My notes of the excellent lectures 7, 8 and 13 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, License: Creative Commons BY-NC-SA.”

The capacitance of two conductors to store energy only depends on their geometry. Adding a dielectric increases the capacitance.


Given two conductors with the same charge, but different polarities. The capacitance to store energy is defined as the charge on one of them divided by the potential difference $$ \newcommand{shaded}[1]{\bbox[5pt,background-color:##F7F7D2]{#1}} \shaded{ C = \frac{Q}{\Delta V} = \frac{\text{charge}}{\text{potential difference}} } \quad \left[\rm F=\frac{\rm C}{\rm V}\right] \nonumber $$

\(U\)Instead of \(\Delta V\), we often use the European symbol for voltage difference: \(U\). The letter ‘u’ stands for “Potentialunterschied”. Do not confuse it with potential energy, that also uses \(U\).

The unit is Farad. Named after the English physicist Michael Faraday (1791-1867).

Example: two parallel plates

According to this definition, what is the capacitance of the parallel plates? $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} C &= \frac{Q}{\Delta V} = \frac{\sigma A}{\oint \vec E\cdot d\vec l} & \left(\vec E\parallelsum d\vec l\right) \\ &= \frac{\sigma A}{E\,d} & \left(E=\sigma/\varepsilon_0\right) \\ &= \frac{\bcancel{\sigma}A\,\varepsilon_0}{\bcancel{\sigma}d} = \frac{A\,\varepsilon_0}{d} \end{align*} $$

The capacitance

  • is linearly proportional with the area of the plates,
  • is inversely proportional with the distance between the plates,
  • is only a matter of geometry

How much energy can it store?

Remember: the potential energy of these plates is

$$ U = \frac{1}{2}\,Q\,\Delta V \nonumber $$

Substituting \(Q=CV\) from the definition $$ U = \frac{1}{2} (C\,\Delta V)\,\Delta V = \frac{1}{2}C\,\Delta V^2 \nonumber $$

We can think of a capacitor as a device that can store electric energy.

Adding a dielectric to a capacitor induces a layers of charge, thereby increasing the capacitance.


Electric fields can induce dipoles in insulators. Electrons are bound to the atoms and molecules.

When I apply an external electric field, then even though the atoms or molecules may be completely spherical, they become a little bit more elongated. The electrons will spend a little more time as before at this top, so that becomes negative, and the bottom becomes positive charged. That creates a dipole.

On the larger picture, you will see a layer of negative charge created at the top and a layer of positive charge at the bottom

No \(E\)-field, charges cancel out
\(E\)-field, charged regions

That is the result of induction. We also call that polarization. We call the substances that do this dielectrics.

Plane capacitor

Given: two planes charged with a certain potential. We remove the power supply, so the charge is trapped on the plates.

We then move in a dielectric. At the top we will see a negative induced layer, and at the bottom a positive induced layer.

The induced charge will produce an electric field in the opposite direction. The free electric field \(E_\rm f\) and the induced electric field \(E_\rm i\) $$ \left. \begin{aligned} E_\rm{f} &= \frac{\sigma_\rm{f}}{\varepsilon_0} \\ E_\rm{i} &= \frac{\sigma_\rm{i}}{\varepsilon_0} \end{aligned} \right\} \begin{aligned} \Rightarrow \vec E_\rm{net} &= \vec E_\rm{f} + \vec E_\rm{i} \\ \Rightarrow E_\rm{net} &= E_\rm{f} – E_\rm{i} \end{aligned} $$

Assume that \(\sigma_\rm i = b\,\sigma_\rm f\), with \(0 \lt b \lt 1\) \(\Rightarrow\) \(E_\rm{i} = b\,E_\rm{f}\), and the net electric field $$ E_\rm{net} = E_\rm{f} \, (1-b) \nonumber $$

We now call this the dielectric constant \(\kappa\) (or K) $$ 1 – b = \frac{1}{\kappa} \nonumber $$

From now in this lecture, \(\vec E\) is always the net electric field.

$$ \shaded{ E = \frac{E_\rm{f}}{\kappa} = \frac{\sigma_\rm{f}}{\varepsilon_0\,\kappa} } \nonumber $$


  • by inserting the dielectric, the \(E\)-field went down by a factor \(\kappa\);
  • e.g., \(\kappa\) for glass is \(5\);
  • by definition \(\kappa=1\) in vacuum


$$ \shaded{ \Delta V = E\,d } \nonumber $$

If the electric field \(E\) goes down, the potential difference between the plates will go down also $$ \Delta V\color{red}{\downarrow}\,= E\color{red}{\downarrow}\,d \nonumber $$

What about Gauss Law?

Does Gauss law still hold?

$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S \vec E \cdot d\vec A = \frac{1}{\varepsilon_0} \sum Q_\rm{inside} \nonumber $$

The \(\sum Q_\rm{inside}\) must include both the free and induced charges \(Q_\rm{f}+Q_\rm{i}\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S \vec E \cdot d\vec A = \frac{1}{\varepsilon_0} \frac{\sum Q_\rm{f}}{\kappa} \nonumber $$

The capacitance is defined as $$ C = \frac{Q_\rm{f}}{\Delta V} \nonumber $$

If you insert a dielectric, then \(\Delta V\) goes down by \(\kappa\) and as a result \(C\) goes up by \(\kappa\).

Remember: the capacitance of parallel plates

$$ C = \frac{A\,\varepsilon_0}{d} \nonumber $$

We now have to amend it for the dielectric constant $$ \shaded{ C = \frac{Q_\rm{f}}{\Delta V} = \frac{A\,\varepsilon_0}{d}\kappa } \nonumber $$

Making a stronger capacitor

To make a large capacitor, $$ \begin{align*} C &= \frac{A\color{red}{\uparrow}\,\varepsilon_0}{d\color{red}{\downarrow}}\kappa\color{red}{\uparrow} \end{align*} $$


  • If you make \(d\) to small you get electric breakdown. For air that would be \(3\times 10^6\,\rm{V}/\rm{m}\).
  • If you take water for \(\kappa=80\), the problem is that water has a very low breakdown electric field. It would be better to take polyethylene, the breakdown electric field is \(18\times 10^6\,\rm{V}/\rm{m}\), and \(\kappa=3\). Mica would be even better.

Permanent dipoles

There are substances that are already dipoles, even in the absence of an electric field.

  • If you apply an electric field to such substances, the dipoles will start to align along the electric field. As they align, they will strengthen the induced electric field.
  • On the other hand, because of the temperature of the substance, these dipoles, these molecules, through chaotic motion will try to disalign.

So on one hand the electric field tries to align them, and the temperature which tries to disalign them.


  • Permanent dipoles are much stronger than any dipole that you can induce by ordinary means in a lab. They have a much higher value for \(\kappa\).
  • Water is a very good example. The electrons spend a little bit more time near the oxygen than near the hydrogen. It has a dielectric constant \(\kappa=80\).

Leyden jar

Pieter van Musschenbroek, a Dutch professor in mathematics, philosophy, medicine and astronomy, discovered the capacitor in 1746. He called it a Leyden jar, after the city he worked in.

it consists of an insulating glass bottle, one conducting beaker inside and one outside.


As an experiment he:

  1. Charged this capacitor using a Wimshurt \(\Longrightarrow\) The dielectric sees an external electric field and it polarizes so you get \(\sigma_{i}\)
  2. Disassemble it, and do whatever to get the free charge off the beakers \(\Longrightarrow\) With the free change gone, the induced charge must also go away.
  3. Then reassemble it.
  4. Short the outside and the inside beaker.
Much to his surprise, there was a strong spark. That means there was energy left.


There must be free charge on the glass. As we will see, it got there through corona discharge.

We will make some simplifying assumptions

  • The capacitor is just two parallel plain plates
  • The Wimshurt produces about \(30\,\rm{kV}\)
  • The airgap between the conductors and the glass is \(1\,\rm{mm}\); the thickness of the glass is \(3\,\rm{mm}\)
  • For the glass, \(\kappa=5\).

We can calculate the electric fields in the air \(E_a\) and glass \(E_g\) $$ \begin{align*} \Delta V &= 2\Delta V_a + \Delta V_{g}, & \left(\Delta V = E\,d\right) \\ &= 2E_a d_a + E_g d_g, & \left(E_g = \frac{E_a}{\kappa}\right) \\ 30\times10^3 &= 2 E_a 10^{-3} + \frac{E_a}{5}(3\times10^{-3}) \\ &= E_a\left(2\times10^{-3} + \frac{3\times10^{-3}}{5} \right) \\ \Rightarrow E_a &\approx 11.5\times 10^6\,\rm{V/m} \\ \Rightarrow E_g &= \frac{E_a}{\kappa} \approx 2.3\times 10^6\,\rm{V/m} \end{align*} $$


However, the electric field in air \(E_a\) can be no larger than \( 3\times10^6\,\rm{V/m}\). In glass it is a bit higher \(10^7\,\rm{V/m}\). When the electric field is higher, you get corona discharge.

The electric field \(E_a\) of \(11.5\times 10^6\,\rm{V/m}\) is far above the breakdown electric field. So, we get corona discharge from the conductor to the glass. It is spraying free charge from the conductor onto the glass. This stores the energy on the glass. It is very hard to remove, because the glass is an insulator.

Before we disassemble it, and after the corona discharge, the maximum \(E_a=3\times10^6\,\rm{V/m}\). That means the electric field in the glass \(E_g\) $$ \begin{align*} E_g &= \frac{\Delta V_g}{d_g} = \frac{\Delta V – 2\Delta V_a}{d_g} \\ &= \frac{\Delta V – 2(E_a d_a)}{d_g} \\ &= \frac{30\times10^3 – 2(3\times10^6)10^{-3}}{3\times10^{-3}} \\ &= 8\times10^6\,\rm{V/m} \end{align*} $$

Due to the corona discharge, the electric field in the glass is much stronger than the field in the air gap. The distribution of fields and potential differences is shown below

It’s no longer the field that is dictated by the external field, and the induced charges. The glass itself now carries free charge. $$ \begin{align*} \frac{\sigma_g}{\sigma_a} &= \frac{E_g \bcancel{\varepsilon_0}\varepsilon_r\kappa}{E_a \varepsilon_0}, & \left(\varepsilon_r\approx 6.5\right) \\ &\approx \frac{3\times \cancel{10^6} \bcancel{\varepsilon_0} (6.5)(5)}{8\times\cancel{10^6} \varepsilon_0} = 12 \end{align*} $$

The free charge on the glass is 12 times more than on the top conductor. So, if you disassemble and remove the free charge on the conductors, it doesn’t make much of a difference.

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