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 $23.9 .$ Show that the electric field at distances $x$ that are large compared with $R$ approaches that of a point charge $Q=\sigma \pi R^{2}$ (Suggestion: First show that $x /\left(x^{2}+R^{2}\right)^{1 / 2}=$ $\left(1+R^{2} / x^{2}\right)^{-1 / 2}$ and use the binomial expansion $(1+\delta)^{n} \approx 1+n \delta$ when $\delta<1 . )$

The electric field along the axis of a uniformly charged disk of radius $R$ and total charge $Q$ was calculated in Example $23.9 .$ Show that the electric field at distances $x$ that are large compared with $R$ approaches that of a point charge $Q=\sigma \pi R^{2}$ (Suggestion: First show that $x /\left(x^{2}+R^{2}\right)^{1 / 2}=$ $\left(1+R^{2} / x^{2}\right)^{-1 / 2}$ and use the binomial expansion
$(1+\delta)^{n} \approx 1+n \delta$ when $\delta<1 . )$

Key Concepts

Electric Field of Continuous Charge Distributions

This concept involves calculating the electric field produced by charges spread continuously over a region rather than being concentrated at a point. It typically requires integrating the contributions of infinitesimal charge elements over the entire distribution, accounting for their relative positions and distances from the point of interest.

Far-Field Approximation

The far-field approximation is used when the distance from the charge distribution to the observation point is much larger than the size of the distribution. In this limit, the spatial details of the charge distribution become insignificant, and the field can be well approximated by simpler expressions, often resembling those of point charges, thereby simplifying the mathematical analysis.

Binomial Series Approximation

The binomial expansion is a mathematical tool used to approximate expressions that involve terms raised to a fractional power. When dealing with small parameters, such as R²/x² when x is much larger than R, one can expand the expression in a series and retain only the leading order terms. This method is instrumental in simplifying complex expressions in physics, especially in electrostatics for approximating square roots and other functions.

Point Charge Approximation

In electrostatics, if an object’s dimensions are negligible compared to the distance from the point of interest, the object can be modeled as a point charge. This approximation significantly simplifies calculations by replacing a distributed charge with a single equivalent charge, allowing the use of Coulomb's law directly to determine the electric field.

Transcript

00:01 So in this problem we should start from the example that was already done in the textbook, which states that actually the result of the example is that electric field, at least the horizontal component of the electric field coming from disk, that is uniformly charged and continually charged along its surface, is in the form 2 pi times electrostatic constant, times the density, surface density of the electric charge, one minus, and then x over x squared plus r which is the radius of the disk or one half.

00:49 So this is the function that basically gives us from, it maps from the set of positions into set of electric fields, field values depending on the distance, and this distance is of course on the x -axis if the x -axis is passing to the center of the disk.

01:19 So like this, this electric field here, x.

01:28 Now, what happens if this distance is very large? well, of course, this will start behaving like a common formula for point -charge, this point -charge electric field, depending on the distance, on the x -axis, of course.

01:44 So first we should do something with the distance term, so this part of the whole function, and see what kind of mathematical approximations we can apply if x is much larger than r.

02:04 So first we'll rewrite it and here we should factor out x somehow.

02:18 So x must go out this x must go out.

02:21 Out in order to cancel out with this x.

02:24 And of course we can do this if we say that this is x over x times then 1 plus and then r squared over x squared and all of this over 1 across.

02:38 We see that if we multiply x with this term, this part of the term here we will obtain this term that we started from.

02:47 So this algebra transformation is completely fine.

02:50 Now this cancel out x and x and we're left with one over one plus r squared over x squared and then for this we can write this in a different manner of course you can say this is one over one plus r over x squared and then this unlock off so finally we can see that we can see that we can also raise this we can raise this on the in the denominator and obtain that x over x plus r squared on one half is equal to 1 plus r over x squared on minus 1 half so next what should we do we need to expand and this was already hinted in the problem, so we need to expand this term and use binomial expansion, which can be used for such terms, of course.

04:18 And how do we expand it? we will only use the first term of the expansion.

04:24 So that means that 1 plus r over x squared on minus 1 half.

04:33 Now binomial expansion means we will take out basically this 1 minus 1.

04:42 Minus 1 half will go next to the r over x term.

04:47 So we'll have 1 plus minus 1 half times r over x squared and then we neglect the rest of the terms...

Please give Ace some feedback