Dipole problem with electric fields

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SUMMARY

The dipole problem in electric fields, specifically solved through mirror imaging, involves understanding the dipole moment's coordinates. The dipole moment, denoted as \vec{p}, can be expressed as \vec{p}=p \cos{\theta} \, \hat{z}+ p \sin{\theta} \, \hat{x}. The electric field \vec{E} at the location (0,0, 2z) can be computed using the image dipole \vec{p}_{image}=-p \cos{\theta} \, \hat{z} +p \sin{\theta} \, \hat{x}. The torque \vec{N} is calculated as \vec{N}=\vec{p} \times \vec{E}, and discrepancies in the book's computations of \vec{p} are noted.

PREREQUISITES
  • Understanding of dipole moments in electromagnetism
  • Familiarity with vector notation and operations
  • Knowledge of electric fields and their calculations
  • Basic algebra for torque computations
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  • Study the concept of mirror imaging in electrostatics
  • Learn how to derive electric fields from dipole moments
  • Explore torque calculations in electromagnetic systems
  • Review the provided resource: Dipole Electric Field PDF
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Students and professionals in physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify the dipole problem and its solutions.

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Dipole problem (which is solved through mirror imaging) has been troubling me with its solution. I understand everything except how the dipole moment's coordinates came to be, since when converted into x-y axis, its doesn't make sense. (problem 4.6)

The screenshot contains the solution which says that dipole moment p can be expressed into the terms shown (at the end of step 2, right before step 3
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)
 

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It helps to start with a good source. Try: http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf ## \\ ## Meanwhile ## \vec{p}_{image}=-p \cos{\theta} \, \hat{z} +p \sin{\theta} \, \hat{x} ## where we will assume the dipole has ## \phi=0 ## for the angle which it points.## \\ ## Using ## \vec{r}=2 z \, \hat{z}##, ##\vec{ E} ## (at location ##(0,0, 2z) ## from ## \vec{p}_{image} ##), can be computed from ## \vec{p}_{image} ##. ## \\ ## Meanwhile, ## \vec{p}=p \cos{\theta} \, \hat{z}+ \sin{\theta} \, \hat{x} ## should make it an easy algebraic exercise to compute the torque.## \\ ## I do think your book made a mistake in computing ## \vec{p} ##. ## \\ ## I got an expression for electric field ## E ## and torque ## \vec{N}=\vec{p} \times \vec{E} ##. I'd be happy to see how your result compares to mine.
 
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