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A physical electric dipole consists of two equal and opposite charges (+/-)q

separated by a distance d. Find the approximate potential at points far

from the dipole.

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- Thread starter asdf1
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- #1

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A physical electric dipole consists of two equal and opposite charges (+/-)q

separated by a distance d. Find the approximate potential at points far

from the dipole.

- #2

- #3

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find the potential at a far away point from the first charge. then find the potential at the same point for the second charge. add'em.

then you use a ton of approximations to get a neat solution. iirc correctly, be prepared to use taylor expansions... and then you can throw away all but a few terms.

- #4

mukundpa

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to it. potential due to perpendicular component is zero and the point is on the axis of the other component. You can use the formula for axil position.

- #5

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opps~

new user...

thank you!

but i'm not too clear on how to make the approximations?

new user...

thank you!

but i'm not too clear on how to make the approximations?

- #6

mukundpa

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-q.....M.......q....................................P

*.......d........* .

.........< ---------------r-------------->

If the point is at distance r from mid point of the dipole along the axis of the depole then the potential is

V = kq/(r - 0.5d) - kq/(r + 0.5d) = kqd/(r^2 - 0.25d^2)

then what if r >> d

*.......d........* .

.........< ---------------r-------------->

If the point is at distance r from mid point of the dipole along the axis of the depole then the potential is

V = kq/(r - 0.5d) - kq/(r + 0.5d) = kqd/(r^2 - 0.25d^2)

then what if r >> d

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- #7

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i see~

that clears a lot of things up!

thank you~

so if r>>d, then it V would be approximately 0?

that clears a lot of things up!

thank you~

so if r>>d, then it V would be approximately 0?

- #8

mukundpa

Homework Helper

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say r =0.20 m and d= 0.001 m then r^2 = 0.04 m^2 and 0.25d^2 = 0.00000025 m^2, so as compared to r^2 we can neglect0.25 d^2, but can't say that the potential at 0.20 m is zeroo. Right?

- #9

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[tex]r-0.5d\approx r[/tex] and [tex]r+0.5d\approx r[/tex]

Then, it seems that the equation becomes

[tex]V= \frac{kq}{r-0.5d}-\frac{kq}{r+0.5d}\approx \frac{kq}{r}-\frac{kq}{r}=0[/tex]

- #10

mukundpa

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- #11

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This is a little video explaining the approximation techniques that you could use here. Taylor series and binomial expansion.

http://physicsstream.ucsd.edu/review/taylor.html [Broken]

http://physicsstream.ucsd.edu/review/taylor.html [Broken]

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