Solve Dipole Problem: Help with E_y = kqd/r^3

[Saladsamurai]

!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??

 

[Saladsamurai]

Anyone??

 

[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.

 

[Saladsamurai]

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?

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.

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

 

[hage567]

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.

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

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

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).

It is to the 1/2 because it is the distance from each

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.

 

[Saladsamurai]

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.

Oh crap-ass! I do crap like this all the time. . . Thanks hage!