# Electric Dipole

1. Apr 25, 2013

### universitypq98

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

It's me again..
I solved this question but i want to be sure about it,so what do you think about my solution bellow?

a dipole consists of a positive charge q at x=d/2 and a negative charge -q at x=-d/2 (d is the distance between the charges). What is the potential produced by the dipole at a point distance y from the center of the dipole(on an axis forming 90 degrees with the dipole)?
a)0
b)kqd/y^2
c)kqd/x^2
d)qd

2. Relevant equations

for a dipole :
V=K.q.d.cos(teta)/x^2

3. The attempt at a solution
so here we have this figure

http://img827.imageshack.us/img827/8777/123sei.jpg [Broken]

teta is equal to 90 no? cos(90)=0
si V=k.q.d.cos(teta)/y^2--->=0 because cos90=0

so A

True?
Evening!! :)

Last edited by a moderator: May 6, 2017
2. Apr 25, 2013

### Staff: Mentor

This formula is only useful for distances x >> d.

But you don't need it. What's the potential at some distance from a point charge? You have two such charges, what must their potentials add to?

3. Apr 25, 2013

### universitypq98

the potential at some distance from a point charge is :
V=k*q/d... :S
Ok i'm lost :/

4. Apr 25, 2013

### universitypq98

why x is not >> d here?
in fact i think in this exercice and by the figure i have drawn y=x
y can be anywhere on the y axix,so y is >>d
no? did i misunterstood something?

5. Apr 25, 2013

### Staff: Mentor

That's all you need. In this formula, "d" is the distance from the charge (not the dipole moment).

How does the potential of one charge compare to that of the other?

6. Apr 25, 2013

### Staff: Mentor

As you state, y can be anywhere, so you cannot assume that y >> d. (In your diagram, y < d!)

7. Apr 25, 2013

### universitypq98

mmmm... but i don't see if i use V=kq/d,how to arrive to one of the 4 answers proposed.. :/
let's try to do it by elimination
b,c are related to the equation i have above which is not applicable here cause y isn't always >> d
d surely not..
so a by elimination ! :P haha
my mind gonna blow tonight!
can't we assume that y>>d and work it out?
i don't know but in class we always used this formula in such cases,that's why i am insisting on it..
Thanks a lot! Doc Al! I do really appreciate your help!
have a nice and peaceful evening!:)

8. Apr 25, 2013

### Staff: Mentor

No need to even use elimination. Since the distances are the same for each charge, and the charges are equal and opposite, you know that the potentials are equal and opposite. Which means that they must add to zero.

You can just plug into the equation and since θ = 90° you'll get "the right answer". But only by luck! That equation just does not apply.

9. Apr 25, 2013

### universitypq98

oh! yeah that's true!! the charges are equals and opposite.. i understood it correctly now! :P
haha thanks a lot!! you have to be a teacher :)
have a nice week-end in advanced! :)