Limit as x becomes much greater than R ie x >> R

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KindofSlow

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Limit as x becomes much greater than R ie x >> R

What does this reduce to when x >> R?

((x^2 + R^2)^(1/2)) – x

I factored out the x^2 to get
((x^2)*(1 + (R^2/x^2)))^(1/2)) – x

I cannot figure out the step(s) that now get to the correct answer of
x*(1 + (R^2/2x^2)) – x

Thank you
 
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KindofSlow said:
What does this reduce to when x >> R?

((x^2 + R^2)^(1/2)) – x

I factored out the x^2 to get
((x^2)*(1 + (R^2/x^2)))^(1/2)) – x

I cannot figure out the step(s) that now get to the correct answer of
x*(1 + (R^2/2x^2)) – x

Thank you
Click to expand...

You are headed in the right direction and that last equation is tantalizing close but incorrect (a typo perhaps?). Lets start with the
((x^2)*(1 + (R^2/x^2)))^(1/2)) – x = [ x2 * ( 1 + (R/x)2)]1/2 - x
= x [ ( 1 + (R/x)2) ]1/2 - x
= x [ (1 + a)1/2 -1 ]
where
a = (R/x)2
Since R << x we have (R/x)2 < R/x << 1 or a << 1. What is the expansion of (1 + a)1/2 when a is much less that 1?

Edit: Sorry, I miss-read the problem. I though that you were just giving another step in the process insteard of the final answer. To get that final answer, we have
(1+a)1/2 ~ 1 + (1/2) a - (1/8) a2 + ...
or, neglecting the a2 term, the answer is, substituting back in for a,
x [ 1 + (1/2) (R/x)2 ] - x
which is what you have. If you wanted to go further, we note that we have, adding back in that a2 term,
x + x [(1/2) (R/x)2 ( 1 - (1/4) a + ...)] - x = \(\displaystyle \frac{R^2}{2 x}\) [ 1 - (1/4) a + ...]
 
Last edited:
Apologies for the non standard.
It is from a physics problem calculating the potential (voltage) at a point x distance from a uniformly charged disc of radius R.
The first equation is the answer to the problem.
Then as x becomes much greater than R, it reduces down to where the disc is like a point charge.
More apologies if this still doesn't make sense.
Thank you for your help.
 
Hi Everyone,
I got some additional help and just wanted to share as I found this very helpful.

Multiply by 1 in the form of conjugate over conjugate so

((x^2 + R^2)^(1/2)) – x times ((x^2 + R^2)^(1/2)) + x
((x^2 + R^2)^(1/2)) + x
The result of this is

(x^2 + R^2) – x^2
((x^2 + R^2)^(1/2)) + x

As x >> R, the numerator reduces to R^2 and the denominator reduces to 2x.
Which is the same thing as this x*(1 + (R^2/2x^2)) – x

Thank you everyone.
 
I confess I did not realize they were both the same until you just told me.

Thank you HallsofIvy and pka and Ishuda.
 
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