The electric field along the axis of a uniformly charged...

The electric field along the axis of a uniformly charged disk of radius $R$ and total charge $Q$ was calculated in Example 23.3. Show that the electric field at distances $x$ that are large compared with $R$ approaches that of a particle with charge $Q=\sigma \pi R^{2} .$ Suggestion: First show that $x /\left(x^{2}+R^{2}\right)^{1 / 2}=(1+$ $\left.R^{2} / x^{2}\right)^{-1 / 2}$ and use the binomial expansion $(1+\delta)^{n} \approx 1+$ $n \delta,$ when $\delta<<1$.

Key Concepts

Electric Field from Continuous Charge Distributions

This concept involves calculating the electric field created by a spread-out charge rather than a discrete point charge. In many physical problems, such as those involving surfaces or volumes uniformly populated with charge, the field is determined by integrating Coulomb's law over the entire charge distribution. This method is fundamental in electromagnetism for handling diverse geometries in which the charge is continuously distributed.

Far-field Approximation

The far-field approximation simplifies a problem by considering the regime where the distance from the charge distribution is much larger than its spatial extent. In this limit, the detailed structure of the distribution becomes negligible, allowing the system to be approximated as a point charge. This approximation is widely used in physics to reduce complex spatial integrals to simpler expressions that capture the leading order behavior at large distances.

Binomial Expansion

The binomial expansion is a mathematical technique used to approximate expressions of the form (1+?)^n when ? is small compared to 1. This method permits the retention of only the leading-order terms in ?, greatly simplifying analyses in physics where higher-order corrections are insignificant. It is particularly useful in approximating factors that arise in the calculation of fields when certain parameters, such as ratios of lengths, are small.

Point Charge Approximation

When the observer is far away from a charge distribution, the entire collection of charges can effectively be regarded as a single point charge with a total charge equal to the sum of the distributed charges. This simplification, justified by the far-field approximation, is central to various analyses in electromagnetism, where it reduces the complexity of the problem by enabling the use of the classical Coulomb field expression for a point charge.

Transcript

00:01 In this problem, we're going to work with the electric field due to a charged disk, which was derived in one of the earlier examples in the book.

00:11 And this is what they got.

00:14 What we want to do here is to see what happens to this as we take x much bigger than r.

00:23 And so we're basically just going to go really far away from this charged disk and see what happens to the e field.

00:29 Well, to do this, we can use what is called binomial expansion.

00:38 So this is what we'll do.

00:41 So this is the term that depends on x, r squared plus x squared, to the one half on the bottom here.

00:52 What we could do here is we could pull out an x squared in the bottom.

00:56 And so what we'll have is on x times r squared over x squared plus 1 all to the 1 half.

01:07 And now these x's on top and bottom cancel out.

01:11 And what we're going to be left with is 1 plus r squared x squared to the 1 half, or negative 1 half actually.

01:21 All right.

01:22 So how do we use the binomial expansion? well, binomial expansion tells us that 1 plus delta to some power is approximately equal to 1 plus and delta, if this delta is really small.

01:36 Well, we have something very similar to that, where we have one plus a number, which is also really small, because what we're saying is that x is much bigger than r, and so we're taking a number and defining it by a really large number, you're going to get a small number out.

01:54 And so this is going to be approximately equal to 1 minus 1⁄2 times r squared over x squared.

02:09 All right, now that we have that, we can just plug this into the electric field formula.

02:16 So this is going to be e equals 2 pi, k e sigma, 1 minus 1, plus 1 1⁄2 r squared over x squared...

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