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 away from a plate of charge density . 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 , measured from the point directly beneath the point where we want to find the electric field. For a ring of width and radius away from the projected point, the electric field will come in at an angle such that . So, the perpendicular contribution will be

What will the E-field from an infinitesimal disc be then? The area of the disc , so . The distance from our point to the ring is , so the complete expression for the E-field contribution from an infinitesimal disc is

If we integrate from to we will cover the entire plane of charge.

]]>Magnetism exerts a force on particles proportional to their charge. The total force on a particle of charge and velocity from an electromagnetic field is:

So supposing you look at a current moving in the direction, with a magnetic field in the y direction. Since no charges are piling up just yet, . So, the force on a free particle of charge is .

We’ve been writing 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 is negative, the velocity of the particles is in the same direction as the current. So, the velocity coefficient is also positive. Since we’re assuming that is positive (and we will get the opposite result if it is pointing in the opposite direction), we can conclude that a force will push the particle in the direction.

When is positive, the velocity of the particles is in the opposite direction as current, making the velocity coefficient *negative*. Since the charge is *also negative*, a force direction will also push a negative charge towards the direction.

No matter what sign charge we use, charges will pile up on the wall of the conductor. There, they will act like a plate with surface charge density, which we will denote with . 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 . We know that the electric field from one such plate is , so from two plates we expect a total electric field of magnitude .

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.

should be zero, so we would expect . We don’t know , however. If we know the density of charged particles , and we know the current , then we can find the velocity of each particle, .

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The potential across the plates can then be calculated.

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Let be a charge density function, dependant on position and time . represents the rate of change of charge density with respect to time. Using the standard basis vectors the spatial partial derivatives 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:

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

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:

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.

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

Other authors often use the notation , but I don’t.

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