Respuesta :
Answer:
If the radius of the disk is much greater than the point where the electric field is calculated, then the field of the disk matches the field of the infinite sheet.
Explanation:
First, we have to calculate the electric field of the disk.
We should choose an infinitesimal area, 'da', on the disk and calculate the E-field of this small portion, 'dE'. Then we will integrate dE over the entire disk using cylindrical coordinates.
According to the cylindrical coordinates: da = rdrdθ
The small portion is chosen at a distance r from the axis. Let's find the dE at a point on the axis and a distance z from the center of the disk.
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2 + z^2}[/tex]
Here dQ can be found by the following relation: The charge density of the disk is equal to the total charge divided by the total area of the disk. The small portion of the disk will have the same charge density, therefore:
[tex]\frac{Q}{\pi R^2} = \frac{dQ}{da}\\dQ = \frac{Qda}{\pi R^2}[/tex]
Furthermore, we need to separate the vertical and horizontal components of dE, because it is a vector and cannot be integrated without separating the components. By symmetry, the horizontal components of dE will cancel out each other, leaving only the vertical components in the z-direction.
[tex]dE_z = dE\sin(\alpha) = dE \frac{z}{\sqrt{z^2+r^2}}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qzda}{\pi R^2(z^2+r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}[/tex]
We have to use a double integral over the radius and the angle to find the total electric field due to a uniformly charged disk:
[tex]E = \int \int dE = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 {\int\limits^R_0 {\frac{1}{(z^2+r^2)^{3/2}}} \, rdr} \, d\theta\\E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[1 - \frac{1}{\sqrt{(R^2/z^2) + 1}}][/tex]
If the radius of the disk is much greater than the point z, R >> z, than the term in the denominator becomes very large, and the fraction becomes zero. In that case electric field becomes
[tex]E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[/tex]
This is equal to the electric field of an infinite sheet.
As a result, the condition for the field of a disk to be equal to that of a infinite sheet is R >> z.
