# Dipole pair electric field

1. Jun 4, 2008

### scholio

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

assume that two dipoles are placed back -t0 - back along axis as shown:

((-) a (+))(( +) a (-)) ------------------x axis

r is oriented between ))(( but above, a distance such that r>>a

compute the strength of the electric field on the axis of the pair at a distance r away from the the center of the dipole pair. use the approximation r>>a to simply your expression. each charge can be assumed to be 1 microcoulomb charge with the sign shown in the diagram

2. Relevant equations

electric potential V = kq/r where k is 9*10^9 constant, q is charge in coulombs, r is distance the charge is from the point r

electric field E = 2kP/r^3 where P is the dipole moment, given as 10^-8 C-m -->derivative of electric potential eq

3. The attempt at a solution

i am not sure where to draw the diagram as shown with lines connecting each charge (4 of them) and using the V equation to sum them up, but r is not given a numeric value

then do i do the derivative of the summation of V?

or do i just use the electric field equation, but how do i involve all four charges, and what do i put in for r in the denominator?

any help appreciated

cheers...

2. Jun 4, 2008

### scholio

sorry for double posting, but i think i may have the wrong electric potential V equation.

should it be:

V =kpcosine(theta)/r^2 where p = 2qa, q = charge, a is distance b/w charges of dipole.

but since the problem involves two dipoles, do i multiply V by 2? how about the derivative (electric field E)?

but in order to find the electric field, i would need to take the derivative, so i should get this:

E = 2kpcosine(theta)/r^3 where r is the distance between pt r and the dipole(s)

do i multiply E by two because the problem involves two dipoles? when i put in p = 2qa with q = 1 microcoulomb and r>>a i got

E = 72000a(cosine(theta))/r^3 newtons/coulomb

any closer?...

Last edited: Jun 4, 2008
3. Jun 4, 2008

### alphysicist

Hi scholio,

I would use either

$$E=\frac{k q}{r^2}$$

or

$$V= \frac{kq}{r}$$

and apply it to all four charges. It doesn't make much difference which one you use since everything is along the same line.

Calculate the exact value of the total field or total potential. You'll have parts like (r+a) and (r-a) in some of the denominators. Then you can use the fact that r >> a to find an approximation for those denominators. After a few cancellations you should get the answer. What do you get?

4. Jan 4, 2012

### jilani

yes we can easily calculate it but my question is what we will do if we have to calculate electric field of a hexapole .....can anyone help me.plz