# Electric Field due to dipole - solution check

## 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.

## Homework Equations

All included in the solution.

## The Attempt at a Solution

Thank you very much!

#### Attachments

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kuruman
Homework Helper
Gold Member
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?

kuruman
Homework Helper
Gold Member
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.

Ivan Hu
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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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.

kuruman