Interaction force, two dipoles

[mathman44]

Homework Statement

A dipole p is directed in the positive z direction at the origin.

The force on a dipole p in an electric field E is [tex]F=(p\cdot\nabla)E[/tex]

If a second dipole is placed at (r, [tex]\theta[/tex]), by what factor must r increase – in terms
of p, [tex]\theta[/tex] and r – so that the net force between the two dipoles decreases by a factor of 64?

---

I don't know how to start this... any hints please? This isn't homework, just a practice test question.

 

[gabbagabbahey]

Homework Statement

A dipole p is directed in the positive z direction at the origin.

The force on a dipole p in an electric field E is [tex]F=(p\cdot\nabla)E[/tex]

If a second dipole is placed at (r, [tex]\theta[/tex]), by what factor must r increase – in terms
of p, [tex]\theta[/tex] and r – so that the net force between the two dipoles decreases by a factor of 64?

---

I don't know how to start this... any hints please? This isn't homework, just a practice test question.

Well, you have a situation where one dipole is exerting a force on another...you already know how to determine the force exerted on a dipole from an external electric field, so...what is the electric field of a dipole? What force does that field exert on a dipole a distance [itex]r[/itex] away from it (do not assume that the dipoles are aligned)?

 

[mathman44]

The electric field of a dipole is

[tex]E=\frac{p}{4\pi\epsilon{r^3}}(2cos\theta\hat{\theta}+sin\theta\hat{r})[/tex]

If I break p apart into [tex]p_r[/tex] and [tex]p_\theta[/tex], dot it with del, then multiply by E, I get a mess of an answer.

 

[mathman44]

This is what I got, by the way:

[tex]F_(r,\theta)=-3\frac{cos\theta}{4\pi\epsilon{r^4}}(2p_\theta+p_r)[/tex]

Any help please?

 

[paranormal]

Sure, let's break down the problem step by step. First, we know that the force between two dipoles is given by F = (p1 · ∇)E2, where p1 is the dipole moment of the first dipole and E2 is the electric field created by the second dipole. Since we are trying to decrease the net force between the two dipoles by a factor of 64, we can write this as:

F' = (p1 · ∇)E2' = F/64

where E2' is the new electric field created by the second dipole and F is the original force. Now, we can use the formula for the electric field created by a dipole, E = (1/4πε0) (3p · r̂ - r · p), where p is the dipole moment and r̂ is the unit vector pointing from the dipole to the point at which the electric field is being measured. Plugging this into our equation above, we get:

F' = (p1 · ∇) [(1/4πε0) (3p2 · r̂ - r · p2)] = F/64

Next, we can expand the dot product using the product rule, (p1 · ∇) [f(r)] = ∇(p1 · f(r)) - f(r) (∇ · p1). Since p1 is directed in the positive z direction at the origin, we can simplify this to:

F' = (∂/∂z) [(1/4πε0) (3p2z - rz · p2)] = F/64

where p2z is the z-component of the dipole moment of the second dipole and rz is the distance from the origin to the second dipole. Now, we can solve for rz by rearranging the equation:

rz = 3p2z - 64F/∂E/∂z

where ∂E/∂z is the derivative of the electric field with respect to z. Finally, we can substitute in the given values for p, θ, and r to get our final answer:

rz = 3p2 cos θ - 64F/∂E/∂z

So, to decrease the net force between the two dipoles by a factor of