## Heart



My notes of the excellent lecture 19 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

The sole purpose of the heart is to pump blood, about 5 to 6 liters of blood every minute when a person is resting. It pumps about seventy times per minute. If the blood flow to your brain would stop for only 5 seconds, you lose consciousness. Four minutes later, permanent brain damage starts.

Our heart has four chambers: two atria and two ventricles

Each heart cell is a mini chemical battery, that pumps ions in or out. In the idle state, the inside of each heart cell is $$-80\,\rm{mV}$$ relative to its outside. When the cells are at $$+20\,\rm{mV}$$, they contract.

## Cardiac cycle phases

Phases of a heart beat:

1. A heart beat starts with pacemaker cells, near the right atrium, changing their potential from $$-80\,\rm{mV}$$ to $$+20\,\rm{mV}$$.
2. Atria
1. The neighboring cells of the atria follow, and the depolarization wave propagates, changing their potential from $$-80\,\rm{mV}$$ to $$+20\,\rm{mV}$$. This causes the atria to contract. (P wave)
2. When the wave reaches the AV node it pauses to allows the atria to contract fully, emptying their blood into the ventricles. (PR interval)
3. The wave continues through to the ventricles (step 3.1), and the atria cells return to $$-80\,\rm{mV}$$ starting a repolarization wave in the atria. (overlaps with QRS complex)
3. Ventricles
1. The depolarization wave continues to the ventricles, changing their potential from $$-80\,\rm{mV}$$ to $$+20\,\rm{mV}$$. This causes these ventricles to contract, and send the blood to the organs and lungs. (QRS complex)
2. After about $$0.2\,\rm{s}$$, the cells return to $$-80\,\rm{mV}$$ starting a repolarization wave in the ventricles. This wave goes from below to above. (T wave)
4. The heart waits for another message from the pacemaker cells.
5. ### At the cellular level

A closer look at the individual heart cells

1. At idle, the inside of a heart cell is at $$-80\,\rm{mV}$$ compared to its outside. That means it has repelled positive ions. so the inside is negative. There is no $$\vec E$$-field outside, because if you put a Gaussian surface around it, there is no net charge inside. But there is an electric field across the cell walls.
2. During depolarization, the potential difference changes from $$-80\,\rm{mV}$$ to $$+20\,\rm{mV}$$. For simplicity, we will assume it changes to $$0\,\rm{mV}$$. The depolarization waves goes from top to bottom. When the depolarization wave is halfway down the cell, the top potential is zero. The cell has pulled the positive ions back inside. You now have a minus layer on top of a positive layer. That creates an electric field, that has roughly the shape of an electric dipole. So, as the wave goes through the cells, only then do they create a dipole.
3. When the wave has passed, the each heart cell is at $$+20\,\rm{mV}$$. (the zero was just to make it easier to explain.) Now the inside has positive ions, and the outside has negative ions. The cell is contracted. The external $$\vec E$$-field is zero.
4. Now, the repolarization waves comes from below to bring the potential difference back to $$-80\,\rm{mV}$$. Again, we will assume the wave is halfway, and it is not $$-80\,\rm{mV}$$, but $$0\,\rm{mV}$$. Once more, we have a minus layer on top of a positive layer. This creates an electric field, with roughly the shape of an electric dipole.
5. After the repolarization wave has passed the heart cells returned to their idle potential difference of $$-80\,\rm{mV}$$.

## Electric fields

The depolarization wave will run down, and leave behind cells at $$+20\,\rm{mV}$$. These cells are contracted. Only the cells where the depolarization occurs, only the ones on the ring, contribute to that electric dipole field.

When the repolarization goes in the other direction, when the heart relaxes, because the cells go back to $$-80\,\rm{mV}$$, then again is there an electric dipole field, but only from the cells through which the repolarization wave propagates.

All the dipoles of the depolarization/repolarization band are in the same direction. As the wave is going, the electric field is spreads around the band.

If there is an electric field, there’s going to be a potential difference between different parts of your body. The integral $$\int \vec E\cdot d\vec l$$ gives you a potential difference. That is the idea behind an electrocardiogram. Typically there are twelve electrodes attached to arms, legs, head and chest to get as much information about the heart as we can. The maximum potential difference between two electrodes is not more than about two to three millivolts.

## Batteries; Kirchhoff’s rules



My notes of the excellent lectures 10 and 12 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

## Batteries

Power supplies maintain a constant potential difference.

In the power supply the electric field opposes the current. Some kind of pump mechanism must force the current to go inside the supply against the electric field.

In the case of common batteries, it is chemical energy that provides the energy.

### Zinc-Copper battery

A zinc and copper plate in a solution of H2SO4

You will get a potential difference between the two plates. To understand that takes quantum mechanics. Instead we will try to do it on a napkin.

The solution is not the key, because if you take two different conductors and touch them together, there will also be a potential difference.

The ions can flow freely from one side to the other. We connect the plates with a resistor, so a current will flow. The electric field must be from plus to minus, so inside the battery it goes against the current.

On the right copper plate, you have SO4 and CU++ ions. On the left zinc plate, you have Zn++SO4 ions

1. On the right, as current start to run, the current carrier SO4, go from the right to the left. Doing so, they travel through an electric field that opposes their motion. Doing so, they engage in a chemical reaction that yields more energy than it costs to go against the electric field.
2. With the SO4 current is flowing from the right to the left, there are fewer SO4 ions at the right. With the liquid remaining neutral, the Cu++ must disappear, and precipitates onto the copper plate.
3. On the left side, you get an increase of SO4. For the liquid to remain neutral, you must also get an increase of Zn++. Some of the zinc will dissolve, to increase the concentration of the zinc ions.

So, the charge carriers in this battery, the SO4 ions, travel through an electric field that opposes their motion. This happens at the expense of chemical energy.

When the copper solution becomes very dilute, as all the copper has been plated onto the copper plate, and the left side becomes concentrated Zn++, then the battery stops.

Now you can force a current to run in the opposite direction using another external power supply, and the chemical reactions will reverse. Copper will go back in the solution, and the zinc will be precipitated onto the zinc plate. Then you can run the battery again.

A car battery is this kind of battery, except that you have lead and lead oxide instead of zinc and copper, but you also have sulfuric acid. Another well know battery is nickel-cadmium.

### Symbol

We will be using the following symbol for batteries

$$U$$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 $$U$$.

### Internal resistance

In reality, batteries are not ideal, and have an internal resistance $$r_i$$ as shown below

If the current starts to flow, it goes through $$r_i$$ and $$R$$. Unloaded $$R=\infty$$, the external battery voltage is the same as the electromotive force (EMF), $$\mathcal{E}$$ $$V_b = \mathcal{E}$$

When you connect resistor $$R$$, a current $$I$$ will and the external battery voltage follows from Ohm’s law \begin{align*} \mathcal{E} &= I\,(r_i + R) \\ \Rightarrow I &= \frac{\mathcal{E}}{r_i + R} & (V_b = I\,R) \\ \Rightarrow V_b &= \mathcal{E}\frac{R}{r_i + R} \end{align*}

### In series

You can put batteries in series, to get a higher potential difference.

That brings the potential difference of the open circuit to $$2\,\mathcal{E}$$.

## Power

If a charge $$dq$$ moves from point A to point B,

With $$V_A \gt V_B$$, the electric field is from $$A$$ to $$B$$.

I move charge from $$A$$ to $$B$$, then the electric field is doing work $$dW$$ $$dW = dq (V_A – V_B) \nonumber$$

The work is positive, if the charge is positive, or negative if the charge is negative. Take the derivative in time on both sides $$\frac{dW}{dt} = \frac{dq}{dt} (V_A – V_B) \nonumber$$

On the left side, we now have work per unit time. That is power in Joules/s. On the right side, $$\frac{dq}{dt}$$ is current in Coulombs/s. The potential difference we simply call $$V$$ $$\shaded{ P = I\,V } \quad \left[\rm W =\frac{\rm J}{\rm s} \right] \tag{Power} \label{eq:power}$$

If you can use Ohm’s law ($$V=I\,R$$), you can write this as $$\shaded{ P = I^2\,R = \frac{V^2}{R} } \quad \left[\rm W =\frac{\rm J}{\rm s} \right] \tag{Power} \label{eq:powerohmslaw}$$

Instead of writing $${\rm J}/{\rm s}$$, we write $$\rm W$$ for Watt, named after the 18th-century Scottish inventor James Watt.

### Examples

#### One

Suppose we have resistance $$R=100\,\Omega$$ and we run a current $$I=1\,\rm A$$ through it,

Then the power dissipated in this resistor, provided by the battery $$P = I^2\, R = 1^2(100) = 1000\,\rm W \nonumber$$

The energy is dissipated as heat. If it gets hot enough, the resistor might produce light. E.g. the filament in a tungsten light bulb heats up to $$\approx 2500\,^{\circ}\rm C$$ and emits light. But the amount of light is produces is no more that 20% of the power it dissipates.

#### Two

If you take a 9 volt battery with an internal resistance of about 2 ohms, and short it

The power dissipated inside the battery is $$P_{max} = \frac{V^2}{r_i} = \frac{9^2}{2} = 40\,\rm W \nonumber$$

#### Three

If you take a 12 volt car battery with an internal resistance of about 0.02 ohms, and short it, the power dissipated inside the battery is $$P_{max} = \frac{V^2}{r_i} = \frac{12^2}{0.02} = 7000\,\rm {W} \nonumber$$

The sulfuric acid will boil, and the case may melt. This is very dangerous.

### kWh

The electric utility company charges you in $$\rm{}kWh$$ $$1 \rm{kWh} = 10^3 (3600)\,\rm J \nonumber$$

## Kirchhoff’s rules

Named after and first described by the German physicist Gustav Kirchhoff in 1845.

### Junction rule

Charge conservation. At each junction, the current that flows in, must flow out. $$\sum_i I_i = 0 \tag{Kirchhoff #1} \label{eq:Kirchhoff1}$$ The direction of the currents can be freely chosen.

### Voltage rule

The sum of the voltage differences around any closed loop in a circuit must be zero. A loop in a circuit is any path which ends at the same point at which it starts. $$\sum_i V_i = 0 \tag{Kirchhoff #2} \label{eq:Kirchhoff2}$$ E.g. as shown below $$-V_1+V_2+V_3=0$$

This is really the same as the work over a close line that we saw earlier $$\oint \vec E\cdot d\vec l = 0 \nonumber$$

### Examples

#### One

Given the circuit shown below. What are the currents through the resistors?

Apply Kirchhoff’s junction rule where the three resistors connect

Current’s $$I_1$$ and $$I_2$$ flow in, and $$I_3$$ goes out. \begin{align*} \sum_i I_i &= 0 \\ \Rightarrow I_1 + I_2 – I_3 &= 0 \\ \Rightarrow I_3 &= I_1 + I_2 \end{align*}

Substitute $$I_3$$ in the schematic

Apply Kirchhoff’s voltage rule on the closed loop on the left \begin{align*} \sum_i V_i &= 0 \\ \Rightarrow V_1+V_3-V_{b1} &= 0 & (V=IR) \\ \Rightarrow I_1R_1 + (I_1+I_2)R_3 – V_{bl} &= 0 \end{align*}

Take a closed loop on the right

Apply Kirchhoff’s voltage rule on the closed loop on the right \begin{align*} \sum_i V_i &= 0 \\ \Rightarrow V_2+V_3+V_{b1} &= 0 & (V=IR) \\ \Rightarrow I_2R_2 + (I_1+I_2)R_3 + V_{b2} &= 0 \end{align*}

We now have two equations with two unknowns, $$I_1$$ and $$I_2$$, that we can solve \begin{align*} I_1R_1 + (I_1+I_2)R_3 – V_{bl} &= 0 \\[0.3em] I_2R_2 + (I_1+I_2)R_3 + V_{b2} &= 0 \end{align*}

When substituting values, the currents may come out negative. That just means we assumed the wrong direction. The answer would still be correct.

#### Two (Lec.12)

Given the circuit given below, what are the currents and potential differences

To calculate the potentials differences, the parallel circuit of resistors $$R_1\ldots R_3$$ can be replaced by $$R_p$$ \begin{align*} \frac{1}{R_p} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_p} \\ \Rightarrow R_p &= \frac{1}{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}} \approx 0.5455\,\rm\Omega \\ \end{align*}

Replacement schematic

The potential differences follow from Kirchhoff’s voltage rule, $$\sum_i V_i=0$$ \begin{align*} \mathcal{E} &= IR_i + IR_p + IR_4 = I(R_i+R_p+R_4) \\ \Rightarrow I &= \frac{\mathcal{E}}{R_i+R_p+R_4} \\ &= \frac{10}{0.1 + 0.5455 + 4} \approx 2.153\,\rm{A} \end{align*}

Back to the original schematic

This gives us potential difference $$V_i$$, $$V_p$$ and $$V_4$$, and indirectly the currents through $$R_1\ldots R_3$$ \left\{ \begin{align*} V_i &= I R_i = 2.153\,(0.1) \approx 0.2143\,\rm V \\ V_4 &= I R_4 = 2.153\,(4) \approx 8.611\,\rm V \\ V_p &= I R_p = 2.153\,(0.5455) \approx 1.174\,\rm V \Rightarrow \end{align*} \\ \right. \Rightarrow \left\{ \begin{aligned} I_1 &= \frac{V_p}{R_1} = \frac{1.174}{1} \approx 1.174\,\rm A \nonumber \\ I_2 &= \frac{V_p}{R_2} = \frac{1.174}{2} \approx 0.587\,\rm A \nonumber \\ I_3 &= \frac{V_p}{R_3} = \frac{1.174}{3} \approx 0.391\,\rm A \nonumber \end{aligned} \right.

## Current; Ohm’s law



My notes of the excellent lecture 9 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

Ohm’s law is the linear relation between the potential and the flow of charge.

## Current

Current is defined charge moved per unit time $$\shaded{ I = \frac{\Delta Q}{\Delta t} } \tag{Current}$$

The convention is that the current $$I$$ is in the direction of positive charges

Keep in mind, that energy moves through electro-magnetic fields, as we will see when we learn about the Poynting vector. Meanwhile, the electric current is an very effective abstraction model.

## Deriving Ohm’s law

### Drift velocity

As we have seen, in static conditions

Charges inside a conductor always move automatically in such a way that they kill the electric field inside. It will reach this equilibrium in a very short time.

If we keep adding electric charge to one end of the conductor, it will will not achieve equilibrium. We do this by applying a potential difference across a conductor. This potential difference then generates an electric field along and inside the conductor. The free electrons (with charge $$\rm e$$) will experience a force $$F$$ $$F = \rm{e} \,E \nonumber$$

The free electrons are accelerated by the electric field, and gain momentum $$p$$ $$\frac{\Delta p}{\Delta t} = -q_e\,E \label{eq:gainrate}$$

We call the average net velocity in the direction opposite to the field, drift velocity $$v_d$$. The collision of free electrons and ions will absorb all this drift momentum. The average momentum is the mass of the electron times the drift velocity $$\frac{\Delta p}{\Delta t} = m_e\,v_d \label{eq:lossrate}$$

The rate of gain $$\eqref{eq:gainrate}$$ and loss of momentum $$\eqref{eq:lossrate}$$ must match \begin{align} -q_e\,E &= m_e\,v_d \nonumber \\ \Rightarrow v_d &= – \frac{q_e E \tau}{m_e} \label{eq:vd} \end{align}

#### For copper

E.g. a copper $$\left(\rm{Cu}\right)$$ conductor at room temperature

• the free electrons in copper have an average speed, $$v_e\approx 10^6\,\rm(m)/\rm{s}$$. A chaotic thermal motion due to the temperature.
• An electron moving through a conductor suffers about $$10^{14}$$ collisions with ions per second. So, the time between collations, $$\tau\approx 3\times 10^{-14}\,\rm{s}$$
• The number of free electrons, $$n\approx 10^{29}\,/\rm{m}^3$$

Apply a potential difference $$V=10\,\rm{V}$$ over length $$l=10\,\rm{m}$$. Then inside the conductor the electric field is $$E = 1\,\rm{V/m}$$, causing a drift velocity \newcommand{numexp}[2]{#1\times 10^{#2}} \begin{align*} v_d &\approx \frac{(\numexp{1.6}{-19})\,(\numexp{3}{-19})}{10^{-30}} \\ &\approx 4\times 10^{-3}\,\rm m/\rm s \approx 0.5\,\rm{cm}/\rm s \end{align*}

The thermal motion is a million of meters per second, but due to the electric field, they only advance along the wire at $$\approx 0.5\,\rm{cm}/\rm s$$. Note that the electrons on the “far end” of the wire start moving around the same time as the ones at the “start” of the wire.

### Ohm’s law on a napkin

When we apply a potential over a conductor, we create an electric field. The free electrons in the conductor will move until the electric field is zero. Because we keep the potential difference, they can’t succeed. The linear relationship between current and potential is called Ohm’s law. To avoid quantum mechanics, we’ll derive Ohm’s law in a crude way.

$$U$$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 $$U$$.

Given a wire with cross section $$A$$, length $$l$$ and potential difference $$V$$.

In time $$\Delta t$$, a free electron moves distance $$\Delta x$$. The shaded wire segment, in the picture below, represents the free electrons passing though cross section $$A$$. The shaded volume is $$A\,\Delta x$$.

Let $$n$$ be the number of free electrons per volume unit, and $$q_e$$ be the charge of one electron. Then, the number of charge crossing through cross section $$A$$ is $$\Delta Q = -q_e\,n\,A\,\Delta x \nonumber$$

Substituting this back in the expression for the current \begin{align*} I &= -\frac{q_e\,n\,A\,\Delta x}{\Delta t} & \left(\frac{x}{\Delta t}=v_d\right) \\ &= -q_e\,n\,A\,v_d & \left(\text{Eq. } \ref{eq:vd}\right) \\ &= q_e\,n\,A\,\frac{q_e\,E}{m_\rm{e}} \tau \\ &= \underbrace{\frac{{q_e}^2\,n\,\tau}{m_\rm{e}}}_{=\sigma} A\, E \end{align*}

Conductivity $$\sigma$$ only depends on the material. Note: we often use resistivity $$\rho=\dfrac{1}{\sigma}$$ \begin{align*} I &= \sigma\,A\,E & \left(E=\frac{V}{l} \right) \\ &= \sigma\, A\, \frac{V}{l} \end{align*}

Rearranging terms \begin{align*} V &= \underbrace{\frac{l}{\sigma\,A}}_{=R} \,\, I \end{align*}

Ohm’s law follows $$\begin{array}{l} \shaded{ \Delta V = I\,R } \quad \left[\Omega\right] \tag{Ohm’s law} \label{eq:ohmslaw} \\ \text{where } R = \frac{l}{\sigma\,A} = \frac{l\,\rho}{A} \end{array}$$

The unit for resistance $$R$$ is volt per ampere. We call that ohm, $$\Omega$$.

Ohm’s law:

• resistance is proportional with the length of the wire (like water flowing through a longer pipe)
• resistance is inversely proportional with the area of the wire (like water flowing through a pipe with a wider cross section).
• Often also holds for insulators.

#### For copper

For copper, the conductivity $$\sigma$$ at 300 K is \newcommand{numexp}[2]{#1\times 10^{#2}} \begin{align*} \sigma_\text{Cu} &= \frac{e^2\,n\,\tau}{m_e} \\ &\approx \frac{\left(\numexp{1.6}{-19}\right)^2\,10^{29}\,(\numexp{3}{-14})}{10^{-30}} \\ &\approx \numexp{7.7}{7} \end{align*}

A $$1\,\rm{mm}^2$$ copper wire of $$10\,\rm{m}$$ has a resistance of roughly \newcommand{numexp}[2]{#1\times 10^{#2}} \begin{align*} R &= \frac{l}{\sigma\,A} \\ &\approx \frac{10}{\numexp{7.7}{7}\,10^{-6}} \\ &\approx 0.13\,\Omega \end{align*} \nonumber

### Conductors vs. insulators

Assume a chunk of material as pictured below. What is the resistance?

For a good conductor, such as $$\rm{Au}, \rm{Ag}, \rm{Cu}$$ \begin{align*} \sigma &\approx 10^8 \\ \Rightarrow \rho &\approx \frac{1}{\sigma} = 10^{-8} \\ \Rightarrow R &\approx \frac{l\,\rho}{A} = \frac{1\,10^8}{10^{-6}} = 10^{-2}\,\Omega \end{align*}

For a good insulator, such as glass, porcelain \begin{align*} \sigma &\approx 10^{-14} \\ \Rightarrow \rho &\approx \frac{1}{\sigma} = 10^{14} \\ \Rightarrow R &\approx \frac{l\,\rho}{A} = \frac{1\,10^{-14}}{10^{-6}} = 10^{20}\,\Omega \end{align*}

You see 22 orders of magnitude difference between good conductors and insulators.

If I make the potential difference $$1\,\rm{V}$$, then the current \begin{align*} I_\text{cond.} &= \frac{V}{R} = \frac{1}{10^{-2}} = 100\,\rm{A} \\ I_\text{ins.} &= \frac{V}{R} = \frac{1}{10^{20}} = 10^{-20}\,\rm{A} \end{align*}

### The limit of Ohm’s law

The conductivity is a strong function of the temperature. If you increase the temperature, then the speed of the free electrons goes up and the time between collisions $$\tau$$ goes down $$\Longrightarrow$$ the conductivity will go down, and the resistance $$R$$ will go up. In other words: when you heat up a substance the resistance goes up.

The moment the resistance becomes a function of the temperature, we call the breakdown of Ohm’s law.

One could say that $$R$$ is $$R(T)$$, but that is a poor way of saving the law, because the temperature $$T$$ itself is a function of the current \left\{ \begin{align*} R &= \frac{V}{I} = f(T) \\ T &= f(I) \end{align*} \right.

Notes:

• It is fortunate that the $$R$$ goes up as the $$T$$ goes up, otherwise a light bulb would self destroy.
• In general we will assume that Ohm’s law hold, and the heat produced does not play an important role.

## Networks of resistors

### Series

Suppose we have the circuit shown below with resistors $$R_1$$ and $$R_2$$ in series, and potential difference $$V$$. What is the current, and potential difference over each resistor?

The current is the same for both resistors. Apply Ohm’s law for each resistor to find the voltage drop over each resistor \left\{ \begin{align*} V_1 &= I\,R_1 \\ V_2 &= I\,R_2 \\ \end{align*} \right.

The total voltage is sum of the voltage drops over each resistor \def\triangleq{\overset{\Delta}{=}} \begin{align*} V &= V_1 + V_2 \\ &= I\,R_1 + I\,R_2 & \text{(Ohm’s law)} \\ \Rightarrow I &= \frac{V}{R_1+R_2} \triangleq I = \frac{V}{R_{eq}} \end{align*}

So $$R_1$$ and $$R_2$$ in series behave as $$\shaded{ R_{eq} = R_1 + R_2 } \tag{R in series} \label{eq:series}$$

### Parallel

Suppose we have the circuit shown below with resistors $$R_1$$ and $$R_2$$ in parallel, and potential difference $$V$$. What is the current through each resistor, and the total current?

The voltage is the same over both resistors. Apply Ohm’s law for each resistor to find the current \left\{ \begin{align*} I_1 &= \frac{V}{R_1} \\ I_2 &= \frac{V}{R_2} \end{align*} \right.

The total current is sum of the current through each resistor \def\triangleq{\overset{\Delta}{=}} \begin{align*} I &= I_1 + I_2 = \frac{V}{R_1} + \frac{V}{R_2} \\ &= V\left(\frac{1}{R_1}+\frac{1}{R_2}\right) \triangleq V\frac{1}{R} \\ \end{align*}

So $$R_1$$ and $$R_2$$ in parallel behave as $$\shaded{ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} } \tag{R in parallel} \label{eq:parallel}$$

Or, taking the reciprocal on both sides $$\shaded{ R_{eq} = \frac{R_1R_2}{R_1+R_2} } \tag{R in parallel} \label{eq:parallel2}$$

## Capacitors



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, https://youtube.com. 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.

## Definition

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.

## Dielectrics

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

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.

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

Note:

• 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*}

Note

• 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.

Note:

• 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.

### Experiment

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.

### Explanation

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*}

Sketch

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.

## Energy density



My notes of the excellent lectures 7 and 12 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

To assemble charges, I have to do work. We called that electrostatic energy, $$U$$.

Here we will evaluate that energy in terms of the electric field.

## Energy in terms of electric field

Continuing the charged plates example. If the plates are close together $$h\ll\sqrt{A}$$, we can approximate the field between these plates as that coming from a pair of infinite planes.

We saw that the electric field in between the plates is $$E=\sigma/\varepsilon_0$$.

We now pull the top plate upwards over distance $$x$$. This creates an electric field, in the shaded area, that wasn’t there before. That field has the same strength $$\sigma/\varepsilon_0$$, because the charges on the plates stayed the same.

These plates are attracted to each other due to the opposite charges. The force on the charge on the top plate is due to the electric field of the bottom plate only ($$\frac{\sigma}{2\varepsilon_0}$$). To move the top plate upwards, we need to overcome this electric field.

The electric field is defined as force per unit charge, an with $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec E \parallelsum \vec F$$, the dot-product becomes a multiplication of scalars \begin{align*} \frac{\sigma}{2\varepsilon_0} &= -\frac{F}{Q} \\ \Rightarrow F &=-\frac{\sigma}{2\varepsilon_0}\,Q & \left(Q=-\sigma\,A\right) \\ &= \frac{1}{2}\frac{\sigma^2}{\varepsilon_0} A & \left(\times\frac{\varepsilon_0}{\varepsilon_0}\right) \\ &= \frac{1}{2}\varepsilon_0\frac{\sigma^2}{\varepsilon_0^2}\,A\ & \left(\frac{\sigma}{\varepsilon_0}=E\right) \\ &= \frac{1}{2}\varepsilon_0\,E^2\,A \end{align*}

The work to move the top plate, follows as the force $$F$$ times the distance $$x$$ \begin{align} W &= F\,x \nonumber \\ &= \frac{1}{2}\varepsilon_0\,E^2\,\underline{A\,x} \label{eq:W} \end{align}

## Field energy density

The term $$Ax$$ in equation $$\eqref{eq:W}$$ is the new volume, in which we created electric field. The work per unit volume $$\shaded{ \frac{W}{\rm{volume}} = \frac{1}{2} \varepsilon_0 E^2 } \quad \left[\frac{\rm{J}}{\rm m^3}\right] \nonumber$$

Since the work created electric field, we call this field energy density.

It can be proven that this applies to the general case, not only for this charge configuration.

## A new expression for energy

This gives us a new way of looking at the energy that it takes to assemble charges. Earlier, we calculated the work to put charges in place. Now, we could calculate that the electrostatic potential energy by integrating over volume $$\shaded{ U = \iiint_\text{all space} \frac{1}{2} \varepsilon_0 E^2\,dV } \nonumber$$

We look at as the energy in the electric field, and we no longer think of it as the work that you have done to assemble the charges.

To calculate the electrostatic energy, we can now choose between

• calculating the work to bring all the charges in place, or
• take the electric field everywhere in space, and integrate over all space.

### Example

The total potential energy of these parallel plates $$U = \iiint \frac{1}{2} \varepsilon_0 E^2\, dV \nonumber$$

Since the outside field is zero everywhere, and the internal field $$\vec E$$ is constant \begin{align*} U &= \int_\text{inside}\ldots+\int_\text{outside}\ldots \\ &= \frac{1}{2} \varepsilon_0 E^2\iiint_\text{inside} dV + 0 \\ &= \frac{1}{2} \varepsilon_0 E^2\,Ah \\ &= \frac{1}{2} \varepsilon_0 EA\,\underline{Eh} \end{align*}

Remember: the potential difference is defined as

\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} V &= \int \vec E\cdot d\vec l & \left(\vec E\parallelsum d\vec l\right) \\ &= \underline{E\,h} \end{align*} \nonumber

Substituting the potential difference between the plates $$V$$, in the expression for $$U$$ \begin{align*} U &= \frac{1}{2} \varepsilon_0 EA\,V & \left(E=\frac{\sigma}{\varepsilon_0}\right) \\ &= \frac{1}{2} \bcancel{\varepsilon_0} \frac{\sigma}{\bcancel{\varepsilon_0}} A\,V & \left(\sigma A=Q\right) \end{align*}

So $$\shaded{ U = \frac{1}{2} Q V } \nonumber$$

## Note

All of these are equivalent:

• the total energy in the field,
• the total work you have to do to assemble these charges,
• the total work you have to do to create the electric fields.

## Complete example (Lec.12)

Two very large conducting plates with a thickness, and separation $$d$$. Charged with density $$+\sigma$$ and $$-\sigma$$.

In the conductor, there is no current, so $$E=0$$. We already derived that the field inbetween the plates using Gauss’ law and is $$E=\frac{\sigma}{\varepsilon_0}$$. The electric field outside the plates is very close to zero. We found that through the superposition principle.

### Potential difference

What is the potential difference $$V_P-V_S$$ over this capacitor?

The potential difference is \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} V_P-V_S &= \int_P^S \vec E\cdot d\vec l & \left(\vec E\parallelsum d\vec l\right) \\ &= E\, d & \left(E=\frac{\sigma}{\varepsilon_0}\right) \\ &= \frac{\sigma}{\varepsilon_0}d \end{align*}

If we had chosen another path, we would have found the same answers, because we’re dealing with conservative fields that are path independent.

What is the potential difference between point $$P$$ and $$T$$? \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} V_P-V_S &= E\, d & \left(E=0\right) \\ &= 0 \end{align*}

### Capacitance

What is the capacitance? It is defined as the charge on one plate divided by the potential difference \begin{align*} C &= \frac{Q}{\Delta V} = \frac{Q}{V_P-V_S}, & \left(Q=\sigma A\right) \\ &= \frac{\bcancel{\sigma} A}{\frac{\bcancel{\sigma}}{\varepsilon_0}d} = \frac{A\,\varepsilon_0}{d} \end{align*}

### Potential energy

What is the electrostatic potential energy? That is the work you have to do to assemble the charges. You can also look at the energy that it takes to create that electric field. For that we derived \begin{align*} U &= \frac{1}{2} Q\,\Delta V, & \left(Q=\sigma A\right) \\ &= \frac{1}{2} \sigma A\,\Delta V, & \left(\Delta V=\frac{\sigma}{\varepsilon_0}d \right) \\ &= \frac{1}{2} \sigma A\,\frac{\sigma}{\varepsilon_0}d \\ &= \frac{\sigma^2\,A\,d}{2\varepsilon_0} \end{align*}

### Charge location

Where is the charge located? On the upper plate: it can’t be inside the plate, because if you make a Gaussian surface around that charge, it will tell met that $$\oint\vec E\cdot d\vec A\neq 0$$ because there is charge inside. But we know the electric field must be zero in the conductor, so the closed surface integral must be zero.

It can’t be on the top surface either, because if I make myself a small pillbox Gaussian surface through the top of the upper plate, the electric field is zero on the top and bottom, and on the sides $$d\vec A\perp \vec E$$. The closed surface integral must be zero, so according to Gauss’ law the contained charge is zero.

That only leaves that nature puts all positive the charge right at the bottom of the top plate, and the negative charge at the top of the bottom plate.

## Example with dielectric (Lec.12)

Aw we saw before, two large parallel plates, that we charge with a power supply. Then we disconnect the power supply. That means the charge is trapped and can never change.

Remember, the formula’s [ 1, 2, 3, 4, 5, 6]

\begin{align} Q_f &= \sigma_f\,A \label{eq:Qf} \\ E &= \frac{\sigma_f}{\varepsilon_0\,\kappa} \label{eq:E} \\ \Delta V &= E\,d \label{eq:dV} \\ C &= \frac{Q_f}{\Delta V} = \frac{A\varepsilon_0}{d}\kappa \label{eq:C} \\ U &= \tfrac{1}{2}Q_F\,\Delta V =\tfrac{1}{2}C\,(\Delta V)^2 \label{eq:U} \end{align}

### Double the distance

With the power supply disconnected, double the distance between the plates, while the dielectric is just air $$\kappa=1$$.

#### Electric field?

What happens with the electric field? $$\vec E$$ can’t change because $$\sigma_f$$ cannot change, since the power supply is disconnected. Based on equation $$\eqref{eq:E}$$, the electric field stays the same.

#### Potential difference?

The potential difference must increase by a factor of 2, based on equation $$\eqref{eq:dV}$$, because we double the distance

#### Capacitance?

The capacitance must decrease by a factor of 2, based on equation $$\eqref{eq:C}$$, because the potential difference doubled.

#### Electrostatic potential energy?

The electrostatic potential energy must increase by a factor of 2, based on equation $$\eqref{eq:U}$$, because the potential difference doubled. When we separate the plates, we do work that increased $$U$$.

### Change the dielectric

Back to the original distance $$d$$, and change $$\kappa=3$$ by moving in a dielectric.

#### Electric field?

What happens with the electric field? $$\vec E$$, based on equation $$\eqref{eq:E}$$, the electric field becomes a third. This is because on the dielectric you introduce opposite charges, that create an opposite induced electric field.

#### Potential difference?

The potential difference must decrease by a factor of 3, based on equation $$\eqref{eq:dV}$$, because the electric field decreased by a factor of 3.

#### Capacitance?

The capacitance must increase by a factor of 3, based on equation $$\eqref{eq:C}$$, because the potential difference decreased by a factor 3.

#### Electrostatic potential energy?

The electrostatic potential energy must decrease by a factor of 3, based on equation $$\eqref{eq:U}$$, because the potential difference decreased by a factor 3. That means that by moving in the dielectric, I did negative work. So in a way, there is a force that pulls it in.

### With power supply connected

What could go through these same exercises, but with the power supply connected. That implies the potential difference can’t change. As a result, you’ll find very different results when increasing $$d$$. E.g., $$E$$ will go down by a factor of two. We leave this as an exercise to the reader.



My notes of the excellent lecture 5 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

The connection between electric potential $$V$$, and electric field, $$\vec E$$.

## Gradient field, $$\vec E=-\nabla V$$

Earlier [1,2,3], we found the following equations \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \label{eq:gauss} \\[1pt] V_P &= \int_P^\infty \vec E\cdot d\vec r \label{eq:electpotential} \\[1pt] V_A – V_B &= \int_A^B \vec E\cdot d\vec l \label{eq:electpotdiff} \end{align}

### Work over closed line

If the electric field is a static electric field (no moving charges) $$\Longrightarrow$$ the forces are conservative forces.

If I take a charge $$q$$ from point $$A$$, all around the conservative field, and back to point $$A$$, then the work based on equation $$\eqref{eq:electpotdiff}$$ is $$V_A-V_A=\int_A^A \vec E\cdot d\vec l = 0 \nonumber$$

We mark the integral with a circle to indicate that it is a closed path. So moving through static electric field, over a closed line, the work is $$0$$ $$\shaded{ \oint \vec E\cdot d\vec l = 0 }$$

### From potentials to electric field

If you look at equation $$\eqref{eq:electpotential}$$ and $$\eqref{eq:electpotdiff}$$, you see that the potential is the integral of the electric field $$\Longrightarrow$$ the electric field must be the derivative of the potential.

Let there be a charge $$Q$$ a distance $$r$$ from point $$P$$

Recall: we derived the electric potential at that location, the scalar

$$V = \frac{Q}{4\pi\varepsilon_0 r} \nonumber$$


Recall: the electric field is defined as the vector

$$\vec E = \underline{\frac{Q\,\hat r}{4\pi\varepsilon_0 r^2}} \nonumber$$

Substituting $$\eqref{eq:pdvV}$$ in this, shows that if you know the potential everywhere in space, you can retrieve the electric field $$\newcommand{dv}[2]{\frac{d #1}{d #2}} \shaded{ \vec E = -\dv{V}{r}\,\hat r } \nonumber$$

That means that if you release a charge at $$0$$ speed, it will always start perpendicular to an equipotential surface, because it always starts to move in the direction of a field line.

#### Going deeper

1. Imagine that I am somewhere in space at position $$P$$. At that position the potential is $$V_P$$. At that position there is an electric field.
2. Now I make a tiny step in the $$x$$-direction. If I measure
• no change in potential, over that little step, it means that the component of the electric field in the direction of $$x$$ is $$0$$,
• a change in potential, it means that the magnitude of the $$x$$-component of the electric field $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} |E_x| = \left|\pdv{V}{x}\right| \nonumber$$ The partial derivative $$\frac{\partial}{\partial x}$$, means you keep the other variables $$y,z$$ constant.
3. Equally for the $$y$$ and $$z$$-directions $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} |E_y| = \left|\pdv{V}{y}\right|,\quad |E_z| = \left|\pdv{V}{z}\right| \nonumber$$

Until now we used the unit $$\tfrac{\rm N}{\rm C}$$ for the electric field, but from now on we will use $$\tfrac{\rm V}{\rm m}$$ because it gives a bit more insight. You move a bit in meters, and see how much the potential changes in volts.

Now we can write the relation between electric field and potential in Cartesian coordinates (instead of just a function of distance $$r$$) \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \vec E &= \left(E_x\hat x+ E_y\hat y+ E_z\hat z\right) \\ &= – \left( \pdv{V}{x}\hat x + \pdv{V}{y}\hat y + \pdv{V}{z}\hat z \right) \\ &= -\left\langle \pdv{V}{x}, \pdv{V}{y},\pdv{V}{z} \right\rangle \end{align*} \nonumber

Using the symbolic $$\nabla$$-operator, this is written as $$\shaded{ \vec E = -\rm{grad}\,V = -\nabla\,V } \tag{gradient} \label{eq:gradient}$$

Note that here in physics the sign is opposite as used in the math notation for a gradient field.

#### Example: given potentials

We have a potential field $$V(x)=10^5 x$$ for $$0\lt x\lt 10^{-2}\rm m$$. What is the electric field in space?

The electric field $$\vec E$$ follows from equation $$\eqref{eq:gradient}$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \vec E &= -\nabla\,V = -\left\langle\pdv{10^5 x}{x},\pdv{10^5 x}{y},\pdv{10^5 x}{z}\right\rangle \\ &= -\left\langle 10^5,0,0\right\rangle = \underline{-10^5\,\hat x} \end{align*}

This is constant and opposite to $$\hat x$$.

#### Example: parallel plates

Continue the charged plates example. If the plates are close together $$d\ll \sqrt{A}$$ where $$A$$ is the area of a plate, we can approximate the field between these plates as those coming from a pair of infinite planes. We saw that the electric field inside is independent of the distance.

Assume this electric field is the $$\vec E=-10^5\hat x$$ from the previous example.

The potential difference follows from equation $$\eqref{eq:electpotdiff}$$, where going from $$A$$ to $$B$$ $$\Rightarrow$$ $$d\vec l=d\vec x$$ \begin{align*} V_A – V_B &= \int_A^B \vec E\cdot d\vec x = \int_A^B \underline{-10^5\,\hat x}\cdot d\vec x \\ &= -10^5 \int_A^B \hat x\cdot\vec x \end{align*}

The unit vector $$\hat x$$ is in the same direction as $$d\vec x$$, so the dot-product becomes a multiplication of scalars $$|\hat x|\,|d\vec x|\,\cos 0=$$$$1.dx.1$$ \begin{align*} V_A – V_B &= -10^5 \int_A^B dx = -10^5\Big[x\Big]_A^B \\ &= -10^5 (x_B – x_A) \end{align*}

If plates $$A$$ and $$B$$ are $$10\,\rm{cm}$$ apart $$V_A – V_B = -10^5\,10^{-2} = -1000\,\rm V \nonumber$$

Note

• $$A$$ is $$1000\,\rm V$$ lower than $$B$$.
• From $$A$$ to $$B$$ the potential increases linearly.

Instead of choosing the $$0$$-potential at $$\infty$$, we could choose it at plate $$A$$. Then plate $$B$$ would be 1000 V, and the potential $$V$$ and the electric field $$\vec E$$ become \shaded{ \left\{ \begin{align*} V &= 10^5 x \\ \vec E &= -10^5\,\hat x \end{align*} \right. } \nonumber

## Electrostatic Shielding

As long as there is no charge moving, and we’re dealing with solid conductors $$\Longrightarrow$$ we have static electric fields (the charges are not heavily moving) $$\Longrightarrow$$ the field inside the conductor is always $$0$$. Because conductors have free electrons, and if these free electrons see an electric field inside, then they start to move until they no longer experience a force $$\Longrightarrow$$ they kill the electric field inside.

So the charge in a conductor always rearranges itself so that the electric field becomes zero.

### Charging a solid conductor

Suppose we bring a plus charge near a solid conductor.

For a very short moment, there will be an electric field inside the conductor. This field will move the free electrons close to the plus charge (induction), leaving net positive charge behind. Once these charges cancel out the electric field, there is no longer an electric field to move them, so the electrons stay still. [MIT]

So, after this very brief moment, the electric field inside is zero. This implies

• This inside of the solid contains no charge. (Assume a small Gaussian surface inside the conductor, then equation $$\eqref{eq:gauss}$$ tell us that this inside surface contains no charge.)
• The entire solid is an equipotential. (Based on equation $$\eqref{eq:electpotdiff}$$.)

The charge has to be on the surface. It however is not uniformly distributed with this odd shape (as we see later). With the solid being an equipotential, the field lines are perpendicular to the surface of the solid.

External electric field lines are perpendicular to the surface. Because if a component of the $$\vec E$$-field was tangent to the surface, charges would flow along the surface until there was no longer any tangential component.

### Charging a hollow conductor

The same shape, but hollow. Once more, we charge it.

Again, for a very short moment, there will be an electric field inside the that moves the free electrons so they cancel out the “external” electric field. After that, there is no longer an electric field to move them.

So, after this very brief moment, the electric field inside is zero. This implies

• This inside of the shell and cavity contains no charge. (Assume a small Gaussian surface $$S$$ just under the surface of the conductor, then equation $$\eqref{eq:gauss}$$ tell us that it encloses no charge.)
• The whole shell including this cavity is an equipotential. (Based on equation $$\eqref{eq:electpotdiff}$$.)

There is never any electric field anywhere inside the shell or cavity. The only electric field is on the outside of the shell. So, again, the charge has to be on the outside.

Pretty amazing: one side becomes negative, the other side becomes positive; the whole things becomes an equipotential; no charge inside.

Earlier we saw that a sphere has no electric field inside. Here, we demonstrated that it doesn’t have to be a sphere (and also can be solid). Any conductor shape will give you an electric field of $$0$$ inside.

An external electric field (such as from a van der Graaff generator), can cause polarization on the outside of the conductor, but inside, even the inside wall will have no charge (no electric field). After a very brief moment, the free moving electrons rearranged themselves, in such a way that the electric field is zero everywhere inside the conductor.

The inside is electrically shielded from the outside world. You would not notice a strong electric field outside. This is called electrostatic shielding. The object will be called a Faraday cage. Named after the English scientist Michael Faraday, who invented them in 1836

### A charge inside a conductor

A hollow metallic sphere is initially uncharged. Now imagine that a positive point charge $$q$$ is placed somewhere (not in the middle) in the sphere without touching the walls.

Now there is positive charge inside, so there has be an electric field inside. For a very short moment the free electrons in the conducting wall will move so that the electric field in the wall becomes zero.

There must be charge on the inside surface, because if we take a Gaussian surface $$S$$ halfway the shell, the electric field is zero at that surface, so the enclosed charge is zero. That means that negative charge must have accumulated on the inside wall of the conductor, to make the net charge inside surface $$S$$ is zero.

Originally the conductor had no charge. On the outside surface we now see charge $$q$$, because the minus charge on the inside surface came from electrons that moved from the outside surface.

The plus charge inside, creates an electric field inside, that creates a negative charge on the inside, and a plus charge on the outside.

Because of the spherical symmetry, the charge on the outside will be uniformly distributed. That is the only way nature can obey the laws of physics. (The conductor must become an equipotential; there can be no electric field inside the conductor; the electric field lines have to perpendicular to the surface; the $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S \vec E \cdot d\vec l$$ must be zero everywhere.)

The uniform outside charge is independent of the position of the charge $$q$$ inside. The outside world will not notice if you move the charge $$q$$ around. It has no way of knowing what happens on the inside. We call that electrostatic shielding, the effect of the Faraday cage.

## Potential



My notes of the excellent lecture 4 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

## Electrostatic Potential Energy

To put two positive charges $$q_1$$ and $$q_2$$ together, we have to do work. This is called electrostatic potential energy.

To place $$q_1$$ in an empty place doesn’t take any work. But to then bring $$q_2$$ from $$\infty$$ to point $$P$$ at distance $$R$$, we have to push with force $$F_{wl}$$ to overcome the force of the electric field $$F_{el}$$ from $$q_1$$ on $$q_2$$.

The force and the direction we’re moving are the same, so we do positive work. The work done to bring $$q_2$$ from $$\infty$$ to $$P$$, is the same as minus the electric force $$\shaded{ W_{wl} = \int_\infty^R \vec F_{wl}\cdot d\vec r = \int_R^\infty \vec F_{el}\cdot d\vec r } \tag{Work}$$

The electric force follows from Coulomb’s law

$$F_{el} = \frac{q_1\,q_2\,K}{r^2} \nonumber$$

$$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec F_{el} \parallelsum d\vec r$$, so the dot-product becomes a multiplication of scalars $$|\vec F_{el}|\,|d\vec r|\,\cos 0=F_{el}\,dr$$ \begin{align*} W_{wl} &= \int_R^\infty \frac{q_1q_2}{4\pi\varepsilon_0r^2}\,dr = \frac{q_1q_2}{4\pi\varepsilon_0} \int_R^\infty \frac{dr}{r^2} \\ &= \frac{q_1q_2}{4\pi\varepsilon_0} \left[-\frac{1}{r}\right]_R^\infty \end{align*} \nonumber

The energy $$U_P$$ ($$E_{pot}$$ in Europe) required to bring $$q_2$$ from $$\infty$$ to point $$P$$ $$\shaded{ U_P = \frac{q_1q_2}{4\pi\varepsilon_0R} } \quad \left[\rm{J}\right] \tag{Potential energy}$$

Notes

Work is the difference between the potential energy at the start point and the end point. $$\shaded{ W = U_1 – U_2 } \tag{Work}$$

## Electrostatic Potential

Assume one charge $$Q$$, a test charge $$q$$ at distance $$\infty$$, and point $$P$$ at distance $$R$$ from $$Q$$.

We just calculated the electrostatic potential energy $$U_P$$, the work required to bring $$q$$ from $$\infty$$ to point $$P$$

$$U_P = \frac{qQ}{4\pi\varepsilon_0 R} \nonumber$$

The electric potential $$V_P$$ ($$\phi$$ in Europe) is defined as the work per unit charge required to bring test charge $$q$$ from $$\infty$$ to point $$P$$ $$\shaded{ V_P = \frac{U_P}{q} = \frac{Q}{4\pi\varepsilon_0 R} } \quad \left[\rm{V=\frac{J}{C}}\right] \tag{Potential}$$

Note

• The unit is Volt named after Alessandro Giuseppe Antonio Anastasio Volta, an Italian physicist, chemist, who invented the battery.
• If $$Q$$ is negative, then $$V_P$$ is negative.
• $$V_P$$ is the potential at distance $$R$$ from charge $$Q$$.
• $$V_\infty = 0$$

### Example: van der Graaff

A van de Graaff generator with radius $$R$$ and charge $$Q$$. What is the electric potential at any point in space?

The potential at point $$P$$ is defined as $$\shaded{ V_P = \int_P^\infty \frac{\vec F_{el}}{q}\cdot d\vec r = \int_P^\infty \vec E\cdot d\vec r } \tag{Potential} \label{eq:electpotential}$$

Both $$\vec E$$ and $$\vec r$$ are radial, so the dot-product becomes a multiplication of scalars $$|\vec E|\,|d\vec r|\,\cos 0=E.dr.1$$; substitute $$\vec E$$ from what we found earlier \begin{align*} V_P &= \int_R^\infty \frac{Q}{4\pi r^2\varepsilon_0} dr = \frac{Q}{4\pi\varepsilon_0} \int_R^\infty \frac{1}{r^2}\,dr \\ &= \frac{Q}{4\pi\varepsilon_0} \left[-\frac{1}{r}\right]_R^\infty = \frac{Q}{4\pi\varepsilon_0 r} & (r\gt R) \end{align*} \nonumber

Note

• This is of course the same as for the point charge.
• For $$r=R$$, $$V=300\,\rm{kV}$$; for $$r=3m$$, $$V=30\,\rm{kV}$$.
• If I bring charge $$q$$ from $$\infty$$ to the surface of the van der Graaff, I do work $$W_R = q\,V_R$$

If bring the charge inside the van der Graaff, where the field is $$0$$, I don’t experience any force $$\Longrightarrow$$ I do no work $$\Longrightarrow$$ the potential must remain constant, $$V_R$$. The absence of an electric field implies that the potential everywhere is the same.

## Equipotential surfaces

Two charges $$Q_1$$ and $$Q_2$$. What is the potential at point $$P$$?

The superposition principle applies $$V_P = V_{P_{Q1\text{ alone}}} + V_{P_{Q2\text{ alone}}} \nonumber$$

### Example: two positive charges

An example of equipotential lines (and field lines)

Note

• If you go far away and close to the charges, the equipotential lines become spheres.
• In between the charges, there is a point where the electric field is $$0$$, but has a positive potential.
• Everywhere the field lines are perpendicular to the equipotential lines. Imagine you purposely move with a charge only perpendicular to the electric field lines $$\Longrightarrow$$ the force on you and the direction in which you move are always at a 90° angle $$\Longrightarrow$$ you never do any work, because $$\vec E\cdot d\vec l=0$$ $$\Longrightarrow$$ the potential remains the same.
These is really 3D, so you should rotate it around the vertical

• The lines are plotted as quantiles, to make it less cluttered around the charges.

### One positive and one negative charge

Another example of equipotential lines (and field lines)

Note

• At one of the lines around $$-1$$ the potential has to be $$0$$, because inside it is negative, and far away it is positive. If you come from far away, and arrive at that $$0$$ line, you have done $$0$$ work. $$E\neq 0$$ there
• Above the $$-1$$, there is a point where the electric field is $$0$$. $$V\neq 0$$ there

In cases where the electric field is very complicated, it is easier to work with the potentials. The change in kinetic energy, only depends on the change in potential.

If you’re only interested in the change of kinetic energy, and not interested in the details of the trajectory, then equipotential lines are very handy.

Do not confuse:

• electrostatic potential energy $$U$$ with unit Joules,
• with electric potential $$V$$ with unit Volts (Joules/Coulomb).
If I have a bunch of charges, $$U$$ has only one value, the work that I have to do to put those charges there. But $$V$$ has a different value everywhere.

In gravity, matter want to go from high potential to a place with low potential. Analogue, positive charges will also go from a high electric potential to a low electric potential. Uniquely for electricity, negative charges will go from a low potential to a high potential.

## Potential difference

Two points in space, with a potential, separated a distance $$R$$

The potentials $$V_A$$ and $$V_B$$ are defined in equation $$\eqref{eq:electpotential}$$ as \begin{align*} V_A &= \int_A^\infty \vec E\cdot d\vec r \\ V_B &= \int_B^\infty \vec E\cdot d\vec r \end{align*}

$$\vec E$$ is force per unit charge, not work.

So, the potential difference between point $$A$$ and $$B$$ \begin{align*} V_A – V_B &= \int_A^B \vec E\cdot d\vec r \\ V_B – V_A &= -\int_A^B \vec E\cdot d\vec r \end{align*} \nonumber

Note

• If there is no electric field between $$A$$ and $$B$$, the potential is the same.

We change the $$d\vec r$$ to a different symbol: element $$d\vec l$$ to indicate any path between $$A$$ and $$B$$

It makes no difference how you march, because we’re dealing with conservative fields (independent of the path).

The potential difference $$\Delta V$$ ($$U$$ in Europe) $$\shaded{ \Delta V \overset{\Delta}{=} V_A – V_B = \int_A^B \vec E\cdot d\vec l } \tag{Potential difference} \label{eq:electpotdiff}$$

Some books write this as $$V_B – V_A = -\int_A^B \vec E\cdot d\vec l \nonumber$$

$$U$$Instead of $$\Delta V$$, we often use the European symbol for voltage difference $$U$$. The letter ‘U’ stands for “Potentialunterschied”.

Let’s assume $$V_A=150\,\rm V$$ and $$V_B=50\,\rm V$$

• If I bring a positive charge $$q$$ to point $$A$$, I have to do work $$q\,V_B = 50\,q$$
• If I bring the charge $$q$$ to point $$B$$, I have to do work $$q\,V_A = 150\,q$$

If I am at point $$A$$ the charge $$q$$ go to the lower potential at point $$B$$, and $$150\,q-50\,q$$ kinetic energy is released. The change in potential energy, when the plus $$q$$ charge goes from $$A$$ to $$B$$, is $$V_A – V_B$$ $$q(V_A-V_B) \nonumber$$

In conservative fields, the sum of potential and kinetic energy stays the same. The conservation of energy. So $$q(V_A-V_B) = K_B – K_A \nonumber$$ where $$K$$ is kinetic energy.

## Conductors

Any piece of metal is an equipotential, as long as there no charge moving inside the metal. Because these charges inside the metal, these electrons, when they experience an electric field, they move immediately in the electric field. They will move until there is no force on them any more $$\Longrightarrow$$ they have made the electric field $$0$$.

Charges inside a conductor always move automatically in such a way that they kill the electric field inside.

Metals, for as long as there is no charge moving inside, will always be an equipotential.

#### Example 1

If I connect point $$A$$ to a metal trash can, and point $$B$$ to a metal soda can, then the trash can will be at $$150\,\rm V$$ and the whole soda can will be at $$50\,\rm V$$. I place the whole thing in vacuum, and release an electron at point $$B$$.

The electron wants to go to a higher potential. The electric potential energy is available, and the electron will start to pick up speed and end up at $$A$$. How it will travel, I don’t know. The field configuration is enormously complicated between the soda can and the trash can.

But if all we want to know is the kinetic energy, the speed, then we can use the work-energy theorem. The available potential energy is charge of the electron time the potential difference between two objects \begin{align*} q (V_A-V_B) &= K_B – K_A \\ \Rightarrow -1.6\times 10^{-19} (100) &= 0 – \tfrac{1}{2}m_e {v_A}^2 \\ \Rightarrow 1.6\times 10^{-19} (100) &= \tfrac{1}{2}9\times10^{-31}\,v{_A}^2 \\ \Rightarrow v_A &\approx 5.96\times 10^6\,\rm\frac{m}{s}\approx 0.02\,c \end{align*}

Using the equipotential, we can quickly calculate the kinetic energy and the speed of the electron as it arrives at $$A$$, without any knowledge of the complicated electric field.

All potentials are defined relative to $$\infty$$ $$\Longrightarrow$$ $$V_\infty=0$$. But it doesn’t matter where you think of you $$0$$, similar as the height with gravity. Only the potential difference matters. Note that electrical engineers always call the potential of the earth $$0$$ when they build their circuits.

#### Example 2 (Lec.12)

Kinetic energy increase due to charges that move over a potential difference.

Two conductors of odd shape that are equipotential (no current inside the conductors) in vacuum. Let’s assume $$V_A\gt V_B$$

Release charge $$q$$ at speed zero, that will go to $$B$$. What is the speed as it reaches $$B$$?

The electric field will do the work $$W_{el}$$ on this charge. Since it is a conservative field, it is path independent. \begin{align*} W_{el} &= \int_A^B \vec F_{el}\cdot d\vec l, & \left(\vec F = q\vec E \right) \\ &= q \int_A^B \vec E\cdot d\vec l \\ \end{align*}

Where $$\int_A^B\vec E\cdot d\vec l$$ is the potential difference $$W_{el} = q\,(V_A – V_B) \nonumber$$

##### Quantitively

A proton traverses a potential difference \begin{align*} m_p &= 1.7\times 10^{-27}\,\rm{kg} \\ q_p &= 1.6\times 10^{-19}\,\rm{C} \\ \Delta V &= 10^6\,\rm V \end{align*}

What is the kinetic energy when the proton reaches $$B$$? \begin{align*} W_{el} &= q\,\Delta V \\ &= 1.6\times 10^{-19}\,(10^6) \\ &= 1.6\times 10^{-13}\,\rm J \end{align*}

This is commonly called one million electro volt, $$1\,\rm{MeV}$$. Where an electro volt is the energy that an electron gains if it moves over a potential difference of one volt.

The speed that it arrives at $$B$$ \begin{align*} W_{el }&= \tfrac{1}{2} m_p\,v^2 \\ \Rightarrow v &= \sqrt\frac{2W_{el}}{m_p} \\ &= \sqrt\frac{2\left(1.6\times 10^{-13}\right)}{1.7\times 10^{-27}} \\ &\approx 1.37\times 10^7\,\rm{m/s} \approx 0.05\,c \end{align*}

One more, that is 5% of the speed of light, comfortably low, so we don’t have to make any relativistic corrections.

## Gauss’ law



My notes of the excellent lectures 3 and 12 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

Gauss’ law relates the electric field through a closed surface with the charge contained within that surface.

## Electric Flux

Electric flux is the amount of electric field passing through a surface.

### Open surface

Bring an open surface in an electric field. The normal vector $$\hat n$$ is perpendicular to the tiny surface $$dA$$ and chosen to be upwards. The electric field at $$dA$$ is $$\vec E$$

The electric flux $$d\phi$$ that goes through $$dA$$ is defined as the scalar \shaded{ \begin{align*} d\phi = \vec E \cdot \hat n\, dA = \vec E \cdot d\vec A = E\,dA\,\cos\theta \end{align*} } \nonumber

The flux for the whole surface $$A$$ follows from the $$\int dA$$.

### Closed surface

By convention, the normal $$\hat n$$ of a closed surface is from the inside to the outside of the surface.

The total flux through region $$R$$ follows from the integral $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi = \iint_R \vec E \cdot d\vec A } \quad \left[\frac{Nm^2}{C}\right] \tag{flux}$$

Notes

• The circle in the integral sign reminds us that it is a closed surface.
• If the flux is $$0$$, then what flows in also flows out.
• If more flows out than in, the flux is positive.

#### Example

A point charge $$+Q$$ in a sphere of radius $$R$$


The electric field $$E_R$$ at distance $$R$$ $$\vec E = \frac{Q}{4\pi\varepsilon_0 R^2}\,\hat r \nonumber$$

Substituting $$E_R$$ in the expression for $$\phi$$ $$\phi = \bcancel{4\pi R^2} \, \frac{Q}{\bcancel{4\pi R^2}\,\varepsilon_0} = \frac{Q}{\varepsilon_0} \nonumber$$

Note:

• This flux $$\phi$$ is independent of the distance $$R$$. This is not surprise, because if you think of it as air flowing out, then all the air has to come out somehow no matter how big the sphere is. This implies that we could have taken any kind of surface around the point charge, and have found the same result.
• The same relation holds for any number of charges inside the surface.

## Gauss’ law

Gauss’ law relates the electric field through a closed surface with the charge contained within that surface.

### Definition

The law was first formulated by Joseph-Louis Lagrange, an Italian mathematician and astronomer (1773), followed by Carl Friedrich Gauss, a German mathematician and physicist (1835).

This brings us to Gauss’ law, also known as Maxwell’s first equation $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi = \oiint_S \vec E \cdot d\vec A = \frac{\sum_i Q_i}{\varepsilon_0} } \tag{Gauss’ law} \label{eq:gauss}$$

Where

• $$Q_i$$ is any charge inside any closed surface $$S$$
• $$\varepsilon_0$$ is the permittivity of free space.

Note

• If the flux is $$0$$, it means there is no net charge inside the surface.
• Gauss’ law always holds, no matter how weird the charge distribution inside the surface is, or how weird the shape of the surface.
• Both Coulomb’s and Gauss’ law make the electric field with a charge $$Q$$.

Gauss’ law holds, but will not help you calculate $$\vec E$$ when the charges are not distributed symmetrically. Here we will use spherical symmetry, cylindrical symmetry and flat planes with uniform charge distributions.

### Example: using spherical symmetry

Consider a sphere with radius $$R$$ and charge $$Q$$ uniformly distributed over the surface. We want to know the electric field at distance $$r$$ inside and outside the sphere.

The key is choosing the Gaussian surface right. We choose a sphere with radius $$r$$.

Symmetry arguments

1. Because of the symmetry the electric field is the same anywhere on the Gaussian surface.
2. The electric field is either pointing radially outwards or radially inwards $$\Longrightarrow$$ $$d\vec A$$ and $$d \vec E$$ are parallel or anti-parallel
3. $$\Longrightarrow$$ so the dot-product becomes a multiplication of scalars.

The surface area of the Gaussian surface is $$4\pi r^2$$.

#### Inside the charged sphere, for $$r\lt R$$

The enclosed charge is $$0$$. Applying equation $$\eqref{eq:gauss}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \\ \Rightarrow E \oiint_S dA &= \frac{0}{\varepsilon_0} \\ \Rightarrow 4\pi r^2 E &= 0 \\ \Rightarrow E &= 0 \end{align*}

That means that anywhere inside the sphere, there is no electric field! This is non-trivial. All distributed charges cancel out on the inside.

#### Outside the charged sphere, for $$r\gt R$$

The enclosed charge is $$Q$$. Applying equation $$\eqref{eq:gauss}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \\ \Rightarrow E \oiint_S dA &= \frac{Q}{\varepsilon_0} \\ \Rightarrow E &= \frac{Q}{4\pi r^2\varepsilon_0} \end{align*}

Note:

• The direction is radially outwards is $$Q$$ is positive, and inwards otherwise.
• This is non-trivial as well, because it behaves as a point charge at the center of the sphere as we have seen in the first example.

#### Conclusion

The electric field

• inside the sphere is $$0$$,
• outside the sphere, it is the same as if there was a point charge at the center of the sphere.

Plot of the electric field, $$E(r)$$

Key to Gauss’ law that the electric field falls of at $$\frac{1}{r^2}$$. A gravitational field also falls of at $$\frac{1}{r^2}$$ $$\Longrightarrow$$ if there was an hollow spherical planet, there would be no gravitational force inside.

### Example: using plane symmetry

Express the electric field anywhere in space for a very large plane with a uniformly distributed charge $$Q$$, with charge density $$\sigma$$ on area $$A$$ $$\sigma = \frac{Q}{A} \quad \left[\frac{C}{m^2}\right] \nonumber$$

If the very large plane is $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum$$ $$xy$$-plane, there is symmetry along the $$x,y$$-axis $$\Longrightarrow$$ physical characteristics will not depend on $$x$$ or $$y$$ $$\Longrightarrow \vec E$$ can only have a $$z$$-component $$\Longrightarrow$$ $$\vec E \perp$$ the plane.

As the Gaussian surface, we choose a closed cylinder through the plane

The Gaussian surface takes advantage of symmetry

1. The rounded wall is $$\perp$$ to $$\vec E \Longrightarrow \phi_\text{wall}=0$$.
2. The flat end plates are $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum$$ to $$\vec E$$.
3. Both end plates are distance $$d$$ from the plane.

For both end plates $$\vec E$$ is pointed away from the plate. Let the surface of each end plate be $$A$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \phi &= \phi_\text{wall} + \phi_\text{top} + \phi_\text{bottom} \\ &= 0 + 2 \oiint_S \vec E \cdot d\vec A = 2 \oiint_S E\,dA \\ &= 2EA \end{align*}

Apply $$\eqref{eq:gauss}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \phi &= \frac{\sum_i Q_i}{\varepsilon_0} \quad \Rightarrow \quad 2E\bcancel{A} = \frac{\sigma \bcancel{A}}{\varepsilon_0} \end{align*}

So, the electric field is independent of distance $$d$$ $$\shaded{ E = \frac{\sigma}{2\varepsilon_0} } \tag{E-field plate}$$

While this is for an infinitely large plane, but also holds if the distance $$d$$ is small compared to the linear size of the plane.

### Example: two charged planes

Two very large plates, with surface charge densities ($$Q/A$$), $$+\sigma$$ and $$-\sigma$$, a distance $$d$$ apart

Sum the electric field vectors using the superposition principle $$\shaded{ E = \frac{\sigma}{\varepsilon_0} } \tag{E-field plates}$$

Picture

Note that, when you get to the edge of the plates, the symmetry argument is no longer true.

### Example: using cylindrical symmetry (lec.12)

A very long cylinder with radius $$R$$ with uniform charge distribution throughout the whole cylinder with positive density $$\rho$$. What is the electric field inside and outside the cylinder?

Let’s start with outside the cylinder, $$r\leq R$$. The gaussian surface is a cylinder with radius $$r$$ and length $$l$$. The ends of the gaussian surface are flat perpendicular to the axis of symmetry

Symmetry arguments:

1. On the curved gaussian surface, the distance $$r$$ is the same everywhere, so the electric field is the same.
2. Given that the curved gaussian surface is a cylinder, the electric field must everywhere be radial, perpendicular to the axis of symmetry.
3. The electric flux through the flat end surfaces must be zero, because $$\vec E\perp d\vec A$$
4. On the curved gaussian surface, $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec E\parallelsum d\vec A$$


Inside the cylinder, $$r\leq R$$, a similar gaussian surface, but inside the cylinder


Plotting $$E(r)$$

If it were a solid conductor, the charge inside would be zero

## Electric field



My notes of the excellent lecture 2 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

If we have a charge positive $$Q$$ and positive test charge $$q$$, separated by distance $$r$$

The force $$\vec F$$ on test charge $$q$$ follows form Coulomb’s law $$\vec F = \frac{q\,Q\,K}{r^2}\,\hat r \nonumber$$

The electric field $$\vec E$$ at location $$P$$ is defined as the force per unit charge $$\shaded{ \vec{E_P} = \frac{\vec F}{q} = \frac{Q\,K}{r^2}\,\hat r } \quad \left[\frac{N}{C}\right] \tag{electric field}$$

Notes:

• by convention the test charge is positive,
• the electric field tells you about the charge configuration.

## Graphical representations

An electrical field is a vector field. The vectors point in the direction that a positive test charge would experience.

Note:
• There are infinite number of vectors, but we only draw a handful
• The length of the vector gives you an idea of the magnitude
• Very close to the charge the arrows are larger, and off at one of $$r^2$$. This makes it hard to see vectors further away from the charge.

## Multiple charges

If we have multiple charges, $$q_1\ldots q_i$$, then each charge will contribute to the electric field $$\vec E_P$$ at point $$P$$

Applying the superposition principle from Coulomb’s law \shaded{ \begin{align*} \vec{E_P} &= \vec{E_1} + \vec{E_2} + \vec{E_3} + \cdots \\ &= \sum_i \vec{E_i} \end{align*} } \nonumber

If I place a charge $$q$$ at location $$P$$ in electric field $$\vec E$$, then the force $$\vec F$$ on that charge $$\shaded{ \vec F = q\,\vec E } \nonumber$$

Note:

• if $$q$$ is negative, the force will be in the opposite direction of $$\vec E$$,
• if $$q$$ is large the force will be large
• if $$q$$ is small the force will be small

## Quantitative

Suppose, two charges: $$q_1=+3$$, and $$q_2=-1$$

Note

• Because of the with inverse $$r^2$$ relationship, just to the right of $$-1$$ that charge will win. Far to the right, the distance between the charges doesn’t mater, the field will be as if there was a $$+3-1=+2$$. Somewhere in between, there is a point $$p$$ where the charges cancel each other out, and $$\vec E=\vec 0$$.
• Anywhere far away from the charges, the vectors are pointing out. There, the configuration behaves as $$+3-1=+2$$.

## Field lines

A positive test particle placed on a field line, will experience a force tangential with that field line. For same charge configuration

Note

• The field lines tell you in what direction a positive charge will experience a force.
• There are infinite field lines (as there are vectors)
• You can think about the field lines as air moving from a source to a sink.
• With field lines we lost the information about the magnitude of the field. But where the density of the lines is high the field is stronger.
• You can visualize field lines using grass seeds suspended in oil.

## Trajectories

Field lines are not trajectories.

$$q$$ in parallel field $$q$$ in non-parallel field
If you release a charge with $$0$$ speed, it will accelerate along the field line and stay on the field line. If you release a charge with $$0$$ speed, it will accelerate tangential to the field line, but immediate abandon that field line.

## Dipole

The special case where the charges are equal in magnitude, but opposite in sign

Let $$Q$$ and $$-Q$$ be point charges, placed at $$(-0.5,0)$$ and ((0.5,0)\). The field on the $$x$$-axis at $$x=r$$ for $$r\gt 0$$ for the charges \begin{align*} E_+ &= \frac{K(+Q))}{\left(r-0.5\right)^2},\quad E_- = \frac{K(-Q))}{\left(r+0.5\right)^2} \nonumber \\ \Rightarrow E &= E_+ + E_- = \frac{KQ}{r^2\left(1-\tfrac{1}{2r}\right)^2} – \frac{KQ}{r^2\left(1+\tfrac{1}{2r}\right)^2} \end{align*}

Since $$\frac{1}{2r}\ll 1$$, approximate using binominal series expansion $$\frac{1}{\left(1\pm \tfrac{1}{2r}\right)^2} \approx 1\mp \frac{1}{r} \nonumber$$

Substitute the expansion \begin{align*} E &= \frac{KQ}{r^2}\left( \left(\bcancel{1}+\tfrac{1}{r}\right) – \left(\bcancel{1}-\tfrac{1}{r}\right) \right) \\ &= \frac{2KQ}{r^3} \end{align*}

So, if you are far away from a dipole, the field falls off at $$\displaystyle\frac{1}{r^3}$$.

## Coulomb’s law



My notes of the excellent lecture 1 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

The electric force between two particles decreases with the inverse square of the distance, just as the gravitational force.

If we take two charges $$q_1$$ and $$q_2$$, separated by a distance $$r$$. Then the force $$\vec F_{1,2}$$ from $$q_1$$ on $$q_2$$ is visualized as

Charles-Augustin de Coulomb, a French physicist, in 1785 published the following relationship $$\shaded{ \vec F_{1,2} = \frac{q_1\,q_2\,K}{r^2}\,\hat r_{1,2} } ,\quad K=\frac{1}{4\pi\varepsilon_0} \tag{Coulomb’s law}$$

Where

• $$\vec F_{1,2}$$, force from $$q_1$$ on $$q_2$$ $$[\rm N]$$
• $$q_1$$ and $$q_2$$, the values of charge $$[\rm C]$$, the unit is named after Coulomb
• $$r$$, distance between the charges $$[\rm m]$$
• $$K$$, Coulomb’s constant $$\approx 9\times 10^9$$
• $$\varepsilon_0$$ is the permittivity of free space.

Note

• The relationship is sign sensitive. If once charge is negative and the other positive the force is in the opposite direction.
• There is a clear parallel with gravity (except that gravity never repels), where $$\vec F_g = \frac{m_1\,m_2\,G}{r^2} \nonumber$$

## Superposition principle

With multiple charges, the superposition principle applies, because it is consistent with all our observations.

Let $$\vec F_{1,2}$$ be the force from $$q_1$$ on $$q_2$$, and $$\vec F_{3,2}$$ be the force from $$q_3$$ on $$q_2$$. Assume $$q_1$$ and $$q_2$$ are positive, and $$q_3$$ is negative. The forces can be shown as

The net force on $$q_2$$ is the vectoral sum of the individual component forces $$\vec F_2 = \vec F_{1,2} + \vec F_{3,2} \nonumber$$

## Electric vs. gravitational force

E.g. charge of a proton/electron $$q_{p^+} = q_{e^-} = 1.6\times 10^{-19}\,\rm C \nonumber$$

In our immediate surroundings, electrical forces are much more powerful than gravitational forces. E.g. two protons repel each other with the forces \left. \begin{align*} F_{el} &= \frac{(1.6\times 10^{-19})^2\ 9\times 10^9}{d^2} \\ F_{gr} &= \frac{(1.7\times 10^{-27})^2\ 6.7\times 10^{-11}}{d^2} \end{align*} \right\} \Rightarrow \frac{F_{el}}{F_{gr}} \approx 10^{36} \nonumber (The nuclear forces hold the protons together)

On the scale of planets, it is gravity that holds our world together. Because the large objects have a very small charge per unit mass. \left. \begin{align*} F_{el} &= \frac{(400\times 10^3)^2\ 9 \times 10^9}{d^2} \\ F_{gr} &= \frac{6\times 10^{24}\ 6.4\times 10^{23}\ 6.7\times 10^{-11}}{d^2} \end{align*} \right\} \Rightarrow \frac{F_{el}}{F_{gr}} \approx 10^{-17} \nonumber