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·dy / (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 ε_r.
In this case, instead of Coulomb's constant k, we have to use 1/(4π·ε_r·ε₀) and the formula for intensity looks like this:
E(h) =
= [σ/(2ε_r·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.