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## Electrical engineering

### Course: Electrical engineering>Unit 5

Lesson 2: Fields, potential, and voltage

# Line of charge

Advanced example: Electric field surrounding a uniformly charged infinite line. Written by Willy McAllister.

## Worked Example: Electric field near a line of charge

We derive an expression for the electric field near a line of charge.
The result will show the electric field near a line of charge falls off as 1, slash, a, where a is the distance from the line.
Assume we have a long line of length L, with total charge Q. Assume the charge is distributed uniformly along the line. The total charge on the line is Q, so the charge density in coulombs/meter is,
mu, equals, start fraction, Q, divided by, L, end fraction
Assume a test charge q is positioned opposite the center of the line, at a distance a.
What is the electric field at the location of q due to (created by) the line of charge?
This derivation will lead to a general solution of the electric field for any length L, and any distance a. Using this general solution, we will solve a particularly useful case where the line is very long relative to the distance to the test charge, L, \gg, a.
First, create and name some variables to talk about.
• a is the distance from the line to the location of our test charge, q.
• d, Q is a tiny amount of charge contained in a tiny section of the line, d, x.
• x is the distance from where a touches the line to d, Q.
• r is the distance from d, Q to the location of the test charge.
• theta is the angle between a and r.
The electric field surrounding some point charge, Q is,
E, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, Q, divided by, r, squared, end fraction
The electric field at the location of test charge q due to a small chunk of charge in the line, d, Q is,
d, E, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, d, Q, divided by, r, squared, end fraction
The amount of charge d, Q can be restated in terms of charge density, d, Q, equals, mu, d, x,
d, E, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, mu, start fraction, d, x, divided by, r, squared, end fraction
The most suitable independent variable for this problem is the angle theta. The analysis is simplified by recasting the equation to sweep d, theta through a range of angles instead of sweeping d, x along the line (this is a change of variable).
After the change of variables, we can redraw the diagram in terms of d, theta,
The change of variables allows us substitute start fraction, d, theta, divided by, a, end fraction for start fraction, d, x, divided by, r, squared, end fraction in the previous equation,
d, E, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, mu, start fraction, d, theta, divided by, a, end fraction
Now we exploit the symmetry of the charge arrangement by figuring out the electric field in just the y direction (the direction going straight from the line through q).
This means we scale the electric field d, E down by the cosine of the angle theta,
d, E, start subscript, y, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, cosine, theta, d, theta
We are ready to integrate (add up) all the contributions from each d, Q to get the electric field,
E, start subscript, y, end subscript, equals, integral, start subscript, minus, theta, end subscript, start superscript, plus, theta, end superscript, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, cosine, theta, d, theta
This is the general solution for the electric field near any length of line, L, at any distance a away from the line. The limits plus minus, theta are the angles to either end of the line.

### Useful case: long line of charge

Now we solve for the useful case where the line of charge is very long relative to the separation a, or L, \gg, a. If you stand at q and turn your head to look in either direction towards each end of this very long line, your head turns (very nearly) plus minus, 90, degrees (plus minus, pi, slash, 2 radians). These become the limits on our integration.
E, start subscript, y, end subscript, equals, integral, start subscript, minus, pi, slash, 2, end subscript, start superscript, plus, pi, slash, 2, end superscript, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, cosine, theta, d, theta
Move anything that doesn't depend on theta outside the integral.
E, start subscript, y, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, integral, start subscript, minus, pi, slash, 2, end subscript, start superscript, plus, pi, slash, 2, end superscript, cosine, theta, d, theta
and evaluate the integral,
E, start subscript, y, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, sine, theta, vertical bar, start subscript, minus, pi, slash, 2, end subscript, start superscript, plus, pi, slash, 2, end superscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, left parenthesis, plus, 1, minus, minus, 1, right parenthesis, equals, start fraction, 2, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction
Finally, the electric field created by a long line of charge at a point a away from the line is,
E, start subscript, y, end subscript, equals, start fraction, mu, divided by, 2, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, 1, divided by, a, end fraction
Well done if you followed this all the way through. The important finding from this exercise is: in contrast to 1, slash, r, squared for a point charge, the field surrounding a line of charge falls off as 1, slash, a.
We did a lot of math to derive this result. It is worthwhile to take a moment to sit with this solution to let it soak in. Now that you have seen the math, does it make intuitive sense that distance has a different exponent, 1, slash, a, compared to a point charge, 1, slash, r, squared?
As an amusing distraction, if you recall the fable of the butter gun from the Inverse Square Law article, can you design a new butter gun for a line of charge, that sprays in a 1, slash, a pattern?