# 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: $$\myv E = \{ \begin{array}{cl} \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\uv{r} & {\text{when}}\ r \gt R\\ \frac{1}{4\pi \epsilon_0 }\frac{qr}{R^3} \uv{r} & {\text{ when} }\ r \lt R \end{array} \right.$$ $E_r$
$r$

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... $V(r)$
$r$

## 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.