Electric Field due to dipole - solution check

  • #1
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Homework Statement


Hi guys, my exam is in four days and my tutor for the electromagnetic module is neither very active nor very competent, so I would like you guys to check my solution for this question. I am afraid I might have messed up some signs or some linear algebra.
2016q4.png


Homework Equations


All included in the solution.

The Attempt at a Solution


2016q4soln.jpg


Thank you very much!
 

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  • #2
kuruman
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Hi Ivan Hu and welcome to PF.

You should not double the potential energy ##U=-\vec p\cdot \vec E~## for the following reason: It costs no energy to bring in the first dipole from infinity because there is no electric field to begin with. When you bring in the second dipole, the work that you do to assemble the system is equal to the change in potential energy of the system from the initial value of zero to the final value ##-\vec p\cdot \vec E##.
 
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Thanks you kuruman for your answer. That makes sense. Is that the only mistake that you’ve picked up?
 
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kuruman
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Thanks you kuruman for your answer. That makes sense. Is that the only mistake that you’ve picked up?
Yes. I did not plug in the numbers. I assume you can double check these on your own.
 
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  • #5
Hi Kuruman,
Thank you for answering Ivan's question!
I'm actually working on the same past paper (Well, a previous one, with identical parameters) and we've produced quite different results that I'd like to verify.
My answer for ## \vec p \cdot \vec r ## was 0, given the two vectors are perpendicular, which produced a bit fat 0 for V, which basically propagated throughout the next few questions.
I can see that Ivan's done some stuff, and it looks like good math to me, but I don't know why my working would be wrong.
X6mRTvD.jpg
 

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  • #6
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My answer for →p⋅→rp→⋅r→ \vec p \cdot \vec r was 0
That's because the dot product is not zero. You can see from my working that the r vector has an x and y component (the variables x and y respectively). The p vector has no x component but a y component of 6.2⋅10-31.
 
  • #7
kuruman
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I can see that Ivan's done some stuff, and it looks like good math to me, but I don't know why my working would be wrong.
Please examine Ivan's working carefully. The y-component of the electric field ##\vec E## at the location of the dipole is the negative derivative of the electric potential ##V## with respect to ##y##. Your mistake is that you found the value of ##V## at a specific point, so naturally when you tried to take the derivative to find the electric field, you got zero. No surprise there because the derivative of a constant is always zero. To handle this, first you need to find a general expression for ##V##, then take the derivative to find a general expression for the electric field with ##y## as a variable and finally evaluate that expression at ##y=0##. That was Ivan's approach.
 
  • #8
Please examine Ivan's working carefully. Your mistake is that you found the value of ##V## at a specific point, so naturally when you tried to take the derivative to find the electric field, you got zero. To handle this, first you need to find a general expression for ##V##, then take the derivative to find a general expression for the electric field with ##y## as a variable and finally evaluate that expression at ##y=0##.
Ivans approach makes plenty of sense - I tried to cut corners and got my comeuppance :D
Thank you for your rigorous breakdown - I believe I have it now.
 

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