Polarizibility, Electric Field & Force

In summary: This was a conversation between two people discussing the dipole approximation and force between a neutral atom and a point charge. In summary, the dipole approximation is used to find the electric field and force between a neutral atom and point charge. The polarizability, α, is related to the dipole moment, p, by p=αE. The electric field, E, is composed of a monopole field from the point charge, EQ, and a dipole field from the induced dipole moment, Edip. The forces exerted by these fields on the atom and the point charge should be equal in magnitude and opposite in direction, according to Newton's third law of motion.
  • #1
Blue Kangaroo
41
1

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.
 

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  • #2
Blue Kangaroo said:
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?
 
  • #3
kuruman said:
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..
 
  • #4
Blue Kangaroo said:
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?
 
  • #5
kuruman said:
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?
 
  • #6
Blue Kangaroo said:
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?
 
  • #7
kuruman said:
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?
 
  • #8
Blue Kangaroo said:
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, EQ and a dipole field from p, Edip. The point charge exerts a force on the atom via EQ and the atom exerts a force on the point charge via Edip. 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 EQ or Edip?
 
  • #9
kuruman said:
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, EQ and a dipole field from p, Edip. The point charge exerts a force on the atom via EQ and the atom exerts a force on the point charge via Edip. 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 EQ or Edip?

So then the E in the equation would be p=αEQ.
 
  • #10
That's it. Now find an expression for EQ and put it in the equation to find the dipole moment p. Make sure that your equations for EQ and p are vectors written in terms of unit vectors.
 
  • #11
kuruman said:
That's it. Now find an expression for EQ and put it in the equation to find the dipole moment p. Make sure that your equations for EQ and p are vectors written in terms of unit vectors.

Would that expression be the dipole approximation? And both EQ and p would point in -y hat?
 
  • #12
Blue Kangaroo said:
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 EQ and what is an expression for it?
 
  • #13
kuruman said:
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 EQ and what is an expression for it?

Q is the origin of EQ. Would EQ=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.
 
  • #14
Look, Q at the origin is a point charge. What is the field of a point charge at distance d on the y-axis?
 
  • #15
kuruman said:
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, EQ=1(4πε0)(Q/d2)yhat
 
  • #16
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.
 
  • #17
kuruman said:
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.
 
  • #18
Blue Kangaroo said:
... I was making it harder than it is.
You were. Can you finish the problem now?
 
  • #19
kuruman said:
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.
 

What is polarizibility?

Polarizibility is a measure of how easily a material can be polarized in response to an electric field. It is a physical property that describes the ability of a material to form an induced dipole moment.

How is electric field defined?

An electric field is a region in space where a charged particle experiences a force. It is defined as the force per unit charge at a given point in space.

What is the relationship between electric field and force?

The force experienced by a charged particle in an electric field is directly proportional to the strength of the electric field. This relationship is described by the equation F = qE, where F is the force, q is the charge of the particle, and E is the electric field.

How does polarizibility affect the strength of an electric field?

The greater the polarizibility of a material, the more easily it can be polarized by an electric field. This means that a material with high polarizibility will have a stronger induced dipole moment and will experience a stronger electric field.

What factors affect the polarizibility of a material?

The polarizibility of a material is affected by its molecular structure, the strength of the electric field, and the presence of other nearby charged particles. In general, materials with larger and more polarizable molecules will have higher polarizibility.

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