5.5 Calculating electric fields of charge distributions  (Page 3/12)

In the case of a finite line of charge, note that for $z\gg L$ , $z^2$ dominates the L in the denominator, so that [link] simplifies to

If you recall that $\lambda L=q$ , the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.

In the limit $L\to \infty$ , on the other hand, we get the field of an infinite straight wire    , which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:

An interesting artifact of this infinite limit is that we have lost the usual $1\text{/}{r}^2$ dependence that we are used to. This will become even more intriguing in the case of an infinite plane.

Electric field due to a ring of charge

A ring has a uniform charge density $\lambda$ , with units of coulomb per unit meter of arc. Find the electric potential at a point on the axis passing through the center of the ring.

Strategy

We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle. We divide the circle into infinitesimal elements shaped as arcs on the circle and use polar coordinates shown in [link] .

Solution

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

$\overrightarrow{\text{E}}\left(P\right)=\frac{1}{4\pi {\epsilon }_0}{\displaystyle {\int }_{\text{line}}\frac{\lambda dl}{r^2}}\widehat{\text{r}}.$

A general element of the arc between $\theta$ and $\theta +d\theta$ is of length $Rd\theta$ and therefore contains a charge equal to $\lambda Rd\theta .$ The element is at a distance of $r=\sqrt{z^2+R^2}$ from P , the angle is $\text{cos}\varphi =\frac{z}{\sqrt{z^2+R^2}}$ , and therefore the electric field is

$\begin{array}{cc}\overrightarrow{\text{E}}\left(P\right)\hfill & =\frac{1}{4\pi {\epsilon }_0}{\displaystyle {\int }_{\text{line}}\frac{\lambda dl}{r^2}}\widehat{\text{r}}=\frac{1}{4\pi {\epsilon }_0}{\displaystyle {\int }_0^{2\pi }\frac{\lambda Rd\theta }{z^2+R^2}\frac{z}{\sqrt{z^2+R^2}}\widehat{z}}\hfill \\ & =\frac{1}{4\pi {\epsilon }_0}\frac{\lambda Rz}{{\left(z^2+R^2\right)}^{3\text{/}2}}\widehat{z}{\displaystyle {\int }_0^{2\pi }d\theta }=\frac{1}{4\pi {\epsilon }_0}\frac{2\pi \lambda Rz}{{\left(z^2+R^2\right)}^{3\text{/}2}}\widehat{z}\hfill \\ & =\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{tot}}z}{{\left(z^2+R^2\right)}^{3\text{/}2}}\widehat{z}.\hfill \end{array}$

Significance

As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of $z>>R$ , we find that

$\overrightarrow{\text{E}}\approx \frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{tot}}}{z^2}\widehat{z},$

as we expect.

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