# 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.

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