int_0^{2\pi} \int_0^{\pi} \int_a^{\infty} \frac{Q^2}{32 \pi^2 r^4 \epsilon_0} r^2 \sin \theta\ dr\ d\theta\ d\phi \\ &= \frac{Q^2}{32 \pi^2 \epsilon_0} \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta\ d\theta\ \int_{\infty}^a -\frac{1}{r^2} dr \\ &= \frac{Q^2}{32 \pi^2 \epsilon_0} 4 \pi \left[\frac{1}{r}\right]_{\infty}^a \\ &= \frac{Q^2}{8 \pi \epsilon_0 a}. \end{align*}

The electric potential at the shell is $\varphi = \frac{Q}{4 \pi \epsilon_0 a}$ . The potential energy can be simplified to:

E_p = \frac{1}{2} Q \varphi.

We can arrive at the same potential energy by calculating the external work required to charge the capacitor:

\begin{align*} E_p &= W_{ext} = \int_0^Q \varphi\ dq \\ &= \int_0^Q \frac{q}{4 \pi \epsilon_0 a}\ dq \\ &= \frac{Q^2}{8 \pi \epsilon_0 a}. \end{align*}

Total Capacitance

There are two ways to connect capacitors:

The first is parallel and the second is serial. For parallel, the voltage across each capacitor is the same - $|U|$ , each capacitor has the capacitance $C_i$ and charge $Q_i$ :

C_i = \frac{Q_i}{|U|} \iff Q_i = |U|C_i.

The capacitors can be replaced by a single capacitor with capacitance $C$ and charge $Q$ :

\begin{align*} Q &= \sum_i Q_i = \sum_i |U|C_i = |U| \sum_i C_i, \\ C &= \frac{Q}{|U|} = \sum_i C_i. \end{align*}

For serial, capacitors with capacitance $C_i$ are connected to a battery with total voltage $|U|$ each capacitor has voltage $|U_i|$ :

The outer plates have charges $\pm Q$ . The inner plates are charged to opposite charge. The voltage of each capacitor $|U_i| = \frac{Q}{C_i}$ and the total voltage is given by:

|U| = \sum_i |U_i| = \sum_i \frac{Q}{C_i}.

Again, the capacitors can be replaced with capacitor with capacitance $C$ :

\begin{align*} |U| = \frac{Q}{C} = \sum_i \frac{Q}{C_i}, \\ \frac{1}{C} = \sum_i \frac{1}{C_i}. \end{align*}

Capacitance Calculator

Parallel connection

C = 0\ F

C_1 =