Charged Disk: Difference between revisions
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<math>\ dq= Q\frac{2πrdr}{πr^2}</math> | <math>\ 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> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math> divided by the area of the disk, <math> πr^2 </math>. | |||
Revision as of 20:06, 2 December 2015
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, as that will change with the infinitely many concentric rings you have. However, r 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].
A Computational Model
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