Electric potential of a dipole moment

[Leporad]

Homework Statement

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?

Homework Equations

p = qr

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.

 

[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

 

[Leporad]

That doesn't really help me.

 

[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

 

[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.

 

[ehild]

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

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?

 

[TheExibo]

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? View attachment 214070

How do I determine what d/2 is?

 

[ehild]

How do I determine what d/2 is?

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.

 

[TheExibo]

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.

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.

 

[TheExibo]

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

 

[ehild]

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.

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.