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Maclaurin Series - Potential due to Dipole (Isaac Physics)
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As it can be assumed that z >> a, does this allow me to remove the a/2 terms from the initial equation, or would this not give an accurate enough approximation?
As it can be assumed that z >> a, does this allow me to remove the a/2 terms from the initial equation, or would this not give an accurate enough approximation?
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#2
(Original post by domm1)
https://isaacphysics.org/questions/m...9-7fa11adbaaa3
As it can be assumed that z >> a, does this allow me to remove the a/2 terms from the initial equation, or would this not give an accurate enough approximation?
https://isaacphysics.org/questions/m...9-7fa11adbaaa3
As it can be assumed that z >> a, does this allow me to remove the a/2 terms from the initial equation, or would this not give an accurate enough approximation?
Hopefully it's obvious you can't remove the a/2 terms without ending up with 0.
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(Original post by DFranklin)
"obtain the first non-zero term in the Maclaurin expansion of V(z)" is the key phrase here.
Hopefully it's obvious you can't remove the a/2 terms without ending up with 0.
"obtain the first non-zero term in the Maclaurin expansion of V(z)" is the key phrase here.
Hopefully it's obvious you can't remove the a/2 terms without ending up with 0.
I initially sorted out the stuff in the brackets, simplified and expanded out and got an expression for V(z) of the form u/v, so applied the quotient rule to find the first derivative, which came out to be not very nice at all. Am I doing the correct thing to carry on finding further derivatives, if not, what should I of done instead?
I am guessing that the assumption z >> a only comes into play on the final expression which helps to simplify it down? Am I wrong in saying this?
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#4
Why did you multiply out? It's easy to differentiate things that are added together and hard to differentiate products/quotients, so it would have been *much* easier if you left the original expression alone.
Getting the first derivative will let you get the z term in the expansion. If that's non-zero you're done, otherwise you need to carry on until you get a non-zero term.
If you start with the original expression it's easy to find multiple derivatives - this shouldn't be an issue.
Getting the first derivative will let you get the z term in the expansion. If that's non-zero you're done, otherwise you need to carry on until you get a non-zero term.
If you start with the original expression it's easy to find multiple derivatives - this shouldn't be an issue.
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(Original post by DFranklin)
Why did you multiply out? It's easy to differentiate things that are added together and hard to differentiate products/quotients, so it would have been *much* easier if you left the original expression alone.
Getting the first derivative will let you get the z term in the expansion. If that's non-zero you're done, otherwise you need to carry on until you get a non-zero term.
If you start with the original expression it's easy to find multiple derivatives - this shouldn't be an issue.
Why did you multiply out? It's easy to differentiate things that are added together and hard to differentiate products/quotients, so it would have been *much* easier if you left the original expression alone.
Getting the first derivative will let you get the z term in the expansion. If that's non-zero you're done, otherwise you need to carry on until you get a non-zero term.
If you start with the original expression it's easy to find multiple derivatives - this shouldn't be an issue.
Thanks anyways
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