In this lecture we take a closer look at the relationship between potential and field.

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

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

Recall from Mechanics in 1-Dimension:

$W=\int Fdx\Longrightarrow\Delta U = -\int_i^f Fdx \Longrightarrow F = -\frac{dU}{dx}$

i.e. the faster the potential energy changes, the larger the magnitude of the force;

this leads in three dimensions to:

$\Delta U = -\int_c\vec{F}\cdot\vec{dl}\Longrightarrow \vec{F} = -\vec{\nabla}U$.

When we want to find the gradient vector field we use an operator $\vec{\nabla}$ which is read as del

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

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 $

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

Here is a simple argument to reinforce the concept and convince yourself of the definition:

$\Delta V = \int_c\vec{E}\cdot\vec{dl}\Longrightarrow dV = -\vec{E}.\vec{dl}$

$dV = \frac{\partial V}{\partial x}dx + \frac{\partial V}{\partial y}dy + \frac{\partial V}{\partial z}dz = \vec{\nabla}V\cdot\vec{dl}$

If one moves along an equipotential line, **$dV$ is zero**. This means that the $\vec{\nabla}V$ is perpendicular to that line, which shows why **electric field lines are perpendicular to equipotentials**.

Brown lines are equipotentials (same height), black lines are potential gradients. If the brown lines are clustered together, the slope is steeper, or the gravitational force is larger.

We can explore the characteristics of the gradient using this applet.

The Gradient is the line of steepest slope.

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.

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).

We can explore the electric field on a metal sphere by observing the response of the Geissler tube in the vicinity of the field around a Van der Graaf generator.

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.

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.

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.

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.

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 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 $1 \mathrm{V}$.

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

Capacitors and electrical energy storage