Electric Fields ( i attached a file)

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SUMMARY

The discussion centers on understanding the electric field generated by electric dipoles, particularly focusing on the contributions from positive and negative charges. The user seeks clarification on the vector components of the electric field, specifically why the expressions for the negative charge include negative signs for both x and y components. The conversation also touches on integrating equations related to electric fields and solving a specific problem involving a uniformly charged insulating rod shaped into a semicircle. Key equations discussed include E = F/q and the relationship between charge, distance, and electric field strength.

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  • Understanding of electric fields and dipoles
  • Familiarity with vector components in physics
  • Knowledge of integration in the context of electric field equations
  • Basic principles of electrostatics, including Coulomb's Law
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Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and dipole interactions.

trah22
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Homework Statement


Hi, my question isn't actually over a particular homework problem, I've attached a file of my notes on electric field, electric dipoles, i don't quite understand why the equation of the electric field is setup that way for the negative charge.

Homework Equations





The Attempt at a Solution

 

Attachments

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can someone help me out:smile:
 
I'd love to help, but I'm unable to open the file
 
weird what error comes up? maybe it requires the latest adobe, my file is in pdf. Argh I am not sure what could be wrong

Is there a specific way to attach files in order for them to be posted correctly for this forum?
 
Last edited:
There's nothing wrong with how you posted your file; attachments just need to be approved before they can be viewed. Done!

Now, what exactly is your question? What about that expression for the field of the dipole is unclear? It's just expressing the field as a vector sum of the components of the field from each charge. Since the field from a negative charge points towards the charge, both components of its field at point P are negative. (Note that E_- stands for the magnitude of the field at point P contributed by the negative charge.)
 
You could also have noted that the components in the x direction cancel each other out by simmetry, without having to worry about cosines ;)
 
Well to find the Electric Field u have to add the electric field for the positive and negative charge. For the xaxis i undestand that its in the positive direction so its Ecos(theta)i but howcome for the y direction its -esin(theta)j, is it because its goin down? Thats for the positive charge electric field. For the negative charge, i don't understand why its minus Ecos(theta)i and then minus Esin(theta)j. Basically could someone explain to me why for the negative charge according to the way its graphed, why its -Ecosthetai-Esinthetaj.:smile:
 
Ok, i actually just figured it out, thnx for the help anyhow,
 
Last edited:
Im just having some trouble on figuring out how some of the components are and how there being used for this part of the notes ( i attached another file),
 

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Last edited:
  • #10
Since E_+ = E_-, E_+ + E_- = E_+ + E_+ = 2 E_+ = 2 kq/D^2.
 
  • #11
ah i c thanks doc
 
  • #12
There was one more thing on that page that i don't quite get, when integrating equation 3 into equation 2, why is the a?d there? because E=F/q=kq/r^2x1/q, I am not quite sure on where and why the a/D is needed.

And then for this one particular problem, its very similar to the above problem, but i ran into a couple problems, 1 during the setup of the electric field equation, another when choosing the correct integral to use and lastly how R-L/Pie is used in the final Electric field equation.

The actual problem is:a uniformly charged insulating rod of length 14 cm is bent into the shape of a semcircle. The rod has a total charge of -7.5uC. Find the magnitude and direction of the electric field at 0, the center of the semicircle.

I wrote down on the scan on the things I am talking. There just 3 parts.
 

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