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## Electrical engineering

### Unit 5: Lesson 2

Fields, potential, and voltage

# 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, sigma, start text, left parenthesis, C, end text, slash, start text, m, end text, squared, right parenthesis.

### What is the electric field due to the plane at a location $\text a$start text, a, end text 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 start text, d, end text, r.
• start text, d, end text, Q is an infinitesimal region of charge in a section of the hoop.
• Line ell goes from the location of start text, d, end text, Q to the location of the test charge.
• start text, d, end text, E is the electric field at point q created by start text, d, end text, Q.
We know the field at location q due to start text, d, end text, Q; it's the definition of the field created by a point charge,
start text, d, end text, E, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, start text, d, end text, Q, divided by, ell, squared, end fraction
To solve the electric field for the whole plane, we have to do two integrations:
• first integration to sweep start text, d, end text, 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 ell away from q, so each start text, d, end text, Q creates the same magnitude field at q. Symmetry tells us the total field contribution from all start text, d, end text, 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 start text, d, end text, Q is exactly cancelled by the start text, d, end text, Q on the opposite side of the hoop. The straight out "a-direction" portion of the electric field start text, d, end text, E, start subscript, a, end subscript is related to start text, d, end text, E,
start text, d, end text, E, start subscript, a, end subscript, equals, start text, d, end text, E, cosine, theta
Which gives this for start text, d, end text, E, start subscript, a, end subscript, the field from a single point charge start text, d, end text, Q,
start text, d, end text, E, start subscript, a, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, start text, d, end text, Q, divided by, ell, squared, end fraction, cosine, theta
Next, express the field contribution from one entire hoop start text, d, end text, E, start subscript, h, o, o, p, end subscript,
start text, d, end text, E, start subscript, h, o, o, p, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, start text, d, end text, Q, start subscript, h, o, o, p, end subscript, divided by, ell, squared, end fraction, start text, c, o, s, end text, theta
start text, d, end text, Q, start subscript, h, o, o, p, end subscript is the total charge contained in one hoop, the sum of the individual point start text, d, end text, 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, sigma, times the area of the hoop,
start text, d, end text, Q, start subscript, h, o, o, p, end subscript, equals, sigma, dot, left parenthesis, 2, pi, r, dot, start text, d, end text, r, right parenthesis
The electric field at the location of q created by a hoop with radius r, containing charge start text, Q, end text, start subscript, h, o, o, p, end subscript is,
start text, d, end text, E, start subscript, h, o, o, p, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, sigma, 2, pi, r, start text, d, end text, r, divided by, ell, squared, end fraction, start text, c, o, s, end text, theta
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 start text, d, end text, r to start text, d, end text, theta.
After the change of variable, the diagram can be redrawn in terms of start text, d, end text, theta and theta,
and the field equation for one hoop becomes,
start text, d, end text, E, start subscript, h, o, o, p, end subscript, equals, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction, start text, t, a, n, end text, theta, dot, start text, c, o, s, end text, theta, start text, d, end text, theta
which can be simplified a bit more,
start text, d, end text, E, start subscript, h, o, o, p, end subscript, equals, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction, start text, s, i, n, end text, theta, start text, d, end text, theta
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 start text, d, end text, E, start subscript, h, o, o, p, end subscript, there is NO dependence on distance. Remarkable.
Almost home. We are ready to perform the integration,
E, equals, integral, start subscript, a, l, l, h, o, o, p, s, end subscript, start text, d, end text, E, start subscript, h, o, o, p, end subscript
where E is the overall electric field from all hoops. Substitute for start text, d, end text, E, start subscript, h, o, o, p, end subscript,
E, equals, integral, start subscript, t, h, e, t, a, s, end subscript, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction, start text, s, i, n, end text, theta, start text, d, end text, theta
What are the angle limits on the integration? The smallest possible hoop is when r is zero; ell coincides with a, and theta is zero. The largest hoop is when r is infinite; line ell comes from way out at the horizon in any direction, and theta is 90, degrees or pi, slash, 2 radians. So the limits on the integration run from theta, equals, 0, start text, space, t, o, space, end text, pi, slash, 2 radians.
E, equals, integral, start subscript, 0, end subscript, start superscript, pi, slash, 2, end superscript, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction, start text, s, i, n, end text, theta, start text, d, end text, theta
E, equals, minus, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction, start text, c, o, s, end text, theta, vertical bar, start subscript, 0, end subscript, start superscript, plus, pi, slash, 2, end superscript, equals, minus, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction, left parenthesis, 0, minus, 1, right parenthesis
The electric field near an infinite plane is,
E, equals, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction start text, n, e, w, t, o, n, s, slash, c, o, u, l, o, m, b, end text

## 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,
1. Charge gives rise to an electric field.
2. The electric field acts locally on a test charge.
Summarizing the three electric field examples worked out so far,
The field due to afalls off at
point charge1, slash, r, squared
line of charge1, slash, r, start superscript, 1, end superscript
plane of charge1, slash, r, start superscript, 0, end superscript
These three charge configurations are a useful toolkit for predicting electric field in lots of practical situations.

## Want to join the conversation?

• where did the sigma came from? dQ​hoop = 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".
• have we taken point charge dQ in the hoop part explanation??
(1 vote)
• Ashley, I don't quite understand your question. Could you rephrase it?
• Why the K be instand by 1/4πϵ​?
(1 vote)
• just to make the expression shorter and easy to remember
(1 vote)
• can you give a video about these science topics
(1 vote)
• 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''