Dipoles and electric field expressions

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SUMMARY

The discussion focuses on calculating the electric field of a dipole, where the positive charge q is positioned at z = a/2 and the negative charge -q at z = -a/2. The electric field along the z-axis is determined by vector addition of the fields from the individual charges, utilizing the formula |E1| = (1/4∏ε0)(q1/r^2). Participants emphasize the importance of restricting calculations to the z-axis and substituting r with z for simplification. The discussion concludes with guidance on how to adjust the distance r to account for the dipole's configuration.

PREREQUISITES
  • Understanding of electric fields and dipole theory
  • Familiarity with Coulomb's law and the formula |E1| = (1/4∏ε0)(q1/r^2)
  • Basic knowledge of vector addition in physics
  • Concept of limits and approximations in calculus (e.g., Taylor series)
NEXT STEPS
  • Study the derivation of the electric field of a dipole in detail
  • Learn about the behavior of electric fields at large distances from dipoles
  • Explore the implications of the approximation 1 / (1+x)^2 ≈ 1 - 2x for small x
  • Investigate the mathematical techniques for vector addition of forces in electrostatics
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand the principles of electric fields generated by dipoles.

melissaisusi
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Okay. So my assignment is due in an hour and I'm stuck on this last question. If I can get some help, I would greatly appreciate it.

So the problem reads: The electric field of a dipole can be calculated by assuming the positive charge q is at z = a/2 and the negative charge -q is at z = -a/2 (x=y=0). The electric field along the z axis found by vector addition of the electric field from the individual charges. FInd an expression for the electric field. (Hint: 1 / (1+x)^2 ) is approximately equal to 1 - 2x for small x.) By what power of z does the electric field fall off?

Okay. So from the problem, I know that I can solve using the equation:
|E1| = (1/4∏ε0)(q1/r^2)

but how do i associate z and r?
I'm super confused. Please help me.
 
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Hello melissaisusi,

Welcome to physics forums!

melissaisusi said:
Okay. So my assignment is due in an hour and I'm stuck on this last question. If I can get some help, I would greatly appreciate it.

So the problem reads: The electric field of a dipole can be calculated by assuming the positive charge q is at z = a/2 and the negative charge -q is at z = -a/2 (x=y=0). The electric field along the z axis found by vector addition of the electric field from the individual charges. FInd an expression for the electric field. (Hint: 1 / (1+x)^2 ) is approximately equal to 1 - 2x for small x.) By what power of z does the electric field fall off?

Okay. So from the problem, I know that I can solve using the equation:
|E1| = (1/4∏ε0)(q1/r^2)

but how do i associate z and r?
I'm super confused. Please help me.

Here are a few hints and guidance.

  1. For this problem, don't bother figuring out the electric field for all space. Restrict yourself calculating the electric field to the z-axis. As a matter of fact, for the second part of this problem, the extreme positive z-axis is all that is necessary. (By that I mean you could break the problem into three parts and calculate the electric field on the positive and negative z-axis, and the portion of the z-axis in between the charges [and maybe you should do that for the first part, where you find an expression for E]. But sticking with the positive z-axis should be sufficient for the second part of this particular problem.)
  2. Since you are only concerned about the z-axis, just set r = z.
  3. And what is r (and thus z)? The distance r is the distance to the origin. But it's not the distance to the charge in this case. Every previous time you have used the
    |E1| = (1/[4πε0])(q1/r2)
    formula, the assumption was that the charge was centered at the origin. All that changes now. You need to find |E1| as a function of the distance from the charge q1. So you'll need to subtract off an a/2 somewhere from the distance to the origin.

Good luck! :smile:
 

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