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Kurs: Fizyka - 12 klasa (Indie) > Rozdział 1

Lekcja 6: Continuous charge distribution

Natężenie pola elektrycznego pochodzącego od jednorodnie naładowanej, nieskończenie długiej, prostej nici

Natężenie pola elektrycznego pochodzącego od nieskończenie długiej, jednorodnie naładowanej, prostej nici. Stworzone przez 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 / 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,

μ = \frac{Q}{L}

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 ≫ 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.
$θ$ ‍ is the angle between $a$ ‍ and $r$ ‍.

The electric field surrounding some point charge,

Q

is,

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

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

The amount of charge

d Q

can be restated in terms of charge density,

d Q = μ d x

d E = \frac{1}{4 π ϵ_{0}} μ \frac{d x}{r^{2}}

The most suitable independent variable for this problem is the angle

θ

. The analysis is simplified by recasting the equation to sweep

d θ

through a range of angles instead of sweeping

d x

along the line (this is a change of variable).

The goal here is to develop an expression for

d θ

in terms of

d x

. Then we will substitute this into the equation for the electric field.

Some useful trig identities, from the definitions of cosine, secant, and tangent,

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

\tan θ = \frac{x}{a} x = a \tan θ

Our goal is to replace

d x

with some form of

d θ

, so take the derivative of

x

with respect to

θ

based on the tangent identity,

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

\frac{d x}{d θ} = a \sec^{2} θ

d x = a \sec^{2} θ d θ

Finally, compute

\frac{d x}{r^{2}}

\frac{d x}{r^{2}} = \frac{a \sec^{2} θ d θ}{a^{2} \sec^{2} θ}

Enjoy some cancellation to end up with an expression for

d θ

and

a

we can swap in for

d x

and

r

\frac{d x}{r^{2}} = \frac{d θ}{a}

Now head back to the main discussion and use our change of variable.

After the change of variables, we can redraw the diagram in terms of

d θ

The change of variables allows us substitute

\frac{d θ}{a}

for

\frac{d x}{r^{2}}

in the previous equation,

d E = \frac{1}{4 π ϵ_{0}} μ \frac{d θ}{a}

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

θ

d E_{y} = \frac{1}{4 π ϵ_{0}} \frac{μ}{a} \cos θ d θ

We are ready to integrate (add up) all the contributions from each

d Q

to get the electric field,

E_{y} = \int_{- θ}^{+ θ} \frac{1}{4 π ϵ_{0}} \frac{μ}{a} \cos θ d θ

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

\pm θ

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 ≫ 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)

\pm 90^{\circ}

(

\pm π / 2

radians). These become the limits on our integration.

E_{y} = \int_{- π / 2}^{+ π / 2} \frac{1}{4 π ϵ_{0}} \frac{μ}{a} \cos θ d θ

Move anything that doesn't depend on

θ

outside the integral.

E_{y} = \frac{1}{4 π ϵ_{0}} \frac{μ}{a} \int_{- π / 2}^{+ π / 2} \cos θ d θ

and evaluate the integral,

E_{y} = \frac{1}{4 π ϵ_{0}} \frac{μ}{a} \sin θ |_{- π / 2}^{+ π / 2} = \frac{1}{4 π ϵ_{0}} \frac{μ}{a} (+ 1 - - 1) = \frac{2}{4 π ϵ_{0}} \frac{μ}{a}

Finally, the electric field created by a long line of charge at a point

a

away from the line is,

E_{y} = \frac{μ}{2 π ϵ_{0}} \frac{1}{a}

Well done if you followed this all the way through. The important finding from this exercise is: in contrast to

1 / r^{2}

for a point charge, the field surrounding a line of charge falls off as

1 / 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 / a

, compared to a point charge,

1 / r^{2}

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 / a

pattern?

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