Dipole magnitude and direction (21.67)

[Calpalned]

Homework Statement

Homework 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 = $P = Ql$

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!

 

[BvU]

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

I didn't get the answer above in part A

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

if the neg charge is on the left and the pos on the right, then the dmp points to the left

No, it points to the right.

 

[Calpalned]

No, it points to the right.

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?

 

[BvU]

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

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