# Electric potential

- Calculating the potential from $\myv E(\myv r)$.
- Calculating the potential from the charge density $\rho(\myv r)$.
- Boundary conditions on $\myv E$ and $V$.
- Differential equations for the potential.

...and a change to the syllabus.

Helmholtz' theorem assures us that an irrotational field, like $\myv E$, can be derived from a scalar potential $\myv E = -\myv \grad V(\myv r)$.

He also tells us how to contruct $V$: $$V(\myv r) = \frac{1}{4\pi} \int \frac{\myv \grad ' \cdot \myv E(\myv r ')}{\rr} d \tau '=\frac{1}{4\pi\epsilon_0} \int \frac{\rho(\myv r')}{\rr} d \tau'.$$

Great. But what if we only know $\myv E$ over a limited portion of space?? and not $\rho$??

### The fundamental theorem for gradients

assures us that... $$\int_{\myv a}^{\myv b} (\myv \grad V) \cdot d \myv l = V(\myv b)-V(\myv a).$$

On the other hand, since $\myv E = -\myv \grad V$, $$\begineq\int_{\myv a}^{\myv b} (\myv \grad V) \cdot d \myv l &=& \int_{\myv a}^{\myv b} (-\myv E)\cdot d \myv l \\ &=& \int_{\cal O}^{\myv b} (-\myv E)\cdot d \myv l - \int_{\cal O}^{\myv a} (-\myv E)\cdot d \myv l.\endeq$$

So, it looks like we could take ** any** reference point ${\cal O}$, and calculate
the potential from:

$$V(\myv r) = -\int_{\cal O}^{\myv r} \myv E\cdot d \myv l.$$

**Units**: Since $\myv E$ has
units of $N / C$, the potential has units of $(N\cdot m)/C$ which is called a **volt**.
Or, since a $N\cdot m$ is a joule, we could also say that a volt is a **joule/coulomb**.

The electric potential does not have units of energy, so the **electric
potential is not potential energy**. However $qV$ does have units of energy!

Notice: $\myv E = \frac{\myv F}{q}$, and $V=\frac{U}{q}$.

### Which reference point?

Changing the reference point from ${\cal O}$ to ${\cal O}'$ just adds a constant to the potential that we come up with... $$\begineq V'(\myv r) &=& -\int_{\cal O'}^{\myv r} \myv E\cdot d \myv l=-\int_{\cal O'}^{\cal O} \myv E\cdot d \myv l+ -\int_{\cal O}^{\myv r} \myv E\cdot d \myv l\\ &=& -\int_{\cal O'}^{\cal O} \myv E\cdot d \myv l + V(\myv r).\endeq$$

Really, what we care about is taking the gradient to find the electric field, and since the gradient of a constant is zero, $$\myv \grad V(r) = \myv \grad V'(r).$$

So let's agree on one. We'll pick...

${\cal O}$ at $\infty$, such that $V(\infty)=0.$

#### Problem 2.21

*Find the potential inside and outside a sphere with a uniform charge density
(total charge $q$).*

You found the field for such a sphere in problem 2.12:
$$\myv E = \{
\begin{array}{cl}
\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\uv{r} &
{\text{when}}\ r \gt R\\
\frac{1}{4\pi
\epsilon_0 }\frac{qr}{R^3} \uv{r} & {\text{ when} }\ r \lt R
\end{array}
\right.
$$
$E_r$

The potential outside the sphere ought to be spherically symmetric, since the field is. Starting at $\infty$ and coming in to a distance $r$ from the origin, which is still outside the sphere... $$V(r) = - \int_\infty^r \myv E \cdot d \myv l'.$$

We approach the sphere along a path where $d \myv l' = -dr'\uv{r}'$, but then $\myv E$ is pointing opposite this direction, so we multiply by another negative sign to get the dot product of the path with the field, to get... $$\begineq V(r) &=& - \int_\infty^r \frac{1}{4\pi \epsilon_0}\frac{q}{r'^2} dr'\\ &=& \frac{1}{4\pi \epsilon_0}\left.\frac{q}{r'}\right|_\infty^r=\frac{1}{4\pi \epsilon_0}\frac{q}{r}.\endeq$$

Once 'inside' the sphere, the potential for $r<R$ is: $$\begineq V(r) &=& V(R) - \int_R^r \frac{q}{4\pi \epsilon_0 R^3}r' dr' \\ &=& V(R)-\frac{q}{4\pi \epsilon_0 R^3} \left.\frac{r'^2}/2 \right|_R^r \\ &=&\frac{q}{4\pi \epsilon_0}\left[\frac{1}{R}-\frac{r^2}{2R^3}+\frac{R^2}{2R^3}\right]\\ &=&-\frac{q}{4\pi \epsilon_0}\frac{1}{2}\left[ \frac{3}{R}- \frac{r^2}{R^3}\right].\endeq$$

Graphing this...
$V(r)$

## Potential of a collection of charge

Gee, wouldn't it be nice if we could calculate $\myv E$ by taking the gradient
of our scalar potential $V$, instead of doing vector integrals over the charge density? Well, if only we could
get the potential based on our charge *distribution*, that might be possible.
Let's see how...

### Superposition

The electric field obeys superposition. So, if we want the potential for a collection of charges, we could integrate $\myv E = \myv E_1 + \myv E_2+...$ as follows: $$\begineq V(\myv r) &=& -\int_\infty^{\myv r} \myv E_\text{total}\cdot d\myv{l}'= -\int_\infty^{\myv r} (\myv E_1+\myv E_2+...)\cdot d\myv{l}' \\ &=&-\int_\infty^{\myv r} \myv E_1\cdot d\myv{l}' -\int_\infty^{\myv r} \myv E_2\cdot d\myv{l}'+...\\ &=&V_1(\myv r)+V_2(\myv r)+... \endeq$$

$\Rightarrow$ So, the potential also obeys a principle of superposition.

### Potential of a point charge

The potential of a single point charge at the origin is... $$\begineq V(\myv r) &=& -\frac{1}{4\pi \epsilon_0} \int_\infty^r \frac{q}{r'^2} dr' =\frac{1}{4\pi \epsilon_0}\left.\frac{q}{r'}\right|_\infty^r\\ &=& \frac{1}{4\pi \epsilon_0}\frac{q}{r}.\endeq$$

Usually we find a charge not at the origin, but at some position $\myv r'$, in which case the potential just depends on the distance $\rr$ away from the charge. $$V(\myv r) = \frac{1}{4\pi \epsilon_0}\frac{q}{\rr}.$$

Since the potential obeys superposition, the potential for several point charges is... $$V(\myv r) = \frac{1}{4\pi \epsilon_0}\sum_{i=1}\frac{q_i}{\rr_i}.$$

It is a small step to the potential based on a continuous distribution of charge:

$$V(\myv r) = \frac{1}{4\pi \epsilon_0}\int \frac{\rho(\myv r')}{\rr} d \tau'.$$

Indeed, the Helmholtz theorem suggested that we could construct the scalar potential as... $$V(\myv r) \equiv \frac{1}{4\pi} \int \frac{D(\myv r ')}{\rr} d \tau ',$$ Where the divergence of our electric field: $D = \myv \grad \cdot \myv E=\rho/\epsilon_0$. So, there ya' go.

This is already an improvement on our integral expression for the electric field which involved keeping track of the vector components of $\uv \rr.$

Notice that positive charge corresponds to "hills" in the potential, and negative charge to "valleys".

#### Problem 2.25

*What's the potential at a height $z$ above the two charges separated by
a distance $d$? Use this to calculate the force as the gradient of the potential.*

Notice how much simpler the diagram is to calculate the potential compared to problem 2.2 ??!!

The potential is just the sum of the two single-particle potentials: $$V(z) = \frac{1}{4\pi \epsilon_0}[q/\rr + q/\rr]=\frac{1}{4\pi \epsilon_0}\frac{2q}{\sqrt{z^2+(d/2)^2}}.$$

Now, to get the electric field we use $\myv E = -\myv \grad V$. But we can still use symmetry to tell us the force has to be in $z$-direction. So, we *know* the gradient points in the $z$-direction. We only need to look at the $z$ component of the electric field,

$$\begineq E_z &=& -\frac{\del}{\del z} V=-\frac{1}{4\pi \epsilon_0}2q\frac{\del}{\del z}\left(z^2+(d/2)^2\right)^{-1/2}\\ &=&-\frac{1}{4\pi \epsilon_0}2q(-1/2)\left(z^2+(d/2)^2\right)^{-3/2}\frac{\del}{\del z}\left(z^2+(d/2)^2\right)\\ &=&-\frac{1}{4\pi \epsilon_0}\frac{2qz}{(z^2+(d/2)^2)^{3/2}}.\endeq$$

Result is: $$E_z = +\frac{1}{4\pi \epsilon_0}\frac{2qz}{(z^2+(d/2)^2)^(3/2)}.$$

### Boundary conditions

Gauss' law is often the most straightforward way to calculate $\myv E.$

If you can't break a problem into pieces solvable by Gauss, the next most productive way of getting to $\myv E$ is often: $$\rho(\myv r')\to V(r) \to - \myv \grad V(\myv r)=\myv E(\myv r).$$

We frequently need to supplement this integrating with **boundary conditions**.

We used this diagram in calculating the (constant) field
above an infinite plane of surface charge $\sigma$ at $z=0$.
$$\myv E = \frac{\sigma}{2\epsilon_0}\frac{|z|}{z}\uv{z}$$

The field had the same magnitude above and below the surface, though it was pointing in both cases perpendicularly away from the surface. So, there was a discontinuity in the perpendicular component "$E_(\perp)$" of: $$\Delta E_{\perp} = E_{\perp}^\text{above} - E_{\perp}^\text{below}=\sigma / \epsilon_0$$

**Any surface** that is varying smoothly looks flat and infinite in the limit as $z \to 0$, so this discontinuity is actually generally true across all smooth surfaces.

Since path integrals of $\myv E$ only depend on the end-points: $$\oint \myv E \cdot d \myv l = 0$$ $$\Rightarrow \myv E_{\parallel}^\text{above} = \myv E_{\parallel}^\text{below}.$$

Again, this holds for all smooth surfaces.

### V is continuous across any boundary

If $a$ is above the surface and $b$ a point below, then the potential difference between these points is $$\lim_{b\to a}(V(b)-V(a)) = \lim_{b\to a}\int_a^b \myv E \cdot d \myv l=0.$$

Of course, that doesn't prevent the *gradient* from being discontinuous. If '$n$' is the perpendicular distance above the surface, then the discontinuity in the perpendicular component of the electric field shows up in the gradient of the potential as:
$$\frac{\del}{\del n}V^\text{above}-\frac{\del}{\del n}V^\text{below} = - \frac{\sigma}{\epsilon_0}$$

### Laplace and Poisson differential equations

These two equations... $$\myv E= - \myv \grad V$$ and $$\myv \grad \cdot E= \rho/\epsilon_0,$$

together imply that... $$\Rightarrow \myv \grad \cdot (\myv \grad V) = \grad^2V= -\rho/\epsilon_0.$$

This is known as **Poisson's equation**.

There are many cases where we're trying to find the field in a region which does not itself contain any charge. So we can restrict ourselves to a special case of Poisson's equation,

In a region without charge:$$\grad^2 V = 0,$$
which has its own name: **Laplace's equation.
**