Why the K be instand by 1/4πϵ?

just to make the expression shorter and easy to remember

have we taken point charge dQ in the hoop part explanation??

Ashley, I don't quite understand your question. Could you rephrase it?

where did the sigma came from? dQhoop = σ⋅(2πr⋅dr)

Lower case "sigma" represents the amount of charge per area of the plane, in units of coulombs/meter^2. This parameter is called the "charge density".

Main content

Electrical engineering

Course: Electrical engineering > Unit 5

Lesson 2: Fields, potential, and voltage

Plane of charge

Google Classroom

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

[full symmetry argument]

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,

[area of a very thin 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

[change of variables]

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,

[with a trig identity]

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

Conclusion

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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francis
Posted 8 years ago. Direct link to francis's post “where did the sigma came ...”
where did the sigma came from? dQhoop = σ⋅(2πr⋅dr)
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- Willy McAllister
  Posted 8 years ago. Direct link to Willy McAllister's post “Lower case "sigma" repres...”
  Lower case "sigma" represents the amount of charge per area of the plane, in units of coulombs/meter^2. This parameter is called the "charge density".
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Toby Eng
Posted 8 years ago. Direct link to Toby Eng's post “can you give a video abou...”
can you give a video about these science topics
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phil.yingkongshi.chang
Posted 8 years ago. Direct link to phil.yingkongshi.chang's post “Why the K be instand by 1...”
Why the K be instand by 1/4πϵ?
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- Naman Jain
  Posted 8 years ago. Direct link to Naman Jain's post “just to make the expressi...”
  just to make the expression shorter and easy to remember
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ashley wilson
Posted 6 years ago. Direct link to ashley wilson's post “have we taken point charg...”
have we taken point charge dQ in the hoop part explanation??
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- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “Ashley, I don't quite und...”
  Ashley, I don't quite understand your question. Could you rephrase it?
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  (2 votes)
dena escot
Posted 10 months ago. Direct link to dena escot's post “1) why does the thickness...”
1) why does the thickness of the hoop not equal to pi*r^2?
2) why does the limits of the last integral not from -90 to 90 degree?
3) I don not understand Cleary this statement "first integration to sweep dQ around its hoop to get the field contribution from one particular hoop)"
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- Willy McAllister
  Posted 10 months ago. Direct link to Willy McAllister's post “1) The area of a very thi...”
  1) The area of a very thin hoop is approximately the same as the area of a long skinny rectangle The long side (width) is the circumference of the hoop 2πr^2 and the skinny height dimension is dr. So the area is 2πr^2 dr.
  
  2. To integrate the hoops you start with the smallest possible hoop which is theta=0 and add up the area of successively larger hoops, which is theta=90.
  
  3. Before doing the integration mentioned in question 2 you have to develop an expression for the total charge contained in a single hoop. This is shown in the last diagram of the article. You do an integration to add up all the little dQ elements around the circumference of one hoop.
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Stuti Rawat
Posted 6 years ago. Direct link to Stuti Rawat's post “3 concentric metallic sph...”
3 concentric metallic spherical shells of radiur R,2R,3R are given charges Q,Q',Q''respectively. The surface charge densitieson the outer surface of the shells are equal. What is the ratio of the charges given to shell Q:Q':Q''
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