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

__Problems 4__

*Problem A*

An infinitesimally thin disk

*of radius*

**α***is electrically charged with uniform density of electric charge*

**R***(coulombs per square meter). What is the*

**σ****intensity**of the electric field produced by this disk at point

*positioned at distance*

**P***(meters) from its surface on a perpendicular through its center?*

**h***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

*instead of an infinite plane as in Problems 2.*

**R**Assume, as in Problems 2, the density of electrical charge of this disk is

*and the point we want to measure the intensity of the field is*

**σ***located above the disk on the perpendicular through its center on height*

**P***above it.*

**h**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

*and outer radius*

**r***, where*

**r+**d**r***is changing from*

**r***to*

**0***:*

**R***d*

**E(h,r) =**

= 2π·k·σ·r·d= 2π·k·σ·r·

**r·h / (h²+r²)**^{3/2}After substitution

**y = 1 + r²/h²**we obtain

*d*

**y = 2r·**d**r/h²**

**2r·**d**r = h²·**d**y***d*

**E(h,y) =**

= π·k·σ·d= π·k·σ·

**y·h³ / (h²+r²)**

= π·k·σ·d^{3/2}== π·k·σ·

**y / (1+r²/h²)**

= π·k·σ·yd^{3/2}== π·k·σ·y

^{−3/2}·**y**which is easy to integrate since it's a plain power function.

The limits of integration for

*are from*

**r***to*

**0***.*

**R**Therefore, the limits of integration for

*are from*

**y***to*

**1***.*

**1+R²/h²**The indefinite integral of

*is*

**y**^{−3/2}*. Therefore, the total vector of intensity at point*

**−2y**^{−1/2}*equals to*

**P***[*

**E(h) =**

= −2π·k·σ·= −2π·k·σ·

*]*

**(1+R²/h²)**^{−1/2}−1*[*

**=**

= 2π·k·σ·= 2π·k·σ·

*]*

**1−(1+R²/h²)**^{−1/2}*[*

**=**

= 2π·k·σ·= 2π·k·σ·

*]*

**1−1/√1+R²/h²**We can rewrite this formula using the

*permittivity*of vacuum

*as*

**ε**_{0}=1/(4π·k)*[*

**E(h) = (σ/2ε**_{0})·*]*

**1−1/√1+R²/h²**Analyzing this formula, we see that the intensity of the electric field of a uniformly charged disk of radius

*at point above its center on the height*

**R***depends on the ratio*

**h***.*

**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

*decreases to zero with a fixed radius*

**h***, the intensity gradually increases to its maximum value*

**R***, 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.*

**2π·k·σ**If the height

*increases to infinity with a fixed radius*

**h***, the intensity gradually decreases to zero.*

**R**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

*, we have to use*

**k***and the formula for intensity looks like this:*

**1/(4π·ε**_{r}·ε_{0})*[*

**E(h) =**

==

*]*

**σ/(2ε**_{r}·ε_{0})*[*

**·***]*

**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.

## No comments:

Post a Comment