# Lecture 7 - Potential distributions

In this lecture we look at a few more examples of calculating potential from different charge distributions and then take a closer look at the relationship between potential and field.

If you need a pdf version of these notes you can get it here

## Potential due to a point charge

If we want to find the change in potential in going from a distance $r_{a}$ to a distance $r_{b}$ from a single point charge we need to evaluate the integral

$V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$

We can recall that the field is

$E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}$

If we integrate along a path which is directly radially outward $\vec{E}$ and $d\vec{l}$ are in the same direction and

$-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=-\frac{Q}{4\pi\varepsilon_{0}}\int_{r_{a}}^{r_{b}}\frac{1}{r^{2}}dr=\frac{Q}{4\pi\varepsilon_{0}}[\frac{1}{r}]_{r_{a}}^{r_{b}}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$

It is common to define the potential as being zero at $\infty$, and if we do this here (setting $r_{a}=\infty$) the potential as a function of the distance $r$ from a point charge is

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$

## Potential due to a continuous distribution of charge

From the previous example we can conclude that for $n$ point charges each which produces a potential $V_{i}$ at point A the potential at point $A$ will be

$V_{A}=\sum\limits_{i=1}^{n}V_{i}=\frac{1}{4\pi\varepsilon_{0}}\sum\limits_{i=1}^{n}\frac{Q_{i}}{r_{ia}}$

Now if we want to instead consider a continuous charge distribution

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}$

## Potential due to a ring of charge

We can now consider a thin ring of charge, total charge $Q$ with radius $a$. We can ask what the potential is at a point on the axis going through the center of the ring a distance $x$ from the center of the ring. This situation is quite easy because all the charge is at the same distance from the point. So that the integral

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}=\frac{1}{4\pi\varepsilon_{0}}\frac{1}{(x^{2}+R^{2})^{1/2}}\int dq=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{(x^{2}+R^{2})^{1/2}}$

when $x>>R$

$V\approx\frac{Q}{4\pi\varepsilon_{0}x}$

## Potential due to a disk of charge

To find the potential due to a disk of charge which carries charge $Q$ we can divide it in to thin rings, each of which carries charge $dq$. The magnitude of the charge in each ring can be determined from the fraction of the area of the disk that each ring occupies.

The area of each ring is $2\pi R\,dR$ (think of it as a rectangle with length $2\pi R$ and breadth $dR$), so

$\frac{dq}{Q}=\frac{2\pi R\,dR}{\pi R_{0}^{2}}\to dq=\frac{2QR\,dR}{R_{0}^{2}}$

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{(x^{2}+R^{2})^{1/2}}=\frac{1}{4\pi\varepsilon_{0}}\int_{0}^{R_{0}}\frac{2QR\,dR}{R_{0}^{2}(x^{2}+R^{2})^{1/2}}$

$=\frac{Q}{2\pi\varepsilon_{0}R_{0}^{2}}\int_{0}^{R_{0}}\frac{R\,dR}{(x^{2}+R^{2})^{1/2}}=\frac{Q}{2\pi\varepsilon_{0}R_{0}^{2}}[(x^{2}+R^{2})^{1/2}]_{R=0}^{R=R_{0}}$

$=\frac{Q}{2\pi\varepsilon_{0}R_{0}^{2}}[(x^{2}+R_{0}^{2})^{1/2}-x]$

For small $z$, $(1+z)^{1/2}\approx 1+\frac{1}{2}z$, so when $x>>R_{0}$

$V=\frac{Q}{2\pi\varepsilon_{0}R_{0}^{2}}[x(1+\frac{1}{2}\frac{R_{0}^{2}}{x^{2}})-x]=\frac{Q}{4\pi\varepsilon_{0}x}$

To understand the approximation we made, see square root for the Taylor series expansion of $(1+x)^{1/2}$.

## Potential due to a long line of charge

For an infinitely long line of charge with a linear charge density $\lambda$

$V=\frac{1}{4\pi\epsilon_{0}}\int \frac{dQ}{r}=\frac{1}{4\pi\epsilon_{0}}\int_{-\infty}^{+\infty} \frac{\lambda\,dy}{(x^{2}+y^{2})^{1/2}}$

This is a doable, but not particularly easy integral. In this case we are better off using Gauss's law to find the field, which is

$E=\frac{1}{2\pi\varepsilon_{0}}\frac{\lambda}{r}$ pointing radially outward

and using

$V_{r_{b}}-V_{r_{a}}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=-\int_{r_{a}}^{r_{b}}\frac{1}{2\pi\varepsilon_{0}}\frac{\lambda}{r}\,dr=-\frac{\lambda}{2\pi\varepsilon_{0}}[\ln(r)]_{r_{a}}^{r_{b}}=\frac{\lambda}{2\pi\varepsilon_{0}}\ln(\frac{r_{a}}{r_{b}})$

## Relating field and potential

We have seen that electric field is a vector while electric potential is a scalar. The electric field can be thought of as the gradient of the potential at a given point. As we are talking about a 3 dimensional space the gradient will depend on the direction. When we want to find the gradient vector field we use an operator $\nabla$ which is read as del

$\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

When applied to a scalar field we can use the word grad, so we would say “$E$ is equal to minus grad $V$”

$\vec{E}=-\nabla V$

$\vec{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}$

## Field from potential for ring of charge

For a point on the x axis

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}=\frac{1}{4\pi\varepsilon_{0}}\frac{1}{(x^{2}+R^{2})^{1/2}}\int dq=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{(x^{2}+R^{2})^{1/2}}$

$E_{x}=-\frac{\partial V}{dx}=\frac{Q}{4\pi\varepsilon_{0}}\frac{x}{(x^{2}+R^{2})^{3/2}}$

which we can compare with the result we got from Coulomb's law in lecture 3.

## Field from potential for a disk of charge

For a point on the x axis

$V=\frac{Q}{2\pi\varepsilon_{0}R_{0}^{2}}[(x^{2}+R_{0}^{2})^{1/2}-x]$

$E_{x}=-\frac{\partial V}{dx}=\frac{Q}{2\pi\varepsilon_{0}R_{0}^{2}}[1-\frac{x}{((x^{2}+R_{0}^{2})^{1/2})}]$

which we can compare to the result we got from Coulomb's law in lecture 3.

## Visualizing the potential and field

If we confine ourselves to considering the potential in a given plane (x,y) it can be useful to visualize the potential as the third dimension (z) and plot the potential as a surface. We can also plot contour lines which represent equipotential lines. Below is surface plot for a positive (red) point charge and a negative (blue) point charge.

The arrows represent the strength and the direction of the field (which is the negative of the gradient of the potential). In order to visualize the field around these very sharp potentials the size of the arrows is proportional to the logarithm of the field. (Also, the z axis has been truncated as the potential diverges at zero).

If you are interested in making plots like this yourself, they are assembled by combining some different plots generated by Maple. The Maple worksheet used can be downloaded from here (Right click and “save link as”).

## Electric dipole potential

We now look at an electric dipole, again restricting ourselves to points within the (x,y) plane

$V=\frac{Q}{4\pi\varepsilon_{0}}\frac{1}{((x-l/2)^{2}+y^{2})^{1/2}}-\frac{Q}{4\pi\varepsilon_{0}}\frac{1}{((x+l/2)^{2}+y^{2})^{1/2}}$

$V=\frac{Q}{4\pi\varepsilon_{0}}\frac{(((x+l/2)^{2}+y^{2})^{1/2}-((x-l/2)^{2}+y^{2})^{1/2})}{((x-l/2)^{2}+y^{2})^{1/2}((x+l/2)^{2}+y^{2})^{1/2}}=\frac{Q}{4\pi\varepsilon_{0}}\frac{\Delta r}{((x-l/2)^{2}+y^{2})^{1/2}((x+l/2)^{2}+y^{2})^{1/2}}$

where $\Delta r$ is the difference in the distance of our point in the (x,y) plane from one end of the dipole to the other. In terms of $l$ and the angle $\theta$, which is the angle between the dipole length and the line pointing from the dipole to the the point in the (x,y) plane $\Delta r=l\cos\theta$. When the distance from the dipole is much greater than it's length

$V\approx\frac{Q}{4\pi\varepsilon_{0}}\frac{\Delta r}{r^{2}}=\frac{Q}{4\pi\varepsilon_{0}}\frac{l\cos\theta}{r^{2}}=\frac{1}{4\pi\varepsilon_{0}}\frac{p\cos\theta}{r^{2}}$

where $p=Ql$ is the magnitude of the dipole moment.

## Gradient of the electric field

The $\nabla$ operator can also be applied to a vector, but as it is a vector itself there are two operations, that can be performed, the scalar product “$\cdot$” and the vector product “$\times$”.

The scalar product of the $\nabla$ operator with a vector field is called the Divergence of the field, for example the divergence of the electric field is

$\nabla\cdot \vec{E}=\frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{z}}{\partial z}$

The divergence of a field gives a measurement of the magnitude of the source or sink of the field at a point, and we will discuss the relationship between this and charge in a moment.

The vector product of the $\nabla$ operator with a vector field,(e.g. $\nabla\times\vec{E}$) is called the Curl of the field, and gives a measurement of the rotation of the field. The physical significance of this quantity in electromagnetism is connected to magnetism, so we'll come back to this when we talk about magnetic fields.

## Divergence of the electric field

The divergence of a vector field gives a measurement of whether it acts a source or sink of flux. We can see this by applying the divergence theorem to Gauss's law.

$\oint_{A} \vec{E}\cdot d\vec{A}=\int_{V}\nabla\cdot \vec{E}\, dV \to \int_{V}\nabla\cdot \vec{E}\, dV=\frac{Q}{\varepsilon_{0}}$

The total amount of charge may also be written as a volume integral so that

$\int_{V}\nabla\cdot \vec{E}\, dV=\frac{1}{\varepsilon_{0}}\int_{V}\rho\, dV$

where $\rho$ is the charge density. As the integration volume on each side of the equation is equivalent and arbitrary we can say that

$\nabla\cdot \vec{E}=\frac{\rho}{\varepsilon_{0}}$

This is referred to as the differential form of Gauss's law.

## Divergence of the field due to a point charge

Gauss's law, $\nabla\cdot \vec{E}=\frac{\rho}{\varepsilon_{0}}$, tells us that wherever space is free of charge the divergence of the field is zero.

As an example we can check this for a point charge field

$\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{3}}\vec{r}=\frac{1}{4\pi\epsilon_{0}}(\frac{x}{(x^{2}+y^{2}+z^{2})^{3/2}}\hat{i}+\frac{y}{(x^{2}+y^{2}+z^{2})^{3/2}}\hat{j}+\frac{z}{(x^{2}+y^{2}+z^{2})^{3/2}}\hat{k}$)

$\frac{\partial E_{x}}{\partial x}=\frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}-\frac{3x^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}=\frac{x^{2}+y^{2}+z^{2}-3x^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}$

$\frac{\partial E_{y}}{\partial y}=\frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}-\frac{3y^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}=\frac{x^{2}+y^{2}+z^{2}-3y^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}$

$\frac{\partial E_{z}}{\partial z}=\frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}-\frac{3z^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}=\frac{x^{2}+y^{2}+z^{2}-3z^{2}}{(x^{2}+y^{2}+z^{2})^{5/2}}$

$\nabla\cdot \vec{E}=\frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{x}}{\partial z}=0$

If we think about this in terms of the integral form of Gauss's law

$\oint_{A} \vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$

it is another way of stating that the net flux does not change as we draw larger and larger gaussian surfaces around an isolated point charge, even though the field changes considerably.

## Earnshaw's theorem

A consequence of Gauss's law, which tells us that the divergence of an electric field in free space is zero, is Earnshaw's theorem. This theorem states that a collection of point charges cannot be held in stable equilibrium by electrostatic forces alone.

$\nabla\cdot\vec{E}=0$

$E=-\nabla V$

$\nabla\cdot(-\nabla V)=-\nabla^{2}V=-\frac{\partial^{2} V}{\partial x^{2}}-\frac{\partial^{2} V}{\partial y^{2}}-\frac{\partial^{2} V}{\partial z^{2}}=0$

This tells us that there are no true minimas of the potential in free space, which would require $\nabla^{2}V>0$ for a positive test charge and $\nabla^{2}V<0$ for a negative charge. Saddle points can exist, but any equilibrium based on solely electrostatic forces must be unstable in at least one direction.

## The electron volt

The electric potential energy gained or lost by a test charge moved through a potential difference $(V_{b}-V_{a})$ is

$\Delta U=(U_{b}-U_{a})=q(V_{b}-V_{a})$

The SI unit for this is the Joule, but this is a very large unit compared to the actual work in moving typical test charge, for example an electron. We can thus define the electronvolt $\mathrm{eV}$ which is the amount of energy required to move an electron through a potential difference of $1 \mathrm{V}$.

$1\mathrm{eV}=1.6\times10^{-19}\mathrm{J}$

## Next Lecture

Capacitors and electrical energy storage

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