Dipole magnitude and direction (21.67)

1. Oct 4, 2015

Calpalned

1. The problem statement, all variables and given/known data

2. Relevant equations
Total electric field = Electric field caused by Q_1 + electric field caused by Q_2
$\vec{e} = \vec{e_1}+\vec{e_2}$
Dipole moment = $P = Ql$
3. The attempt at a solution

**For Part A**
$\vec{E} = \vec{E_1}+\vec{E_2}$
The charges are opposite so it will be a difference, not a sum.
$\frac{KQ}{(R+\frac{l}{2})^2} - \frac{KQ}{(R-\frac{l}{2})^2}$
Define $A=(R+\frac{l}{2})^2$ and $B=(R-\frac{l}{2})^2$
Then $\frac{KQ}{A} - \frac{KQ}{B}$
$\frac{BKQ-KQA}{AB}$
$\vec{E}=\frac{KQ(B-A)}{AB}$
After simplification, $AB=R^4 + \frac {(R^2) (l^2)}{2}-R^2 l^2 + \frac{l^4}{4}$ and $B-A=-2Rl$
I didn't get the answer above in part A. Maybe my algebra is wrong or something else?

**Now for Part B**
I feel that $\vec{E}$ should be in the opposite direction of the dipole moment vector (DMP). In the picture above, if the neg charge is on the left and the pos on the right, then the dmp points to the left, but the field at the point points to the right, is this correct? Something similar will happen if we switch the positions of the two charges?

THANK you very much!

2. Oct 5, 2015

BvU

Must say I can't understand the problem statement. You have the correct answers in front of you ?

For R >> l you can ignore terms like R2l2 aside R4.

No, it points to the right.

3. Oct 5, 2015

Calpalned

Yes you are right. I read the textbook and the Dipole Moment points from the neg charge to the positive one.
So I understand part B. But what about part A?

4. Oct 5, 2015

BvU

Well, what do you get if you only keep the R4 in the denominator of $| \vec{E}\, | =\frac{KQ(B-A)}{AB}$ ?

 by the way, $(r+l/2)^2(r-l/2)^2 = (r^2-l^2/4)^2 = r^4 -r^2l^2/2+l^4/4$ 2 never mind, that's just what you wrote....