Homework Help: Aproximation question

1. Jun 16, 2009

transgalactic

why the "a" disappears
i cant understand the process of the approximation .

if r1 goes to be equal r2 =r1=r
that "a " should be 0 .

http://i42.tinypic.com/119ndqt.gif

2. Jun 16, 2009

Cyosis

Yes if r2=r1 then a=0, exactly. But we're making an approximation. If r is way bigger than a then r2 is almost equal to r1, because r2-r1 is a very small part of r2. Therefore you can approximate r1r2 as r^2.

3. Jun 16, 2009

transgalactic

but why "a" disappears
?

i agree that if r>>a that r1r2=r^2
because the are close to each other
but there a multiplication by 'a'
i cant understand mathematically
why "a" disappears
?

4. Jun 16, 2009

Cyosis

What is the definition of a dipole moment?

5. Jun 16, 2009

transgalactic

electric dipole moment is p=qd

where p is the moment created by the charges separated by a distance d. (on each end of the first body). It's an induced dipole moment.

6. Jun 16, 2009

Cyosis

Yes and what is the electrical dipole moment of this particular configuration?

7. Jun 16, 2009

transgalactic

"electric field created by two point charges +q and and -q, separated by a (small) distance d.

If you reduce the distance and increase the charge so that qd remains constant, the field stays approximately the same, except for the region close to the charges.

A dipole is the limiting case of this, as d goes to zero and q goes to infinity but qd remains finite."

in my case how qd is a charge of one ball q
d is 'a'
and if there multiplication stays the same
then the field has the value
still it doesnt explain how 'a' dissapears

8. Jun 16, 2009

Cyosis

You didn't answer my question, you just cited a book. I want you to calculate the electrical dipole moment for the configuration in your problem. Can you do that?

9. Jun 16, 2009

transgalactic

p=aq

10. Jun 16, 2009

Cyosis

Almost, but not quite. Can you describe the configuration quantitatively to me using terms such as positive charge = ? ,negative charge =? separation distance =?

11. Jun 16, 2009

transgalactic

positive charge = +q
negative charge =-q
separation distance =a

12. Jun 16, 2009

Cyosis

Nope the separation distance is not a, I suggest you take a good look at the picture. After that calculate the correct electrical dipole moment.

13. Jun 16, 2009

transgalactic

separation distance =2a

14. Jun 16, 2009

Cyosis

I would appreciate if you would reply to my entire post. Calculate the correct electrical dipole moment now that you have found the correct separation distance.

15. Jun 16, 2009

transgalactic

p=q2a

16. Jun 16, 2009

Cyosis

Yes can you see now why the a "disappears"?

17. Jun 16, 2009

transgalactic

lol its being substituted by dipole moment sign
thanks

18. Jun 16, 2009

Cyosis

Indeed, I am glad you finally see it!

19. Jun 16, 2009

transgalactic

i cant understand why they change the drawing in that way.
i know that if we go far far away than our dipole will look like a single charge.
and there will not be any r1 and r2 only r.
so i cant see why they change the drawing in such way
?

20. Jun 16, 2009

Cyosis

Because the drawing represents a real physical dipole. They then show you how to calculate it when r>>a by using an approximation. Their intend most likely is that you realise the dipole potential is an approximation for a physical dipole. The dipole you're talking about is a mathematical dipole in the limit a->0, they don't exist in reality.

21. Jun 16, 2009

transgalactic

the distance between r2 an r1 lines
is not what they are calculating it the other side of the triangle (the blue line)
why they take the cosine

http://i39.tinypic.com/90vvc9.jpg

22. Jun 17, 2009

transgalactic

why they take this side
and not the other
??

23. Jun 17, 2009

Staff: Mentor

I'm not sure what your question is. The page you scanned shows that
$$V \approx \frac{k_e q}{r_1 r_2}2a cos \theta \approx \frac{k_e p cos \theta}{r^2}$$

All they have done is replace q*2a with p, and $r_1 r_2$ with $r^2$.

Last edited: Jun 17, 2009
24. Jun 17, 2009

transgalactic

but they say that $$cos \theta =r_2 - r_1$$

i think that the difference from the lines should be sine of this angle
the shortest way between two lines in the perpendicular lines between them
??

25. Jun 17, 2009

Cyosis

You should really review your trigonometry. r2-r1 is the adjacent side of the angle theta and 2a is the hypotenuse.

$$\cos \theta=\frac{adjacent side}{hypotenuse}=\frac{r_2-r_1}{2a}$$