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

__Problems 2__

*Problem A*

An infinite infinitesimally thin plane

*is electrically charged with uniform density of electric charge*

**α***(coulombs per square meter).*

**σ**What is the

**intensity**of the electric field produced by this plane at point

*positioned at a distance*

**P***(meters) from its surface?*

**h***Solution*

The magnitude of the field intensity at point

*produced by any infinitesimal piece of plane*

**P***is inversely proportional to a square of its distance to point*

**α***and directly proportional to its charge, and the charge, in turn, is proportional to its area with density*

**P***being a coefficient of proportionality.*

**σ**Let's define a system of cylindrical coordinates in our three-dimensional space with the origin at the projection

*of the point*

**O***onto our electrically charged plane*

**P***with Z-axis along segment*

**α***and polar coordinates on plane*

**OP***with*

**α***for radial distance*

**r***between any point*

**OA***on this plane and the origin of coordinates*

**A***and*

**O***for a counterclockwise angle from the positive direction of some arbitrarily chosen base ray*

**φ***within plane*

**OX***, originated an point*

**α***, to radius*

**O***.*

**OA**Then the coordinates of point

*, where we have to calculate the vector of electric field intensity, are (*

**P***). The coordinates of any point*

**0,0,h***on plane*

**A***will be (*

**α***).*

**r,φ,0**From the considerations of symmetry, the vector of intensity of the field, produced by an entire infinite electrically charged plane

*, at point*

**α***outside of it should be directed along a perpendicular*

**P***from point*

**OP***to a plane. Indeed, for any infinitesimal area of plane*

**P***there is an area symmetrical to it relatively to point*

**α***, which produces the intensity vector of the same magnitude, the same vertical (parallel to*

**O***) component of it and opposite horizontal (within a plane*

**OP***) component. So, all horizontal components will cancel each other, while vertical ones can be summarized by magnitude.*

**α**Therefore, we should take into account only projections of all individual intensity vectors from all areas of a plane onto a perpendicular

*from point*

**OP***onto plane*

**P***, as all other components will cancel each other.*

**α**The approach we will choose is to take an infinitesimal area on a plane in a form of a ring centered at point

*of infinitesimal width*

**O***d*with inner radius

**r***and outer radius*

**r***and calculate the vertical component of the intensity vector produced by it. Then we will integrate the result from*

**r+**d**r***to infinity.*

**r=0**This choice is based on a simple fact that for every small piece of this ring its distance to point

*is the same, as well as an angle between its vector of intensity and Z-axis is the same, hence the vertical component of the field intensity vector produced by it will be the same as for any other such piece of this ring, if it has the same area, while the horizontal component of the intensity vector will be canceled by a symmetrical piece of this ring lying diametrically across it.*

**P**Since any infinitesimal part of this ring has exactly the same vertical component of the intensity vector as any other part having the same area, to get the total vertical component of intensity for an entire ring, we can use its total charge that depends on its total area and charge density

*.*

**σ**The area of a ring equals to

*area*[

**(r,**d**r) = π***]*

**(r+**d**r)²−r²**

**=**

= 2π·r·d= 2π·r·

**r + π·(**d**r)²**We can drop the infinitesimal of the second order

*and leave only the first component - infinitesimal of the first order, that we plan to integrate by*

**π·(**d**r)²***from 0 to infinity.*

**r**The charge concentrated in this ring is

*d*

**Q(r) = 2π·σ·r·**d**r**To find the magnitude of the intensity vector produced by this ring we need to know its charge (calculated above) and the distance to a point, where the intensity is supposed to be calculated. This distance can be calculated using the Pythagorean Theorem:

**L² = h² + r²**The magnitude of the intensity vector produced by this ring is, therefore,

*d*

**E(h,r) = k·**d**Q(r)/L² =**

= 2π·k·σ·r·d= 2π·k·σ·r·

**r / L² =**

= 2π·k·σ·r·d= 2π·k·σ·r·

**r / (h² + r²)**We are interested only in vertical component of this vector, which is equal to

*d*∠

**E**d_{z}(h,r)=**E(h,r)·sin(**

**PAO) =**

=d=

**E(h,r)·h/L =**

= 2π·k·σ·r·d= 2π·k·σ·r·

**r·h / L³ =**

= 2π·k·σ·r·d= 2π·k·σ·r·

**r·h / (h² + r²)**^{3/2}Integrating this by

*from 0 to ∞ can be done as follows.*

**r**First of all, let's substitute

**x = r/h**Then

**r = h·x***d*

**r = h·**d**x**The limits of integration for

*are the same, from 0 to ∞.*

**x**Now the expression to integrate looks like

*d*

**E**

= 2π·k·σ·h³·x·d_{z}(h,x) == 2π·k·σ·h³·x·

**x / h³(1 + x²)**

= 2π·k·σ·x·d^{3/2}== 2π·k·σ·x·

**x / (1 + x²)**^{3/2}__Before going into details of integration, note that this expression does not depend on distance__

*from point***h***to an electrically charged plane***P***. This is quite remarkable!***α**No matter how far point

*is from plane*

**P***, the intensity of electric field at this point is the same.*

**α**To integrate the last expression for a projection onto Z-axis of the intensity of electric field produced by an infinitesimal area of plane

*, introduce another substitution:*

**α**

**y = x² + 1**Then

**x·**d**x =**d**y/2**The limits of integration for

*are from 1 to ∞.*

**y**The expression to integrate becomes

*d*

**E**d_{z}(y) = π·k·σ·**y / y**

= π·k·σ·yd^{3/2}== π·k·σ·y

^{−3/2}·**y**This is easy to integrate. The indefinite integral of

*is*

**y**^{n}*. Using this for*

**y**^{n+1}/(n+1)*, we get an indefinite integral of our function*

**n=−3/2**

**−2·π·k·σ·y**^{−1/2}+ CUsing the Newton-Leibniz formula for limits from 1 to ∞, this gives the value of the magnitude of the total intensity of a charged plane

*:*

**α***∫*

**E =**_{[1,∞]}

**π·k·σ·y**d^{−3/2}·**y = 2π·k·σ**Let's note again that this value is independent of the distance

*of point*

**h***, where we measure the intensity of the electric field, from an electrically charged plane*

**P***. It only depends on the density*

**α***of electric charge on this plane.*

**σ**As for direction of the intensity vector, as we suggested above, it's always perpendicular to the plane

*.*

**α**Hence, we can say that the electric field produced by a uniformly charged plane is

**uniform**, at each point in space it is directed along a perpendicular to a plane and has a magnitude

*, where*

**E=2π·k·σ***represents the density of electric charge on a plane and*

**σ***is a Coulomb's constant.*

**k***Problem B*

An infinite infinitesimally thin plane

*is electrically charged with uniform density of electric charge*

**α***(coulombs per square meter).*

**σ**What is the

**work**needed to move a charge

*(coulombs) from point*

**q***positioned at a distance*

**M***(meters) from its surface to point*

**m***positioned at a distance*

**N***(meters) from its surface?*

**n***Solution*

Notice that positions of points

*and*

**M***are given only in terms of their distance to a charged plane*

**N***, that is in terms of vertical displacement. Distance between them in the horizontal direction is irrelevant since any horizontal movement will be perpendicular to the vectors of field intensity and, therefore, require no work to be done.*

**α**So, our work only depends on the distance along the vertical and can be calculated as

**W**

= 2π·k·σ·q·(n−m)_{MN}= E·(n−m)·q == 2π·k·σ·q·(n−m)

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