# Homework Help: Electric potential of a dipole moment

1. Mar 10, 2013

1. The problem statement, all variables and given/known data
The dipole moment of a water molecule is 6.29E-30Cm. What is the electric potential's magnitude 1.43nm from a water molecule along the axis of the dipole?

2. Relevant equations
p = qr

3. The attempt at a solution
I don't have values for q or r. I just know the magnitude of the dipole moment and the distance away from the molecule in which I have to figure out the electric potential. I don't know where to go from here.

And even if given q, how would I figure out V.

2. Mar 10, 2013

### ehild

Find the electric potential of two equal and opposite charges d distance apart symbolically, at distance x from the centre of the dipole, and then take into account that x>>d.

ehild

3. Mar 10, 2013

That doesn't really help me.

4. Mar 10, 2013

### ehild

You know the formula for the potential of a point charge? A point charge q is at x1=d/2, and an other one is at x2=-d/2. What is the potential at the point with coordinate x?

ehild

5. Oct 30, 2017

### TheExibo

Has anyone figured out how to do this question? I have it too but I can't find the procedure anywhere online for it.

6. Oct 30, 2017

### ehild

Do you know what is the electric potential of a point charge at distance D from it?
You have two charges, arranged along the x axis, both at distance d/2 from the origin. What are their potential at the point P, distance x from the origin?

7. Oct 30, 2017

### TheExibo

How do I determine what d/2 is?

8. Oct 31, 2017

### ehild

You do not need to determine it. The dipole momentum is given, and P=qd. At the end, you will find that the potential is proportional to qd, that is, the dipole moment.

9. Oct 31, 2017

### TheExibo

So I have P/(2q)=d/2 so far. I have also made the equations V=qk/(r-d/2) and V=qk/(r+d/2) and r is the distance from the origin to the point at which potential is measured.

What is next? I don't understand how to substitute them since d/2, V, and q are unknown.

10. Oct 31, 2017

### TheExibo

Nevermind. I have found that the equation V=kp/r^2 works, although I don't know how this equation was derived.

11. Oct 31, 2017

### ehild

The two charges have opposite signs. The positive charge causes V+=qk/(r-d/2) potential at distance r from the origin, and the negative charge contributes to the potential by V-=qk/(r+d/2). The potential at P is the sum of V+ and V-: $V(r)=qk\left(\frac{1}{r-d/2}-\frac{1}{r+d/2}\right)$. Bring the fractions to common denominator and use that d/r << 1.