5.5 Calculating electric fields of charge distributions (Page 2/12)

University physics volume 2 Page 2 / 12

Electric field of a line segment

Find the electric field a distance z above the midpoint of a straight line segment of length L that carries a uniform line charge density

λ

Strategy

Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length dl , each of which carries a differential amount of charge

d q = λ d l

. Then, we calculate the differential field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify the calculation ( [link] ). Finally, we integrate this differential field expression over the length of the wire (half of it, actually, as we explain below) to obtain the complete electric field expression.

A long, thin wire is on the x axis. The end of the wire is a distance z from the center of the wire. A small segment of the wire, a distance x to the right of the center of the wire, is shaded. Another segment, the same distance to the left of center, is also shaded. Point P is a distance z above the center of the wire, on the z axis. Point P is a distance r from each shaded region. The r vectors point from each shaded region to point P. Vectors d E 1 and d E 2 are drawn at point P. d E 1 points away from the left side shaded region and points up and right, at an angle theta to the z axis. d E 2 points away from the right side shaded region and points up and r left, making the same angle with the vertical as d E 1. The two d E vectors are equal in length. — A uniformly charged segment of wire. The electric field at point P can be found by applying the superposition principle to symmetrically placed charge elements and integrating.

Solution

Before we jump into it, what do we expect the field to “look like” from far away? Since it is a finite line segment, from far away, it should look like a point charge. We will check the expression we get to see if it meets this expectation.

The electric field for a line charge is given by the general expression

\vec{E} (P) = \frac{1}{4 π ε_{0}} \int_{line} \frac{λ d l}{r^{2}} \hat{r} .

The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal ( x )-components of the field cancel, so that the net field points in the z -direction. Let’s check this formally.

The total field $\vec{E} (P)$ is the vector sum of the fields from each of the two charge elements (call them ${\vec{E}}_{1}$ and ${\vec{E}}_{2}$ , for now):

\vec{E} (P) = {\vec{E}}_{1} + {\vec{E}}_{2} = E_{1 x} \hat{i} + E_{1 z} \hat{k} + E_{2 x} (− \hat{i}) + E_{2 z} \hat{k} .

Because the two charge elements are identical and are the same distance away from the point P where we want to calculate the field, $E_{1 x} = E_{2 x},$ so those components cancel. This leaves

\vec{E} (P) = E_{1 z} \hat{k} + E_{2 z} \hat{k} = E_{1} cos θ \hat{k} + E_{2} cos θ \hat{k} .

These components are also equal, so we have

\begin{matrix} \vec{E} (P) & = \frac{1}{4 π ε_{0}} \int \frac{λ d l}{r^{2}} cos θ \hat{k} + \frac{1}{4 π ε_{0}} \int \frac{λ d l}{r^{2}} cos θ \hat{k} \\ = \frac{1}{4 π ε_{0}} \int_{0}^{L / 2} \frac{2 λ d x}{r^{2}} cos θ \hat{k} \end{matrix}

where our differential line element dl is dx , in this example, since we are integrating along a line of charge that lies on the x -axis. (The limits of integration are 0 to $\frac{L}{2}$ , not $- \frac{L}{2}$ to $+ \frac{L}{2}$ , because we have constructed the net field from two differential pieces of charge dq . If we integrated along the entire length, we would pick up an erroneous factor of 2.)

In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given. In this case, both r and $θ$ change as we integrate outward to the end of the line charge, so those are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that

r = {(z^{2} + x^{2})}^{1 / 2}

and

cos θ = \frac{z}{r} = \frac{z}{{(z^{2} + x^{2})}^{1 / 2}} .

Substituting, we obtain

\begin{matrix} \vec{E} (P) & = \frac{1}{4 π ε_{0}} \int_{0}^{L / 2} \frac{2 λ d x}{(z^{2} + x^{2})} \frac{z}{{(z^{2} + x^{2})}^{1 / 2}} \hat{k} \\ = \frac{1}{4 π ε_{0}} \int_{0}^{L / 2} \frac{2 λ z}{{(z^{2} + x^{2})}^{3 / 2}} d x \hat{k} \\ = \frac{2 λ z}{4 π ε_{0}} {[\frac{x}{z^{2} \sqrt{z^{2} + x^{2}}}] |}_{0}^{L / 2} \hat{k} \end{matrix}

which simplifies to

\vec{E} (z) = \frac{1}{4 π ε_{0}} \frac{λ L}{z \sqrt{z^{2} + \frac{L^{2}}{4}}} \hat{k} .

Significance

Notice, once again, the use of symmetry to simplify the problem. This is a very common strategy for calculating electric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals and may need to be calculated numerically by a computer.

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Source: OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3

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