Quick yes or no charged disc e field equation question

[charlies1902]

Homework Statement

the e field is given by (2*pi*k*omega)[1-(1+(R^2)/(z^2))^-.5]



I was wondering if the capital R in that equation is the radius of the charged disk? And if so, why is it capitalized?

 

[collinsmark]

Homework Statement

the e field is given by (2*pi*k*omega)[1-(1+(R^2)/(z^2))^-.5]

Yeah, that looks about right to me! Good job.

But I would use the Greek letter sigma, rather than omega, for the surface charge density. There's no particular reason for that except it's more conventional.

$$\vec E = 2 \pi k \sigma \left(1 - \frac{1}{ \sqrt{1+ \frac{R^2}{z^2}}} \right) \hat z$$

where $\hat z$ is a unit vector in the z direction. It assumes the disk is on the x-y plane, with the disk's center at the origin.

I was wondering if the capital R in that equation is the radius of the charged disk?

Yes.

And if so, why is it capitalized?

So as not to confuse it with r, the radius component of the position vector (in cylindrical coordinates), the distance to the z-axis. Often you'll find that it is somewhat conventional in physics to use lower case letters for variables and capital letters for constants. (Of course some fundamental constants use Greek letters, or special lowercase letters -- yeah, I know it might be confusing, but it's a loose convention and not cut in stone. You'll often come across exceptions too). It's assumed that the radius of the disk is not dynamically changing, so it's treated as a constant.

The above equation happens to work in both Cartesian coordinates (x, y, z) and cylindrical coordinates (r, θ, z). So in the above equation, R is the radius of the disk, not to be confused with r, the distance from the z-axis. But the equation, as it has been solved, is only valid under certain restrictions.

You can't just pick any old r, θ and z and expect the above equation to be valid. There are restrictions on at least one variable. (The equation is not valid in all of space; rather it's valid only on a single, specific line within 3-dimensional space.) Since this is the homework forum, I'll let you ponder what the restriction is, and leave the rest to you.