How Does the Electric Field Approximation Change When x is Much Smaller Than a?

Click For Summary

Homework Help Overview

The problem involves two positive charges positioned along the y-axis, with the objective of demonstrating how the electric field behaves when the distance x is much smaller than the distance a. The original poster has already derived an expression for the electric field and is now questioning the implications of the approximation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand why the electric field does not approach zero as x becomes very small, despite the presence of x in the numerator of the derived expression.
  • Some participants discuss the approximation of the denominator and the implications of ignoring terms when x is much smaller than a.
  • Questions arise regarding the manipulation of the denominator and the treatment of the numerator in the context of limits.

Discussion Status

Participants are actively engaging with the mathematical expressions and exploring the implications of approximations. There is a recognition of the need to clarify how small values affect both the numerator and denominator, but no consensus has been reached on the interpretation of the results.

Contextual Notes

There is an emphasis on the relative sizes of x and a, with discussions about the acceptable level of approximation and the potential errors introduced by ignoring certain terms. The original poster's confusion reflects a common challenge in understanding limits and approximations in physics problems.

charlies1902
Messages
162
Reaction score
0

Homework Statement


2 positive charges are placed on the why axis. one is at y=+a the other at y=-a. show that the e field where x is much smaller than a is approximately equal to (2kqx)/a^3. The 1st part of the question was to show that the e field = (2kqx/(x^2+a^2)^1.5) I did that part already so we can just use that equation.



The Attempt at a Solution


okay so, using the equation (2kqx/(x^2+a^2)^1.5) to do this part where x is much smaller than a means that x is approaching 0. so shouldn't the whole e field be approx = to 0 instead of like the equation they gave (2kqx)/a^3. where they only got rid of the x at the bottom and not the one at the top?
 
Physics news on Phys.org
Write the expression x2+a2 in the form a2[1+(x/a)2] If x is much smaller than a its square is even much smaller. For example, x/a = 0.01, then (x/a)2=0.0001. Ignoring it causes 0.01% error only. What you can call "small" depends on the accuracy of a. If it is given with 3 digits, the error of a is greater than the error the approximation brought in.

ehild
 
Last edited:
i don't understand how you got a[1+(x/a)^2] out of the denominator.
 
charlies1902 said:
i don't understand how you got a[1+(x/a)^2] out of the denominator.

My bad... I did not take the power 1.5 into account, and forgot a square. Now I edited my original post. The denominator becomes

[a2(1+(x/a)2]1.5=a3(1+(x/a)2)1.5,
and the relative error when ignoring x/a is 1.5 (x/a)2.

ehild
 
Last edited:
sorry but I'm still confused why that gives the answer. when you're doing the denominator, you're assuming x approaches 0 right? which gives you the denominator with only a in it. But the answer STILL has x in the numerator. shouldn't what was done in the denominator be also applied to the numerator which gives an answer of 0?
 
Yes there is x in the numerator. As it stands alone (nothing added) you can not compare it to anything and ignoring with respect to. x approaches zero, but is not zero, only very small with respect to a. Assume a=1 and evaluate x/[1+x2]1.5 for x=0.1, 0.01, 0.001. You will get what the approximation means.


ehild
 

Similar threads

Replies
4
Views
4K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
1K