Polarizibility, Electric Field & Force

[Blue Kangaroo]

Homework Statement

A neutral atom with known polarizability α is located at the origin. A point charge Q is situated on the y-axis a large distance d from the atom. (The atom therefore becomes polarized due to the electric field of the point charge.)
(a) Find the electric field due to the atom. Use the dipole approximation and write the answer using spherical coordinates.
(b) What is the force on the point charge due to the atom? Write the direction of the force using rectangular coordinates.
(c) What is the force on the atom due to the point charge?

Homework Equations

p=qd
F=(p⋅∇)E

The Attempt at a Solution

I've uploaded a picture of what I believe is the dipole approximation as well as a sketch of the problem, along with what I have so far for part a. However, I feel as if I've simplified it too much and made a mistake.

For anyone who may have trouble reading my answer, I got E(r)= (Q/(4πε₀r³))(r hat +2d)

Then I think the answers for (b) and (c) will be equal and opposite, but I'm not quite sure of how to get them yet.

 

[kuruman]

For anyone who may have trouble reading my answer, I got E(r)= (Q/(4πε0r3))(r hat +2d)

If this is the answer to part (a), it is incorrect. Starting with the dipole approximation of the electric field is correct but your expression is not a vector because of the $\hat{r}+2d$ thing which a mixture of a unit vector and a scalar distance. Furthermore, you need to work the polarizability $\alpha$ into the picture. What is its definition?

 

[Blue Kangaroo]

If this is the answer to part (a), it is incorrect. Starting with the dipole approximation of the electric field is correct but your expression is not a vector because of the $\hat{r}+2d$ thing which a mixture of a unit vector and a scalar distance. Furthermore, you need to work the polarizability $\alpha$ into the picture. What is its definition?

According to my book, polarization is the dipole moment per unit volume, in units of C/m². I believe polarizability is the ease with which something such as an atom is polarized. The polarizability is related to the dipole by p=αE..

 

[kuruman]

The polarizability is related to the dipole by p=αE.

Correct. What do you think is an expression for E in the above expression as related to this particular problem?

 

[Blue Kangaroo]

Correct. What do you think is an expression for E in the above expression as related to this particular problem?

Well p=qd=αE, so E=qd/α?

 

[kuruman]

Well p=qd=αE, so E=qd/α?

Nope. In the equation p = αE, E is the polarizing field in which the atom finds itself. Where does that field originate and what is it at the location of the atom?

 

[Blue Kangaroo]

Nope. In the equation p = αE, E is the polarizing field in which the atom finds itself. Where does that field originate and what is it at the location of the atom?

The field originates at the point charge Q and at the atom it is E=p/α I believe. Or am I just going in circles?

 

[kuruman]

Or am I just going in circles?

You have not understood the meaning of p = αE and you did not read #6 carefully enough. Here is the big picture. Please read what follows carefully to clear your understanding.

A neutral atom is placed on the y-axis. If that's all there is, the atom has no net charge, no dipole moment, therefore there is no electric field anywhere in space. Now an additional point charge Q is brought in and placed at the origin. This point charge Q generates an electric field everywhere in space. The electric field due to Q slightly separates spatially the atom's positive and negative charges thereby inducing a dipole moment p. How much of a dipole moment is induced depends on α. Now, the induced electric dipole contributes an additional electric field in that region of space so there are two fields, a monopole field from Q, E_Q and a dipole field from p, E_dip. The point charge exerts a force on the atom via E_Q and the atom exerts a force on the point charge via E_dip. It's these two forces that you have to find in parts (b) and (c) and if the spirit of Sir Isaac Newton walks with you, these two forces must come out as having equal magnitudes and opposite directions as required by his third law of motion.

So one more more time, what is the electric field E as in p = αE? Is it E_Q or E_dip?

 

[Blue Kangaroo]

You have not understood the meaning of p = αE and you did not read #6 carefully enough. Here is the big picture. Please read what follows carefully to clear your understanding.

A neutral atom is placed on the y-axis. If that's all there is, the atom has no net charge, no dipole moment, therefore there is no electric field anywhere in space. Now an additional point charge Q is brought in and placed at the origin. This point charge Q generates an electric field everywhere in space. The electric field due to Q slightly separates spatially the atom's positive and negative charges thereby inducing a dipole moment p. How much of a dipole moment is induced depends on α. Now, the induced electric dipole contributes an additional electric field in that region of space so there are two fields, a monopole field from Q, E_Q and a dipole field from p, E_dip. The point charge exerts a force on the atom via E_Q and the atom exerts a force on the point charge via E_dip. It's these two forces that you have to find in parts (b) and (c) and if the spirit of Sir Isaac Newton walks with you, these two forces must come out as having equal magnitudes and opposite directions as required by his third law of motion.

So one more more time, what is the electric field E as in p = αE? Is it E_Q or E_dip?

So then the E in the equation would be p=αE_Q.

 

[kuruman]

That's it. Now find an expression for E_Q and put it in the equation to find the dipole moment p. Make sure that your equations for E_Q and p are vectors written in terms of unit vectors.

 

[Blue Kangaroo]

That's it. Now find an expression for E_Q and put it in the equation to find the dipole moment p. Make sure that your equations for E_Q and p are vectors written in terms of unit vectors.

Would that expression be the dipole approximation? And both E_Q and p would point in -y hat?

 

[kuruman]

Would that expression be the dipole approximation?

It would not. Please read #8 one more time to understand what the picture is and what you need to do in terms of it. What is the origin of E_Q and what is an expression for it?

 

[Blue Kangaroo]

It would not. Please read #8 one more time to understand what the picture is and what you need to do in terms of it. What is the origin of E_Q and what is an expression for it?

Q is the origin of E_Q. Would E_Q=Qd/α, or p=Qd? I feel like there's something I'm just struggling with in this section that I haven't in previous ones.

 

[kuruman]

Look, Q at the origin is a point charge[/color]. What is the field of a point charge at distance d on the y-axis?

 

[Blue Kangaroo]

Look, Q at the origin is a point charge. What is the field of a point charge at distance d on the y-axis?

OK, E_Q=1(4πε₀)(Q/d²)yhat

 

[kuruman]

That's right. It would help if you used LaTeX to type your equations. It is easy to learn and use and you will find all the info you need if you click the LaTeX link next to the question mark near the bottom of the page and above the galley of thumbnail pictures.

Anyway, now you can calculate the induced dipole moment of the atom. See post #8 if you have forgotten how to do that.

 

[Blue Kangaroo]

That's right. It would help if you used LaTeX to type your equations. It is easy to learn and use and you will find all the info you need if you click the LaTeX link next to the question mark near the bottom of the page and above the galley of thumbnail pictures.

Anyway, now you can calculate the induced dipole moment of the atom. See post #8 if you have forgotten how to do that.

Thanks for the help. Now that I take a step back and look at it, I was making it harder than it is.

 

[kuruman]

... I was making it harder than it is.

You were. Can you finish the problem now?

 

[Blue Kangaroo]

You were. Can you finish the problem now?

Yep, I've already done it. The rest went quickly once I knew what I was doing. Thanks.