Show Fx,Fy when x>>a: 65 Character Title"Show Fx,Fy When x Much Greater Than a

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Homework Help Overview

The discussion revolves around deriving expressions for the forces \( F_x \) and \( F_y \) generated by an electric field when the variable \( x \) is much greater than \( a \). The original poster presents the equations for the forces and seeks to simplify them under the condition \( x >> a \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Taylor series to expand terms in the force equations and identify negligible terms. There is a focus on understanding the expansion process and its implications for simplifying the forces.

Discussion Status

Several participants have provided guidance on using Taylor series for expansion, with some expressing uncertainty about the original poster's understanding of the concept. The discussion is ongoing, with participants exploring the mathematical reasoning behind the expansions without reaching a consensus.

Contextual Notes

The original poster expresses difficulty in understanding the concept of expansion and seeks further clarification on the steps involved. There is mention of limits and the nature of functions that can be expanded, indicating a need for foundational understanding in this area.

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Show that if x>>a...

Homework Statement



I've got the following force generated by an electric field

{F_x} = - KQq\frac{1}{{x\sqrt {{x^2} + {a^2}} }}

{F_y} = \frac{{KQq}}{a}\left( {\frac{1}{x} - \frac{1}{{\sqrt {{x^2} + {a^2}} }}} \right)

Homework Equations



I need to show that when x>>a:

{F_x} = - \frac{{KQq}}{{{x^2}}}

{F_y} = \frac{{KQqa}}{{2{x^3}}}

The Attempt at a Solution



I think I'm on the right track but I'm stuck here:

\frac{1}{{\sqrt {{x^2} + {a^2}} }} = \frac{1}{x}{\left( {1 + \frac{{{a^2}}}{{{x^2}}}} \right)^{ - \frac{1}{2}}}
 
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Have you heard of Taylor series? Why don't you try expanding the above term using taylor series and see which terms are negligible?
 


You are indeed on the right track. When x >> a, then \delta := \frac{a}{x} is very small. So you can expand
x \sqrt{1 + a^2 / x^2} = x \sqrt{1 + \delta^2} around \delta = 0.
 


Thanks for your responses anirudh215 and CompuChip, but I really don't know what you mean by "expand". If you could show some steps I might have a clue.
 


Do you know limits?

ehild
 


baseballman said:
Thanks for your responses anirudh215 and CompuChip, but I really don't know what you mean by "expand". If you could show some steps I might have a clue.

Certain functions can be "expanded" around points, i.e. if you have a function f, and it is "expandable", then you can equate the value of that function about a point in terms of its derivatives.

The square root function above is one such function which is expandable. The Taylor series would be something like

\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{4.2!}...

You can read more about Taylor series here:
http://en.wikipedia.org/wiki/Taylor_series

Now, substitute \frac{a^{2}}{x^{2}} as y, and expand similar to above. See what you can do from here. Note: the number of terms I have used in the expansion above is enough for you to complete the sum. Just look at what is and isn't negligible. I'm just doubtful whether you calculated Fy properly.
 
Last edited:

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