Electric field of discrete point charges

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Homework Help Overview

The problem involves calculating the electric field at the center of a circle formed by five equal negative point charges placed symmetrically around it. The subject area pertains to electrostatics and electric fields generated by point charges.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the electric field components based on the symmetry of the charge arrangement. Some participants question the accuracy of the calculations, particularly regarding the number of charges considered and the distance squared in the calculations. Others suggest using symmetry arguments to simplify the problem.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations and suggesting alternative approaches. There is an acknowledgment of potential errors in the original calculations, and some guidance has been offered regarding symmetry considerations.

Contextual Notes

Participants note that the distances in the x and y directions may not be equal to the radius R, which could affect the calculations. The original poster expresses confusion regarding the treatment of the charges and their contributions to the electric field.

mborn
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Five equal negative point charges (-q) are placed symmetrically around a circle of radius R. Calculate the electric field at the center of the circle.

My Answer:

Each one will be place 72 degrees from the other one (360/5),
Each field line is directed toward the center (charges are negative),
I had;

E_x = (kq/R^2)[-cos 72 + cos 36 + cos 36 - cos 72]
= kq/R^2

E_y = (kq/R^2)[- sin 72 - sin 36 + sin 36 - sin 72]
= 0 N/C

My book says that both E_x and E_y are zero, What did I do that was wrong?

mbron
 
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why do you only have 4 charges in your calculations? plus, i don't think you are calculating the distance squared properly, think about it some more

anyways, i would just make a symmetry argument. hey, works for MIT
 
Remember that the distance between the centre and any point is R. However, the distance in the x and y direction is not R.
 
DarkEternal,
Thanks! I missed the one on the x-axis. This gives me -kq/R^2 which will cancel the other equal but positive term.

Parth Dave,
x- and y- have nothing to do with the problem as I am asked to find E at the center. Thank you for caring to answer me.

mborn
 
Last edited:

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