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Kurs: Elektrotechnika > Rozdział 5

Lekcja 2: Pole, potencjał i napięcie elektryczne

Plane of charge

Advanced example: Electric field generated by a uniformly-charged infinite plane. Written by Willy McAllister.

Example: Electric field near a plane of charge

We investigate the next interesting charge configuration, the electric field near a plane of charge.

The result will show the electric field near an infinite plane of charge is independent of the distance away from the plane (the field does not fall off).

Imagine we have an infinite plane of charge.

The total charge on the plane is of course infinity, but the useful parameter is the amount of charge per area, the charge density,

σ (C / m^{2})

What is the electric field due to the plane at a location $a$ ‍ away from the plane?

We exploit the symmetry of the problem to set up some variables:

$a$ ‍ is a perpendicular line from the plane to the location of our test charge, $q$ ‍.
Imagine a hoop of charge in the plane, centered around where $a$ ‍ touches the plane. The radius of the hoop is $r$ ‍, and its infinitesimal thickness is $d r$ ‍.
$d Q$ ‍ is an infinitesimal region of charge in a section of the hoop.
Line $ℓ$ ‍ goes from the location of $d Q$ ‍ to the location of the test charge.
$d E$ ‍ is the electric field at point $q$ ‍ created by $d Q$ ‍.

We know the field at location

q

due to

d Q

; it's the definition of the field created by a point charge,

d E = \frac{1}{4 π ϵ_{0}} \frac{d Q}{ℓ^{2}}

To solve the electric field for the whole plane, we have to do two integrations:

first integration to sweep $d Q$ ‍ around its hoop to get the field contribution from one particular hoop, and a
second integration to add up the contributions from all possible hoops (from zero radius to infinite radius).

Sweep around a hoop to get the field contribution from one particular hoop

The hoop construct cleverly allows us to duck the first integral. All parts of the hoop are the same distance

ℓ

away from

q

, so each

d Q

creates the same magnitude field at

q

. Symmetry tells us the total field contribution from all

d Q

's in a hoop has to point straight away from the plane, along line

a

. Why? Because any sideways component of the field from a particular

d Q

is exactly cancelled by the

d Q

on the opposite side of the hoop. The straight out "

a

-direction" portion of the electric field

d E_{a}

is related to

d E

The symmetry argument is clearer with a slightly different view of the plane and test charge. In these diagrams, the infinite plane on the right side of the diagram, is shown edge-on. From this view, the hoop appears as a vertical blue line.

$d Q_{1}$ ‍ is a tiny bit of charge positioned at the top of the hoop.
$d Q_{2}$ ‍ is a tiny bit of charge positioned symmetrically across the hoop, at the very bottom.
The electric fields contributions from the two $d Q$ ‍'s are $d E_{1}$ ‍ and $d E_{2}$ ‍.

Test charge

q

is distance

ℓ

from both

d Q_{1}

and

d Q_{2}

, therefore,

d E_{1}

and

d E_{2}

have the same magnitude (but different direction).

The e-field vectors can be decomposed into orthogonal components pointing in the "

a

" direction and "

r

" direction. The "

r

" component of the two e-field vectors are equal in magnitude and point in opposite directions. Therefore, they cancel each other and the net electric field in the "

r

" direction is zero.

What remains is the "

a

" component of the two e-field contributions. These components are equal in magnitude and point in the same direction. Therefore, these contributions reinforce each other and add to

d E_{1 + 2}

The next diagram is a little simpler. We look at just one of the

d E

's and drop the 1 suffix. We explicitly see the electric field decomposed into two orthogonal components,

d E_{r}

and

d E_{a}

By similar triangles, the component of the field in the

a

direction,

d E_{a}

, is related to

d E

in the same proportion as distance

a

is related to

ℓ

. From the definition of cosine,

\frac{a}{ℓ} = \cos θ

Embracing the similar triangles reasoning, this means,

\frac{d E_{a}}{d E} = \cos θ

So the component of the electric field in the

a

direction due to

d Q

is therefore,

d E_{a} = d E \cos θ

d E_{a} = d E \cos θ

Which gives this for

d E_{a}

, the field from a single point charge

d Q

d E_{a} = \frac{1}{4 π ϵ_{0}} \frac{d Q}{ℓ^{2}} \cos θ

Next, express the field contribution from one entire hoop

d E_{h o o p}

d E_{h o o p} = \frac{1}{4 π ϵ_{0}} \frac{d Q_{h o o p}}{ℓ^{2}} cos θ

d Q_{h o o p}

is the total charge contained in one hoop, the sum of the individual point

d Q

's making up the hoop. This can be computed without doing an integral. The total charge on a hoop is the charge density of the plane,

σ

, times the area of the hoop,

d Q_{h o o p} = σ \cdot (2 π r \cdot d r)

The electric field at the location of

q

created by a hoop with radius

r

, containing charge

Q_{h o o p}

is,

d E_{h o o p} = \frac{1}{4 π ϵ_{0}} \frac{σ 2 π r d r}{ℓ^{2}} cos θ

Now we know the field contributed by a single hoop.

Integrate the contributions from all possible hoops

The next step is to sum up all possible hoops. Unfortunately, we can't sneak out of doing this integral. Just like we did for Line of Charge example, we perform a change of variables, from

d r

d θ

For this change of variables, the goal is to develop an expression for

d θ

in terms of

d r

, allowing

d r

to be eliminated.

Some useful identities based on the definitions of trig functions,

\cos θ = \frac{a}{ℓ} ℓ = a \frac{1}{cos θ} = a sec θ ℓ^{2} = a^{2} {sec}^{2} θ

\tan θ = \frac{r}{a} r = a tan θ

To generate an expression for

d θ

, take the derivative of

r

with respect to

θ

using the tangent identity,

\frac{d r}{d θ} = \frac{d}{d θ} a tan θ

\frac{d r}{d θ} = a {sec}^{2} θ

d r = a {sec}^{2} θ d θ

A reminder of the expression for

d E_{h o o p}

d E_{h o o p} = \frac{1}{4 π ϵ_{0}} \frac{σ 2 π r d r}{ℓ^{2}} cos θ

Plug in the identities for

r

d r

, and

ℓ^{2}

d E_{h o o p} = \frac{1}{4 π ϵ_{0}} \frac{σ 2 π (a \tan θ) (a \sec^{2} θ d θ)}{(a^{2} {sec}^{2} θ)} cos θ

After simplification and some spectacular cancellation, the change of variable is complete,

d E_{h o o p} = \frac{σ}{2 ϵ_{0}} tan θ \cdot cos θ d θ

(Notice all of the

a

terms cancelled out.)

After the change of variable, the diagram can be redrawn in terms of

d θ

and

θ

and the field equation for one hoop becomes,

d E_{h o o p} = \frac{σ}{2 ϵ_{0}} tan θ \cdot cos θ d θ

which can be simplified a bit more,

d E_{h o o p} = \frac{σ}{2 ϵ_{0}} sin θ d θ

Something very interesting just happened. As a result of the change of variable and cancellation, all the

r

's and

a

's vanish! Wait, What! In the resulting expression for

d E_{h o o p}

, there is NO dependence on distance. Remarkable.

Almost home. We are ready to perform the integration,

E = \int_{a l l h o o p s} d E_{h o o p}

where

E

is the overall electric field from all hoops. Substitute for

d E_{h o o p}

E = \int_{t h e t a s} \frac{σ}{2 ϵ_{0}} sin θ d θ

What are the angle limits on the integration? The smallest possible hoop is when

r

is zero;

ℓ

coincides with

a

, and

θ

is zero. The largest hoop is when

r

is infinite; line

ℓ

comes from way out at the horizon in any direction, and

θ

90^{\circ}

π / 2

radians. So the limits on the integration run from

θ = 0 to π / 2

radians.

E = \int_{0}^{π / 2} \frac{σ}{2 ϵ_{0}} sin θ d θ

E = - \frac{σ}{2 ϵ_{0}} cos θ |_{0}^{+ π / 2} = - \frac{σ}{2 ϵ_{0}} (0 - 1)

The electric field near an infinite plane is,

E = \frac{σ}{2 ϵ_{0}}

newtons/coulomb

Podsumowanie

This the electric field (the force on a unit positive charge) near a plane. Amazingly, the field expression contains no distance term, so the field from a plane does not fall off with distance! For this imagined infinite plane of charge, it doesn't matter if you are one millimeter or one kilometer away from the plane, the electric field is the same.

This example was for an infinite plane of charge. In the physical world there is no such thing, but the result applies remarkably well to real planes, as long as the plane is large compared to

a

and the location is not too close to the edge of the plane.

Review

Using the notion of an electric field, the analysis technique is,

Charge gives rise to an electric field.
The electric field acts locally on a test charge.

Summarizing the three electric field examples worked out so far,

The field due to a	falls off at
point charge	$1 / r^{2}$ ‍
line of charge	$1 / r^{1}$ ‍
plane of charge	$1 / r^{0}$ ‍

These three charge configurations are a useful toolkit for predicting electric field in lots of practical situations.

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