# Dipole problem help

1. Feb 18, 2008

!!!dipole!!!

Okay. So I am trying to do this kind of the old fashioned way. I broke the E's down into x and y component and am adding them together component-wise.

I have found that the x components cancel out. Now I am trying to widdle down the form I have for the y components. As of now I have that:

$$E_y=k[\frac{-|q_1|-|q_2|}{[(d/2)^2+r^2]^{1/2}}*\frac{d/2}{[(d/2)^2+r^2]^{1/2}}$$

The text has it in the form: $$kqd/r^3$$ which I am having a hard time getting.

How do I get rid of the negative sign? AND how do I make use of the fact r>>d??

2. Feb 18, 2008

Anyone??

3. Feb 18, 2008

### hage567

q1 and q2 are equal.

In your first term you have the denominator to the 1/2. That shouldn't be there. Check your work.

I'm not sure what to say about the negative sign because I'm not sure how you came up with it.

If r>>d then you can ignore your (d/2)^2 term in the denominator because it will be negligible compared to r^2.

Last edited: Feb 18, 2008
4. Feb 18, 2008

I got the negative sign when I drew the free body diagram of my test charge. Since q_o is +, both fields are in the negative y direction.

Or should I just be considering their magnitude of 2q in this case and forget which direction they point?

It is to the 1/2 because it is the distance from each q to p $c^2=b^2+a^2$
Thanks!

Last edited: Feb 18, 2008
5. Feb 18, 2008

### hage567

OK, that's just a matter of convention then. You could have made it positive if you wanted.

Since the y components are in the same direction (and equal magnitude), you can just say that E = 2*Ey (where Ey is the y component from one of the charges).

The general equation is $$E = \frac{kq}{x^2}$$ right?

So you found the length of the hypotenuse using Pythagorus, so $$x^2 = (\frac{d}{2})^2 + r^2$$

So there is no need to take the square root to get just x, since you need x^2 in your equation anyway. Does that make sense? Just the first term in your denominator is wrong, the second one is right.

6. Feb 18, 2008