Dipole moment, electric potential

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Cocoleia
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Homework Statement


I am given this picture
upload_2017-1-29_15-56-52.png

and I know that |q1|=2nC, |q2|=5nC, d=1mm
I need to first find the total dipole moment of the system. Then I need to find an equation that represents the electric potential due to this net dipole moment for all (everywhere)

Homework Equations


p=qd

The Attempt at a Solution


I saw a formula which was p=qd (vectors)
so I thought I would separate it into two different dipole moments, one for q1 and one for q2, and then add them together. For p1 I got 2x10^-12 in the y direction, and for p2 I got 5x10^-12 in the x direction. Would this be correct?

Also, would I be using this formula for the potential?
V = kpcosθ/r^2
 
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haruspex said:
What happened to the 3/2?
What is the direction of a dipole moment vector?
Ok, so 7.5x10^-12 in the x direction. And for the potential?
 
haruspex said:
Assuming a certain definition of θ, yes.
I'm not quite sure what θ would be in this case
 
haruspex said:
Maybe better to put the answer in vector form. If ##\vec p## is the dipole at the origin and ##\vec r## is a point in space, what is the potential at ##\vec r##?
I don't know what the formula for potential would be,except I had V=kpcosθ/r^2 in my notes but I don't know when to use it
 
Cocoleia said:
I don't know what the formula for potential would be,except I had V=kpcosθ/r^2 in my notes but I don't know when to use it
That makes for not terribly useful notes!
It arises from the dot product of the two vectors.
 
haruspex said:
That makes for not terribly useful notes!
It arises from the dot product of the two vectors.
So the dot product of p and r?
 
haruspex said:
Yes. Can you write the whole expression?
I'm not really sure what you want, I assume r will have an x and y component, so r cosθ and r sinθ
so then
(7.5x10^-12)(rcosθ)+(2x10^-12)(rsinθ)

Sorry I don't really have any idea what I am doing because I have no notes/ teaching on this stuff.
 
haruspex said:
Yes.
What will the r hat be ?
 
haruspex said:
Putting a hat on a vector is a standard notation meaning the unit vector in that direction. So for any vector ##\vec x##, ##\hat x=\frac{\vec x}{|\vec x|}##
Here they don't specify a length or direction for r, it just says "everywhere"
 
Cocoleia said:
Here they don't specify a length or direction for r, it just says "everywhere"
The potential is necessarily a function of position. You cannot say what it is "everywhere" without involving a variable to specify position.