In the same way it was useful to define electric field as the electric force divided by the charge it is also useful to define electric potential.

If we define the electric potential energy at a point as $U$ then the electric potential at that point $V$ is

$V=\frac{U}{q}$

In the example we just considered where a charge was moved a distance $d$ by a field $E$

$\Delta U=-W=-qEd\rightarrow\Delta V=-Ed$

We should note that we have been careful to consider differences in potential, as with other forms of potential energy, the zero of electric potential energy, and hence electric potential, can be set arbitrarily.

The unit for electric potential is the volt $V$ and we can see that as the volt is derived from dividing energy by charge $1V=1\frac{J}{C}$

We can also see from the equation above that an equivalent unit for field to $\mathrm{N/C}$ is $\mathrm{V/m}$.

We should note that potential is defined with positive charges in mind, ie. a positive charge at a high potential with respect to some other point has a high potential energy with respect to that point, whereas a negative charge at a high potential with respect to some other point has a low potential energy with respect to that point.

This information is contained in the equation

$\Delta U = q\Delta V$

The electric potential and field are easily related.

$U_{b}-U_{a} = -\int_{a}^{b}\vec{F}\cdot\,d\vec{l} $

$V_{b}-V_{a}=\frac{U_{b}-U_{a}}{q}=-\int_{a}^{b}\vec{E}\cdot\,d\vec{l} $

In the special case of a uniform electric field where the distance between points $a$ and $b$ is $d$

$V_{b}-V_{a}=-\int_{a}^{b}\vec{E}\cdot\,d\vec{l}=-E\int_{a}^{b}dl=-Ed $

When the field is not uniform we can obtain the potential from the field by integration. Note again that the path of the line integral has not been specified: we are free to choose the most convenient path to facilitate the calculation.

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

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

We just showed that the potential due to a point charge as a function of the distance from the charge is

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

If I have another charge $q$ that I want to move from a distance $r_{a}$ to a distance $r_{b}$ then the change in the potential is

$V_{b}-V_{a}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$

and the work done is

$W=q(V_{b}-V_{a})=q(\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}}))$

We should note that here we have taken into account the fact that the field through which are moving the charge is not uniform, we cannot simply say that $W=qEd$. And again, we have conveniently chosen a path where we first travel radially from $r_{a}$ to $r_{b}$ and then travel at the same radius to the specified final location. Travel along the latter path involves no work and no change in potential and is called an **equipotential line (or surface)**. We will have more to say about equipotentials later today and in the next lecture.

If we consider a conducting sphere of radius $r_{0}$ that carries a total charge of $Q$ on it's surface we can easily determine from Gauss's law that

$E=0$ $[r<r_{0}]$

$E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}$ $[r\geq r_{0}]$

As we did in the case of a point charge the integral $V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$ can be done along a radial path so that $\vec{E}$ and $d\vec{l}$ are in the same direction.

If we restrict our integral to points outside the sphere

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

and as before define the potential as being zero at $\infty$

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$ $[r>r_{0}]$

This implies that at $r=r_{0}$

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

and as inside the sphere $E=0$ the integral $-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=0$ which implies that everywhere inside the sphere

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r_{0}}$ $[r\leq r_{0}]$

Indeed, we can say generally that in addition to the field inside a conductor being everywhere zero, the potential at every point in conductor is everywhere the same.

Whenever we have determined the electric field as a function of position we should be able to calculate the electric potential. Sometimes however the vector nature of the field can make this tricky. One of the great appeals of electric potential is that it is a scalar quantity, and therefore it can be easier to find the potential due to multiple charges or a distribution of charges than it is to find the field.

For example if I want to calculate the field due to equal positive and negative charges separated by some distance I need to pay attention to directions of the field, whereas to find the potential I simply add together the potential of each charge. We denote the negative charge as $A$ and the positive charge as $B$ and we can then say that the potential is given by

$V=\frac{1}{4\pi\varepsilon_{0}}(-\frac{Q}{R_{A}}+\frac{Q}{R_{B}})$

We can note that anywhere on the plane that lies in between the two charges the potential will be equal to zero. As each point on this plane has the same potential we call it an equipotential surface. Although the equipotential surface we have just considered is the most trivial to identify we could in fact identify an infinite number of equipotential surfaces (however, none of the others will be flat!).

We can explore the field distribution around various collections of charges using this applet.

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

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

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 $x>>R$

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

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 $(1+x)^{1/2}$.

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

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

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 field distribution around various collections of charges using this applet.

We previously talked about the direction of electric field lines, where they originate and terminate, and also the density of field lines and how that relates to the charges responsible for the field and the magnitude of the field at various locations. Now let us study the potential lines in a similar fashion. Note the characteristic angle between field lines and potential lines when they intersect. Note the relationship between the density of field lines and the density of potential lines.

Another neat way to study equipotentials is the Falstad's 3D applet. (Note: it looks like this applet cannot plot the surfaces in 3D, but we can look at slices).

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.