What Is the Electric Field Produced by a Uniform Line Charge on the x-axis?

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Homework Help Overview

The problem involves a uniform line charge distributed along the x-axis with a linear charge density β. The task is to use Gauss' Law to determine the electric field produced by this charge distribution and subsequently calculate the work done on a charge Q moving along the y-axis from y = a to y = b.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Gauss' Law using a cylindrical surface to find the electric field, with some questioning the calculation of the enclosed charge. There is a focus on clarifying the meaning of "density" in this context, specifically distinguishing between linear and volumetric charge density.

Discussion Status

Some participants have provided clarifications regarding the charge enclosed and the correct expression for the electric field. There are ongoing questions about the integration needed to calculate work done on the charge Q, with suggestions to consider the force required to move the charge and the potential difference between the two points.

Contextual Notes

Participants are navigating the definitions of charge density and the implications for calculating the electric field and work done. There is an emphasis on ensuring the correct interpretation of the problem statement and the mathematical relationships involved.

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Homework Statement


Charge is uniformly distributed along the x-axis with density ß. Use Gauss' Law to find the electric field it produces, and use this to calculate the work done on a charge Q that moves along the y-axis from y = a to y = b.


Homework Equations



\phi=\int\vec{E}*\hat{n}dA

\phi= \frac{Q}{\epsilon}

The Attempt at a Solution



I used a cylinder for my surface since the normal vector will always align with the electrical field. So the first part, the equation ends up
\phi=E\intdA
\phi=E(2∏rh)

(r is the radius from the axis to the edge of the cylinder and h is the length of the cylinder.)

and if I remember right, Q is the density times the are of enclosure.
so Q = β(2∏rh)
I set the two \phi equations equal to each other and get
\frac{β}{ε}=E

I don't think that's right though. What did I do wrong?
 
Last edited:
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Since the charge is along the x-axis, then β is a linear charge density.
So, to get the charge enclosed, you multiply β by the length h, and not the volume.
 
Hi Colts! :smile:
Colts said:
Charge is uniformly distributed along the x-axis with density ß.

and if I remember right, Q is the density times the are of enclosure.
so Q = β(2∏rh)

No, "density" here means the line density (in coulombs per metre, not per metre3)

So Q = βh. :wink:

(of course, sometimes "density" means surface density, and occasionally it actually means density! :rolleyes:)

EDIT: ap123 beat me to it! :biggrin:
 
So the electrical field is

E = \frac{β}{2πrε}
 
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Is the last question asking me to integrate E from a to b?
 
Colts said:
Is the last question asking me to integrate E from a to b?
Integrate the force, FExt, that would need to be exerted on a charge, Q, to move the charge from a to b . (Actually integrate the work the force does.)
 
\int\frac{βxdx}{2πrε}

x is the distance

does that look right? and the integral would be from a to b
 
Colts said:
\int\frac{βxdx}{2πrε}

x is the distance

does that look right? and the integral would be from a to b
Not correct.

What is the force on a charge Q located on the y-axis , a distance y from the x-axis ?

You (or some outside agent) must apply what force on Q to move it, at a constant rate, when the charge is located on the y-axis ?

Added in Edit:

Another way to do this is to find the potential difference from y = a to y = b .

To do that, you do integrate -E .
 
Last edited:

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