How Does Taylor Series Approximation Determine Electric Potential Over Distance?

In summary: So if you have something like (1+a^2/R^2)^1/2, you can think of it as (1+(a/R)^2)^1/2 ~ 1+(1/2)(a/R)^2.So just remember that rule and learn to recognize square roots as 1/2 powers, and you'll be able to do most taylor expansions in physics.In summary, for large distances R, the electric potential V at a distance R along the axis perpendicular to the center of a charged disc with radius a and constant charge density d can be approximated as V = pi*a^2*d/R. This can be obtained by using the binomial approximation for the square root and simplifying the expression
  • #1
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The electric potential V at a distance R along the axis perpendicular to the center of a charged disc with radius a and constant charge density d is give by

V = 2pi*d*(SQRT(R^2 +a^2) - R)

Show that for large R

V = pi*a^2*d / R



This is what I have done so far...

V = 2pi*d * (SQRT(R^2 * (1+ a^2/R^2)) - R)
V = 2pi*d*R (SQRT(1 + a^2/R^2) - R)
V = 2pi*d*R * ( 1 + (1/2)(a^2/R^2) + (1/2)(1/2 -1)/2! * (a^2/R^2)^2 + ... - R)

V = 2pi*d*R * (1/2) (1/2 + a^2/r^2 - (1/2)/2! * (a^2/R^2)^2 + ... - R)
V = pi*d* R ( 1/2 + a^2/R^2 - (1/4)*a^4/R^4 + ...-R)
V = pi * d * R/R^2 (1/2 + a^2 - (1/4)a^4/R^2 + ...-R)
V = pi * d *R (1/2 + a^2 - (1/4)a^4/R^2 + ...-R)
V = pi *d * a^2/R (1/2a^2 + 1 - (1/4) + ...-R)

Am I doing this correctly. How do I simplify what is in the parenthesis to get 1, which multiply to give me pi*d*a^2/R?
 
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  • #2
How did you get from your first line to your second? :smile:

Other than that, when you are using Taylor series to create approximations, you usually just handwave away all but the constant and linear terms (and perhaps the quadratic term if there is no linear term).
 
  • #3
Hi, I haven't read all your steps, but this is the way to do it,

You know that for small [itex]\epsilon[/itex], [itex]\sqrt{1+\epsilon}\approx 1+\frac{1}{2}\epsilon[/itex], right? Then you want to get V under this form. so take R² out of the root and set [itex]\epsilon =(a/R)^2[/itex].

In physics, when doing Taylor expansions, you almost always only want to keep just two terms. And 90% of taylor approximations you make are binomial. So just remember that (1+x)^n ~ 1+(x/n).
 

Related to How Does Taylor Series Approximation Determine Electric Potential Over Distance?

1. What is the Taylor series and how is it used in science?

The Taylor series is a mathematical representation of a function as an infinite sum of terms. It is used in science to approximate complex functions and make calculations easier.

2. What are the applications of Taylor series in physics?

Taylor series is used in physics to model physical phenomena, such as motion, using mathematical equations. It is also used in calculus to solve problems related to rates of change.

3. Can the Taylor series be used to find the value of a function at a specific point?

Yes, the Taylor series can be used to approximate the value of a function at a specific point by using a finite number of terms in the series. The more terms used, the more accurate the approximation will be.

4. How do you determine the number of terms needed in a Taylor series for a desired level of accuracy?

The number of terms needed in a Taylor series depends on the desired level of accuracy and the complexity of the function being approximated. Generally, the more terms used, the higher the accuracy will be.

5. Are there any limitations to using Taylor series?

Yes, Taylor series can only be used to approximate smooth and continuous functions. It may also not be accurate for functions with complicated behavior, such as oscillations or singularities.

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