Electric Field On A Point Charge Due To A Uniformly Charged Rod

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Homework Statement


This is regarding setting up an integral to calculate the electric field on a point charge that is at a distance "a" from a uniformly charged rod of length "L". I have attached a picture of my work, which includes a diagram of the problem, and wanted to know if my thought process is correct.


Homework Equations





The Attempt at a Solution


Finding a general expression for the electric field produced by an infinitely small "piece" of charge and then adding them all up. The charged rod is parallel to the y-axis and it's center is at the origin.

We know that the linear charge density is

λ=(Q/L).

For an infinitely small piece of charge, we say λ=(dQ/dL), so dQ=λ*dL


The formula for the Electric field on a point charge is E=q/((4∏ε)*r^2)

To find an expression for the Electric field produced by the small piece of charge, we can replace q with dQ

dE=dQ/((4∏ε)*r^2)

(substitue λ*dL for dQ)

dE=λ*dL / ((4∏ε)*r^2)


The part that I am unsure about is finding an expression for the "r". Would it be reasonable to say that "r = (a+y)", where y is the variable that I am integrating, and my limits of integration would be [(-L/2),(L/2)]? The reason I am saying (a+y) is because if I plug "-L/2" in for y, then (a-(L/2)) is the distance from the point charge "a" to the end of the rod that is above the origin. If I plug in "L/2" for y, then (a+(L/2)) takes care of the distance from the point charge "a" to the end of the rod below the origin. If I plug in 0 for y, then I simply get "a" which makes sense, since that is the distance from the origin to the point charge.

I hope my post was formatted correctly, please let me know if It is not, and I will be sure to make changes in the future
 

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  • #2
tiny-tim
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Hi Unix! :smile:

(try using the X2 button just above the Reply box :wink:)

Yes that all seems ok so far …

what is worrying you about that?​

(btw, are you sure the diagram is correct? these questions usually have the point charge perpendicular to the rod)
 
  • #3
SammyS
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Homework Statement


This is regarding setting up an integral to calculate the electric field on a point charge that is at a distance "a" from a uniformly charged rod of length "L". I have attached a picture of my work, which includes a diagram of the problem, and wanted to know if my thought process is correct.

Homework Equations




The Attempt at a Solution


Finding a general expression for the electric field produced by an infinitely small "piece" of charge and then adding them all up. The charged rod is parallel to the y-axis and it's center is at the origin.

We know that the linear charge density is

λ=(Q/L).

For an infinitely small piece of charge, we say λ=(dQ/dL), so dQ=λ*dL

The formula for the Electric field on a point charge is E=q/((4∏ε)*r^2)

To find an expression for the Electric field produced by the small piece of charge, we can replace q with dQ

dE=dQ/((4∏ε)*r^2)

(substitue λ*dL for dQ)

dE=λ*dL / ((4∏ε)*r^2)


The part that I am unsure about is finding an expression for the "r". Would it be reasonable to say that "r = (a+y)", where y is the variable that I am integrating, and my limits of integration would be [(-L/2),(L/2)]? The reason I am saying (a+y) is because if I plug "-L/2" in for y, then (a-(L/2)) is the distance from the point charge "a" to the end of the rod that is above the origin. If I plug in "L/2" for y, then (a+(L/2)) takes care of the distance from the point charge "a" to the end of the rod below the origin. If I plug in 0 for y, then I simply get "a" which makes sense, since that is the distance from the origin to the point charge.

I hope my post was formatted correctly, please let me know if It is not, and I will be sure to make changes in the future
The distance from y=a to any arbitrary y on the charged rod is a - y .

Therefore, r = (a - y) .
 
  • #4
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Thank you for your quick responses! :)

Tiny-tim: All of the examples in my physics book did in fact consist of charges that were perpendicular to the rod. I hadn't tried a problem where the charge was on the same axis as the rod so I thought I would give it a shot and make sure I could reason it out if I saw it on an exam (I mainly posted up here to get confirmation on my thought process :) ).

SammyS:
I plugged in a few numbers into (a-y) and it makes sense now. By having (a-y), then the negative sign accounts for points that are below the axis such as (a-(-L/2)) correct?
 
  • #5
SammyS
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Thank you for your quick responses! :)
...

SammyS:
I plugged in a few numbers into (a-y) and it makes sense now. By having (a-y), then the negative sign accounts for points that are below the axis such as (a-(-L/2)) correct?
Yes.
 

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