The Electric Field from an Infinite Flat Plate

When you are much closer to touching a plate than you are to its edges, it is approximately an infinite plate.  Suppose the plate has a charge density \sigma, what is the electric field that you experience?

For a plate that is truly infinite, with no granularities, every point looks the same.  Since the plate goes on forever in two directions, there’s no means of measuring your distance to the plate.  It would make sense if the electric field was constant throughout the space that we are considering.

I will evaluate this using both the integral approach and through Gauss’s law.  First the integral approach.

Suppose we want to know the electric field a distance d away from a plate of charge density \sigma.  We could try an integral over all of the plate, but we could be even smarter about it.

Let’s consider the contributions  to the electric field from each ring of a radius r, measured from the point directly beneath the point where we want to find the electric field.  For a ring of width dr and radius r away from the projected point, the electric field will come in at an angle \theta such that \tan{\theta} = \frac{r}{d}.  So, the perpendicular contribution \vec{E}_{\perp} will be

E \left( \cos{\arctan{\frac{r}{d}}} \right) \hat{z} = E\frac{d}{\sqrt{d^2 + r^2}}\hat{z}

What will the E-field from an infinitesimal disc be then?  The area of the disc dA = 2\pi r dr, so dQ = \sigma dA = 2 \pi r \sigma dr.  The distance from our point to the ring is \sqrt{r^2 + d^2} , so the complete expression for the E-field contribution from an infinitesimal disc is

\vec{dE} = \frac{1}{4 \pi \epsilon_0} \frac{ 2 \pi r \sigma dr}{\left( r^2 + d^2 \right)^{\frac{3}{2}}}

If we integrate from r = 0 to r = \infty we will cover the entire plane of charge.

\begin{aligned}    \vec{E}_{\perp} & = \int_0^{\infty} \frac{ 2 \pi r \sigma dr}{\left( r^2 + d^2 \right)^{\frac{3}{2}}} \\    & = \frac{ 2 \pi \sigma}{4 \pi \epsilon_0} \int_0^{\infty} \left( \frac{\partial}{\partial_r} \frac{ - d}{\sqrt{r^2 + d^2}} \right) \\    & = \frac{\sigma}{2 \epsilon_0} \left[ \frac{-d}{\sqrt{\infty^2 + d^2}} - \frac{-d}{\sqrt{0^2 + d^2}} \right] \\    & = \frac{\sigma}{2 \epsilon_0} \left( 0 + \frac{d}{d} \right) \\    \vec{E}_{\perp} & = \frac{\sigma}{2 \epsilon_0}    \end{aligned}

The Hall Effect

Suppose you have a conductor.  We’ve already seen that it contains individual charges.  It is easy to measure the current flowing through the conductor.   But, it is not necessarily easy to tell if a positive current corresponds to positive charges moving forwards, or negative charges moving backwards.  It might be possible to find out using magnetism.

Magnetism exerts a force on particles proportional to their charge.  The total force on a particle of charge q and velocity \vec{v} from an electromagnetic field \vec{E}, \vec{B} is:

\vec{F} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)

So supposing you look at a current \vec{J} = J\hat{x} moving in the \hat{x} direction, with a magnetic field \vec{B}=B\hat{y} in the y direction.  Since no charges are piling up just yet, \vert \vec{E} \vert = 0.  So, the force on a free particle of charge q is \vec{F} = q\vec{v} \times \vec{B} = qvB\left( - \hat{z} \right).

We’ve been writing q as positive, but there’s no need for it to be anything other than zero.  Let’s look at the positive and negative cases separately, and see what kind of behavior we find.

When q is negative, the velocity of the particles is in the same direction as the current.  So, the velocity coefficient v is also positive.  Since we’re assuming that \vec{B} is positive (and we will get the opposite result if it is pointing in the opposite direction), we can conclude that a force F = qvB\left( - \hat{z} \right) will push the particle in the -\hat{z} direction.

When q is positive, the velocity of the particles is in the opposite direction as current, making the velocity coefficient v negative.  Since the charge is also negative, a force F = qvB \left( -\hat{z} \right) direction will also push a negative charge towards the - \hat{z} direction.

No matter what sign charge we use, charges will pile up on the -\hat{z} wall of the conductor.  There, they will act like a plate with surface charge density, which we will denote with \sigma.  Since the overall material is electrically neutral, there will be a corresponding pileup of not-electrons on the other side of the conductor, corresponding to a surface charge density -\sigmaWe know that the electric field from one such plate is \vec{E}_{\perp} = \frac{\sigma}{2 \epsilon_0}, so from two plates we expect a total electric field of magnitude \frac{\sigma}{\epsilon_0}.

We could run the current and let the charge build up until the charges reach a steady state.  In this configuration, the electric force from the charged walls of the conductor exactly cancels out the magnetic force from the external magnetic field.

F_\perp = q \left( E_\perp + \vec{v} \times \vec{B} \right) should be zero, so we would expect \frac{\sigma}{\epsilon_0} = \vec{F}_{E} = \vec{F}_{B} = vB.  We don’t know v, however.  If we know the density of charged particles \rho, and we know the current J, then we can find the velocity of each particle, v = J \frac{1}{\rho q}.

\sigma = \frac{\epsilon_0 J B}{\rho q}.

The potential across the plates can then be calculated.

 

 

What was Maxwell Thinking?

Conservation is one of the basic assumptions of physics.  The basic assumption is that for any kind of “stuff,” the amount of stuff going into any region is equal to the increase in stuff in that region. Using the formalism of vector calculus there is a clear way of stating this conservation law for the specific case of charge.

Let \rho be a charge density function, dependant on position \vec{r} and time t. \partial_t \rho represents the rate of change of charge density with respect to time. Using the standard basis vectors x_1,x_2,x_3 the spatial partial derivatives \partial_{x_1} \rho, \partial_{x_2} \rho, \partial_{x_3} \rho represent the rate of change in the respective directions.

The basic statement of the Continuity Equation makes a statement about any region.  It says that the sum of all of the partial derivatives is zero, or:

\partial_{x_1} \rho + \partial_{x_2} \rho + \partial_{x_3} \rho + \partial_t \rho = 0.

If we call time x_0, then we can write this a simpler way:

\sum_{\mu = 0}^3 \partial_{x^\mu} \rho = 0

If we assume that any greek index is being summed from 0 to 3, and any latin index is being summed from 1 to 3, we can get a more compact continuity equation:

\partial_{\mu} \rho = 0

Now, let’s look at the laws of nature that Maxwell learned in his physics classes.

The first one was that the partial derivatives of the electric field coming out of a region is proportional to the charge density.

\partial_1 E_1  + \partial_2 E_2 + \partial_3 E_3 \propto \rho

Using our new, compact notation where latin indices represent summing over all the spatial variables (1 to 3), we can express this as:

\partial_i E_i \propto \rho

Other authors often use the notation div \vec{E} \propto \rho, but I don’t.