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Griffiths 4.7

  • Thread starter ehrenfest
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1,996
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1. Homework Statement
This question refers to Griffiths E and M book.
See the solution here http://www.physics.gatech.edu/academics/Classes/fall2005/3122/hw9solution.pdf

I don't understand why they say the torque we exert is clockwise (I assume that means in the minus z direction).

2. Homework Equations



3. The Attempt at a Solution
 
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Answers and Replies

1,996
1
anyone?
 
1,860
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Solution aside, how would you approach the problem? If your answer is I don't know then try some of the problems before and see if you can get those ones, and if you can't get those then look over the section again.
 
1,996
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OK. Well what I did was take the negative gradient of eqn 4.6 and showed that it was equal to eqn 4.5 using product rule 4. Then I was checking my answer with this solution and it was completely different, so I was trying to understand it. Can you read and help me? I think I am missing something obvious (again).
 
1,860
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Hmm, in your way after you crank through it all you find that [itex]\mathbf{E} \times(\nabla \times \mathbf{p})=-(\mathbf{E}\cdot\nabla)\mathbf{p}[/itex]? If that is true then you are okay, and it seems like an acceptable solution.

As far as the other solution goes, it's probably the picture that is screwing you up, in the picture zhat is pointing out of the page, and the electric field wants to torque the the dipole to line up with it, so it wants to torque the dipole into the zhat direction. "We" are rotating the dipole clockwise, the way it wants to go, so we are doing negative work. This is all really justification though. We could have also gone the other way, and done positive work.

What really matters is the integral they show. Since an angle of zero is the z axis, and we start on the x axis, we integrate from that point to wherever we go along the rotation.

Make sense?
 
1,996
1
Hmm, in your way after you crank through it all you find that [itex]\mathbf{E} \times(\nabla \times \mathbf{p})=-(\mathbf{E}\cdot\nabla)\mathbf{p}[/itex]? If that is true then you are okay, and it seems like an acceptable solution.
Both sides of that equation are clearly 0 because p is a constant.
 
1,996
1
Hmm, in your way after you crank through it all you find that [itex]\mathbf{E} \times(\nabla \times \mathbf{p})=-(\mathbf{E}\cdot\nabla)\mathbf{p}[/itex]? If that is true then you are okay, and it seems like an acceptable solution.

As far as the other solution goes, it's probably the picture that is screwing you up, in the picture zhat is pointing out of the page, and the electric field wants to torque the the dipole to line up with it, so it wants to torque the dipole into the zhat direction. "We" are rotating the dipole clockwise, the way it wants to go, so we are doing negative work.
You mean the way it DOES NOT want to go?

I guess I don't really understand why "we" are doing this.

Also, I don't understand the integral (i.e. the equation U = ). Is that how you calculate the work done to a rigid body by rotation? That is probably something I can find in my mechanics book.
 
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1,996
1
anyone?
 
1,996
1
anyone?
 
511
1
The solution is post possibly is not correct.

Start from a general expression for U(r)
 
1,996
1
The solution is post possibly is not correct.
Can you say why it is not correct?
 
1,996
1
please?
 
1,996
1
OK. In general, how to you calculate the work done to an object rotating about its axis in a force field?
 
nrqed
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Homework Helper
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1. Homework Statement
This question refers to Griffiths E and M book.
See the solution here http://www.physics.gatech.edu/academics/Classes/fall2005/3122/hw9solution.pdf

I don't understand why they say the torque we exert is clockwise (I assume that means in the minus z direction).

2. Homework Equations



3. The Attempt at a Solution

the torque exerted by the E field is p cross E so it is in the positive z direction. Therefore if you slowly bring the dipole from its initial position to its final position at constant speed, you must exert a torque in the negative z direction. That's all there is to it as far as I can see.
 
1,996
1
yes that makes sense. But can you answer my last question? How co you calculate the work?
 

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