# Lecture 5 - Calculating Electric Fields and Introduction to Electric Potential

## Charged metal sphere

We can now look at a situation we discussed earlier, a hollow metal sphere with a net positive charge.

Because of the spherical symmetry for any spherical Gaussian surface of radius r

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\oint EdA=E\oint dA=EA=E4\pi r^{2}$

and Gauss's law gives us

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Any spherical Gaussian surface we draw inside the sphere of charge does not include any charge and so the net electric flux through the sphere must be zero. By contrast if we draw a spherical gaussian surface outside the charge we can show that the field outside the sphere is $E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}$

## Solid uniformly charged sphere

If we now consider an insulating sphere of radius $r_{0}$ that has total charge $+Q$ uniformly distributed inside we need to take in to account the fact that when we draw a spherical Gaussian surface inside the sphere not all the charge will be included inside the sphere.

The amount of charge included inside a Gaussian sphere of radius $r$ within the sphere of charge is $Q_{enc}=Q\frac{4/3\pi r^{3}}{4/3\pi r_{0}^{3}}=Q\frac{r^{3}}{r_{0}^{3}}$

As before the field has spherical symmetry

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\oint EdA=E\oint dA=EA=E4\pi r^{2}$

and Gauss' law gives us

$\Phi_{E}=\frac{Q_{enc}}{\varepsilon_{0}}$

The field for $r<r_{0}$ is therefore $E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}\frac{r^{3}}{r_{0}^{3}}=\frac{Qr}{4\pi\varepsilon_{0}r_{0}^{3}}$

For $r>r_{0}$ all the charge is included inside the Gaussian sphere and so the field is again $E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}$. From the outside the fields due to a charged conducting sphere and a charged insulating sphere are identical!

## Long uniform line of charge

We will now revisit one of the field distributions we derived from Coulomb's law, the field due to an infinitely long line of charge. As before the charge per unit length is $\lambda$ and we can reason that the field should always be perpendicular to the line of charge. To reflect the symmetry of the problem we take a cylindrical Gaussian surface of length $l$ and radius $R$ where $R$ is the distance at which we wish to know the field.

To evaluate the total flux passing through the surface we should take into account the ends of the cylinder as well as it's curved surface, but as the field is parallel to the ends of the cylinder the flux through the ends is zero. Gauss's law is then

$\oint\vec{E}.d\vec{A}=E(2\pi R l)=\frac{Q}{\varepsilon_{0}}=\frac{\lambda l}{\varepsilon_{0}}$

which gives

$E=\frac{1}{2\pi\varepsilon_{0}}\frac{\lambda}{R}$

## Infinite plane of charge

We showed previously, based on taking the limit of our result for the field of a uniformly charged disk, that the field due to an infinite plane of charge is $E=\frac{\sigma}{2\varepsilon_{0}}$. The same result can be obtained very easily from Gauss's law.

We again draw a cylindrical Gaussian surface, but this time it is the flat surfaces that have flux passing through them. The flat surfaces have area $A$, and we can see that in fact we could have taken any kind of prism to be our surface, so long as the sides of the prism are parallel to the field.

$\oint\vec{E}.d\vec{A}=-E(-A)+EA=E(2A)=E(2A)=\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}$

$E=\frac{\sigma}{2\varepsilon_{0}}$

## Planar conductor

Suppose we now consider a thin metal sheet, which has a charge density $\sigma$ on both the front and back sides. From symmetry we should expect that the field on opposite sides of the sheet should point in opposite directions, but have the same magnitude. If we draw a Gaussian surface that includes both sheets of charge we can now find that

$\oint\vec{E}.d\vec{A}=-E(-A)+EA=E(2A)=\frac{Q}{\varepsilon_{0}}=\frac{\sigma 2A}{\varepsilon_{0}}$

$E=\frac{\sigma}{\varepsilon_{0}}$

The same result can also be obtained by drawing a surface that ends within the metal, in this case the flux is only through the top surface of the cylinder, as no field can exist in the metal

$\oint\vec{E}.d\vec{A}=E(A)=\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}$

$E=\frac{\sigma}{\varepsilon_{0}}$

## Gaussian surfaces within conductors

In order to exclude electric field from a conductor any Gaussian surface that lies entirely within the conductor must contain no net electric charge. This fact can be used to easily explain the charge distribution we discussed previously for a hollow metal sphere with a charge inside.

## Infinite slab of charge

What is the field in the center of this slab of thickness $d$ with uniform charge density $\rho$?

What is the magnitude of the field outside this slab?

## Electric field from charge distributions

The field from a point charge is a vector $\vec{E}$ which has magnitude

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

and when $Q$ is positive points radially outwards from the charge.

You could write this as

$\vec{E}=E\hat{r}$

where $\hat{r}$ is a unit vector pointing away from the charge.

For a negative charge, Q is negative so the multiplication of the negative scalar with the outward $\hat{r}$ vector results in a field $\vec{E}$ that points inward towards the charge.

## Field from an element of charge

The field from an element of charge,

$\vec{dE}=dE\,\hat{r}$

which also points radially outwards from the charge element has magnitude

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^{2}}$

If we want to find the field directly from a distribution of charge we can integrate over the charged distribution, but we have to be careful to take into account the vector nature of the field! In a 3 dimensional coordinate system with no symmetry this would be 3 integrals, one for each vector in our coordinate system.

We will not do general problems like this because they are way too hard.

We will only tackle problems using direct integration of charge to obtain field where there is cancellation of components of the field from different parts of the charge distribution.

## Integration over a charge distrbution

Suppose we want to integrate over a distribution of charge. What does $\int dq$ really mean?

For a line of charge $\int dq = \int \lambda\,dx$
For a sheet of charge $\int dq = \int \sigma\,dA$
For volume of charge $\int dq =\int \rho\,dV$

If the charge density is uniform with respect to the variable being integrated over, then it can be taken out of the integral, if not then it needs to be included in the integral.

Depending on the symmetry of the problem we might choose to integrate over Cartesian coordinates or polar coordinates, so for example,

for a rectangular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma \,dx\,dy$
for a circular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma r\,dr\,d\theta$

Typically we have considered situations where one of these integrals is already done, for example when considering a uniform disk we first did the integral over $\theta$ for a given radius and then integrated this result over the radius for a disk.

## Gauss's Law Applications

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Gauss's law will often give us a better way to evaluate the field for a distribution of charge, because on the right hand side we only need to evaluate the total amount of charge, we don't need to consider the direction or the distance of the field from the charge to the point at which we need to know the field.

The most important step is to choose an appropriate Gaussian surface which reflects the symmetry of the charge. We should remember that the Gaussian surface must be a closed surface, and so it may be composed of more than one surface. If possible we want to choose the surfaces so that the field on each surface is uniform and we don't have to actually any integration over the surface. It's important to remember that flux has a sign - it is positive when it is leaving the volume, and negative when entering (as the $\vec{A}$ always points out).

In this course we will consider only 3 kinds of symmetry when applying Gauss's Law.

## Spherical symmetry

If the field points out radially from the center of the sphere

$\oint\vec{E}.d\vec{A}=\oint E\,dA$

and if it is uniform over the surface

$\oint E\,dA=E \oint dA=E4\pi r^{2}$

To find the enclosed charge

$Q=\int\int\int\rho\,dV$

In cases where $\rho$ is either constant or only depends on r

For a spherical volume the integral $\int\int\int\rho\,dV=\int\rho A dr=\int\rho 4\pi r^{2}dr$

## A spherical example

A insulating sphere of radius $r_{1}$ has a total positive charge Q which is evenly distributed throughout, except for in the center of the sphere where there is a spherical cavity of radius $r_{0}$ which contains no charge. Let $r$ be the distance of a point from the center of the sphere. Write your answers in terms of $Q$, $r$, $r_{1}$, $r_{0}$ and $k$ or $\varepsilon_{0}$.

• What is the electric field $E(r)$ for points $r<r_{0}$?

$E=0$

• What is the electric field $E(r)$ for points $r_{0}<r<r_{1}$?

$E4\pi r^2=\frac{1}{\varepsilon_{0}}\int_{r_{0}}^{r}\rho 4\pi r^{2}\,dr=\frac{\rho}{\varepsilon_{0}}\frac{4\pi}{3}(r^{3}-r_{0}^3)$

$E=\frac{\rho}{\varepsilon_{0}}\frac{1}{3r^{2}}(r^{3}-r_{0}^3)$

$\rho=\frac{Q}{\frac{4}{3}\pi r_{1}^{3}-\frac{4}{3}\pi r_{0}^{3}}$

$E=\frac{1}{4\pi\varepsilon_{0}r^{2}}Q\frac{r^{3}-r_{0}^{3}}{r_{1}^3-r_{0}^{3}}=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r_{1}^{3}-r_{0}^{3}}(r-\frac{r_{0}^{3}}{r^{2}})$

• What is the electric field $E(r)$ for points $r>r_{1}$?

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

• Make a rough sketch of the electric field $E(r)$ against $r$. Make sure to label $r_{0}$ and $r_{1}$ on your sketch.

## Cylindrical symmetry

If we consider a field that has cylindrical symmetry so that the flux through the end circular surfaces is zero and for the curved surface

$\oint\vec{E}.d\vec{A}=\oint E\,dA$

As the field is uniform over the surface

$\oint E\,dA=E \oint dA=E2\pi rl$

To find the enclosed charge

$Q=\int\int\int\rho\,dV$

In cases where $\rho$ is either constant or only depends on r

$Q=\int\int\int\rho\,dV=\int\rho A dr=\int\rho 2\pi r l dr$

## Electrical potential energy

In PHY 141, the conceptual foundations of the conservation of energy was introduced. The potential energy for a conservative force was defined to be

$\Delta U = -\int_{1}^{2}\vec{F}\cdot\,d\vec{l} = - W_{Net}$

Note that we did not specify the path; as was discussed in 141, the integral is independent of path for conservative forces. Both the gravitational force and the electromagnetic force are conservative forces.

In the case where a charge is moved a distance $d$ by a uniform electric field $E$ the work done is

$W=Fd=qEd$

we can note here that this became a simple product because the charge moved in the direction of the field (as we can expect it should do if the only force acting is the electric force)

and the change in the electric potential energy is

$\Delta U=-W=-qEd$

## Electric potential

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 $\mathrm{V/m}$.

## Signs

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

$\Delta U = q\Delta V$

## Electric potential and field

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.