Charged Disk

We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

The Main Idea

In this section, we will do a step-by-step process of calculating the electric field of a uniformly charged disk. This is especially important because two oppositely charged metal disks collectively are known as a capacitor, a concept seen in several places in physics and the real world.

A Mathematical Model

Before we begin our calculations, take a look at this image of a uniformly charged disk:

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a ring. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

[math]\displaystyle{ \ E= \frac{1}{4π\epsilon_0}\frac{qz}{sqrt((R^2+z^2)^3)} }[/math]

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring [math]\displaystyle{ r }[/math], as that will change with the infinitely many concentric rings you have. However, [math]\displaystyle{ r }[/math] is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

[math]\displaystyle{ \ dq= Q\frac{2πrdr}{πr^2} }[/math]

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, [math]\displaystyle{ 2πrdr }[/math], (if you roll out the ring, it is a rectangle with height [math]\displaystyle{ dr }[/math] and width [math]\displaystyle{ 2πr }[/math]) divided by the area of the disk, [math]\displaystyle{ πr^2 }[/math]. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

[math]\displaystyle{ \ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{sqrt((R^2+z^2)^3)} }[/math]

Now we take the integral of this equation, and the result is this:

[math]\displaystyle{ \ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{sqrt(R^2+z^2)}] }[/math]

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. [math]\displaystyle{ 0 \lt \lt z \lt \lt R }[/math]), the equation becomes this:

[math]\displaystyle{ \ E = \frac{Q}{2\epsilon_0A} }[/math]

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors.

A Computational Model

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