Electric Field due to dipole - solution check

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SUMMARY

The forum discussion focuses on the calculation of the electric field due to a dipole, specifically addressing common mistakes in applying the potential energy formula and the dot product in vector analysis. A key error identified is the incorrect assumption that the dot product of two perpendicular vectors results in zero, which led to incorrect calculations for the electric potential. The correct approach involves deriving a general expression for the electric potential before evaluating it, as demonstrated by user Ivan Hu. This method ensures accurate results when calculating the electric field.

PREREQUISITES
  • Understanding of electric dipoles and their potential energy, specifically the formula ##U=-\vec p\cdot \vec E##.
  • Knowledge of vector operations, particularly the dot product and its implications in physics.
  • Familiarity with electric potential and its relationship to electric fields.
  • Ability to perform calculus operations, including differentiation of functions.
NEXT STEPS
  • Study the derivation of electric potential due to dipoles in detail.
  • Learn how to calculate electric fields from potential functions using differentiation.
  • Review vector calculus, focusing on dot products and their physical interpretations.
  • Explore common pitfalls in electromagnetic theory to avoid similar mistakes in future calculations.
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Students preparing for exams in electromagnetism, physics educators seeking to clarify concepts, and anyone interested in mastering the principles of electric fields and dipoles.

Ivan Hu
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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.
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Homework Equations


All included in the solution.

The Attempt at a Solution


2016q4soln.jpg


Thank you very much!
 

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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##.
 
Thanks you kuruman for your answer. That makes sense. Is that the only mistake that you’ve picked up?
 
Ivan Hu said:
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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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.
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Matthew Strasiotto said:
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.
 
Matthew Strasiotto said:
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.
 
kuruman said:
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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