Taylor expansion of an electrostatics problem

In summary, the problem involves finding the electric field along the x-axis for six charges arranged in a regular hexagon. The goal is to use a Taylor series expansion with a small quantity a/x to approximate the field. This approximation is accurate as long as a/x is close to 0.
  • #1
mmpstudent
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Homework Statement



The problem has six charges that are at the corners of a regular hexagon in the xy plane, each charge a distance a from the origin. I have already solved for the electric fields in the x and y direction and now am trying to apply an approximation for the field on the x-axis at where x>>a. the field along the x-axis is

[itex]E_{x}= \displaystyle\sum_{n=1}^{6} \frac{x-acos\frac{k\pi}{3}}{[x^{2}-2axcos\frac{k\pi}{3}+a^{2}]^{3/2}}[/itex]

I am supposed to use a power series in the small quantity a/x using the method of taylor series to get to

[itex]E_{x}= \frac{1}{4\pi\epsilon_{0}}[{\frac{6q}{x^{2}}+\frac{9qa^{2}}{2x^{4}}}][/itex]

I haven't done this level math in a long time and I am sure that it is not too difficult, but I don't know what it means in the small quantity a/x. Do I just set x>>a in the efield equation and go about my business, if so, I don't understand where the a^2 term comes from in the numerator.
 
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  • #2
mmpstudent said:

Homework Statement



The problem has six charges that are at the corners of a regular hexagon in the xy plane, each charge a distance a from the origin. I have already solved for the electric fields in the x and y direction and now am trying to apply an approximation for the field on the x-axis at where x>>a. the field along the x-axis is

[itex]E_{x}= \displaystyle\sum_{n=1}^{6} \frac{x-acos\frac{k\pi}{3}}{[x^{2}-2axcos\frac{k\pi}{3}+a^{2}]^{3/2}}[/itex]

I am supposed to use a power series in the small quantity a/x using the method of taylor series to get to

[itex]E_{x}= \frac{1}{4\pi\epsilon_{0}}{\frac{6q}{x^{2}}+\frac{9qa^{2}}{2x^{4}}}[/itex]

I haven't done this level math in a long time and I am sure that it is not too difficult, but I don't know what it means in the small quantity a/x. Do I just set x>>a in the efield equation and go about my business, if so, I don't understand where the a^2 term comes from in the numerator.

You need to use [noparse] [ tex ] [ /tex ] or [ itex ] [ /itex ] [/noparse] (without the spaces). In other words, your slashes were oriented the wrong way in your closing tags. I fixed them for you. Note that you can also surround your LaTeX code with double dollar signs or double pound signs on either side as shorthand for tex and itex tags respectively.

Taylor series expansion: let's say you have some function f(u) of some independent variable u. The Taylor series expansion of the function around some point "b" is given by $$f(u) = \sum_{n=0}^\infty \frac{1}{n!} \left[ \frac{d^n}{du^n}f(u)\right]_{u=b} (u - b)^n $$ $$ = f(b) +\left. \frac{df}{du}\right|_{u = b} (u - b) + \frac{1}{2}\left. \frac{d^2f}{du^2}\right|_{u=b}(u - b)^2 +~\textrm{H.O.T.}$$ where H.O.T. means "higher order terms." If you think about it, if you truncate the infinite series after n terms, then this amounts to finding a (n-1)th -order polynomial approximation to the function. This approximation is exactly correct at the point u = b, but it begins to diverge away from it as u moves away from b.

In this case, your independent variable is u = a/x, and you're taking the Taylor series expansion around the point b = 0. So it's okay to neglect higher order terms, and still have a pretty accurate approximation, as long as a/x is very close to 0. (u is very close to b). That is the situation you have in this problem.
 
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  • #3
Okay, so my thinking was not too far off. Now I just need to get to work. Thanks for the clarification on both the tex and the problem.
 

1. What is the Taylor expansion of an electrostatics problem?

The Taylor expansion of an electrostatics problem is a mathematical technique used to approximate the solution of an electrostatics problem at a particular point. It involves expressing the electrostatic potential or electric field as an infinite sum of terms, where each term represents a higher order approximation of the solution.

2. Why is the Taylor expansion useful in electrostatics?

The Taylor expansion is useful in electrostatics because it allows us to approximate the solution of a problem at a specific point without having to directly solve the entire problem. This can save time and computational resources, especially for complex problems. It also helps us understand the behavior of the electric field or potential near a particular point.

3. How is the Taylor expansion calculated for an electrostatics problem?

The Taylor expansion is calculated using the derivatives of the electrostatic potential or electric field at a given point. The first term in the expansion is the value of the function at that point, and each subsequent term is the value of the corresponding derivative multiplied by the appropriate power of the distance from the point. The higher the order of the derivative, the more accurate the approximation will be.

4. What are the limitations of the Taylor expansion in electrostatics?

The Taylor expansion is only accurate for points near the point of expansion. As we move farther away from the point, the approximation becomes less accurate. Additionally, the Taylor expansion assumes that the electrostatic potential or electric field is continuous and differentiable at the point of expansion, which may not always be the case in real-world problems.

5. Can the Taylor expansion be used for any type of electrostatics problem?

Yes, the Taylor expansion can be used for any type of electrostatics problem, as long as the electrostatic potential or electric field can be expressed as a continuous and differentiable function. It is particularly useful for problems with complex geometries or boundary conditions, as it allows us to approximate the solution without having to solve the entire problem numerically.

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