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

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The discussion focuses on calculating the electric field produced by a uniform line charge along the x-axis using Gauss' Law. The initial attempt involved using a cylindrical surface to derive the electric field, but there was confusion regarding the charge enclosed, which should be calculated using the linear charge density (β) multiplied by the length (h) rather than the area. The correct expression for the electric field is E = β/(2πrε). Participants also discussed the need to integrate the force exerted on a charge Q moving along the y-axis and suggested that finding the potential difference between two points could be an alternative approach. The conversation emphasizes the importance of correctly interpreting charge density in this context.
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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?
 
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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 .
 
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