## Monday, February 10, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 4

Notes to a video lecture on http://www.unizor.com

Problems 4

Problem A

An infinitesimally thin disk α of radius R is electrically charged with uniform density of electric charge σ (coulombs per square meter). What is the intensity of the electric field produced by this disk at point P positioned at distance h (meters) from its surface on a perpendicular through its center?

Solution
Please refer to Problems 2 of "Electric Field", as we will use its results. The Problem A from Problems 2 is about intensity of the electric field produced by an infinite plane.

Here we will consider a finite electrically charged infinitely thin disk of the radius R instead of an infinite plane as in Problems 2.
Assume, as in Problems 2, the density of electrical charge of this disk is σ and the point we want to measure the intensity of the field is P located above the disk on the perpendicular through its center on height h above it.

Going through exactly the same logic as in Problems 2, dividing our disk into concentric rings, we will come to a formula for intensity produced by an infinitely narrow ring of inner radius r and outer radius r+dr, where r is changing from 0 to R:
dE(h,r) =
= 2π·k·σ·r·
dr·h / (h²+r²)3/2

After substitution
y = 1 + r²/h²
we obtain
dy = 2r·dr/h²
2r·dr = h²·dy
dE(h,y) =
= π·k·σ·
dy·h³ / (h²+r²)3/2=
= π·k·σ·
d/ (1+r²/h²)3/2 =
= π·k·σ·y−3/2·
dy

which is easy to integrate since it's a plain power function.
The limits of integration for r are from 0 to R.
Therefore, the limits of integration for y are from 1 to 1+R²/h².
The indefinite integral of y−3/2 is −2y−1/2. Therefore, the total vector of intensity at point P equals to
E(h) =
= −2π·k·σ·
[(1+R²/h²)−1/2−1] =
= 2π·k·σ·
[1−(1+R²/h²)−1/2] =
= 2π·k·σ·
[1−1/√1+R²/h² ]

We can rewrite this formula using the permittivity of vacuum ε0=1/(4π·k) as
E(h) = (σ/2ε0[1−1/√1+R²/h² ]

Analyzing this formula, we see that the intensity of the electric field of a uniformly charged disk of radius R at point above its center on the height h depends on the ratio R/h.
If the height remains the same, but the radius increases to infinity, the formula transforms into the one we obtained in the Problem 2 for infinite charged plane.
If the height h decreases to zero with a fixed radius R, the intensity gradually increases to its maximum value 2π·k·σ, which is the same as for an infinite plane in Problems 2. So, for a small height the uniformly charged disk acts like an infinite plane.
If the height h increases to infinity with a fixed radius R, the intensity gradually decreases to zero.
All conclusions are intuitively obvious.

Another parameter from which the intensity depends is the medium around a charged object. Knowing from a previous lecture about permittivity, we can consider the space around the charged disk to be not only vacuum, but any media with known dialectic constant εr.
In this case, instead of Coulomb's constant k, we have to use 1/(4π·εr·ε0) and the formula for intensity looks like this:
E(h) =
[σ/(2εr·ε0)]·[1−1/√1+R²/h² ]

From the above formula for any media filling the space around a charged disk we see that the greater dielectric constant for this media - the smaller intensity of the field around it.