Guass's Law over the x axis

[Colts]

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

[itex]\phi[/itex]=[itex]\int[/itex][itex]\vec{E}[/itex]*[itex]\hat{n}[/itex]dA

[itex]\phi[/itex]= [itex]\frac{Q}{\epsilon}[/itex]

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
[itex]\phi[/itex]=E[itex]\int[/itex]dA
[itex]\phi[/itex]=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 [itex]\phi[/itex] equations equal to each other and get
[itex]\frac{β}{ε}[/itex]=E

I don't think that's right though. What did I do wrong?

 

[ap123]

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.

 

[tiny-tim]

Hi Colts!

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. ​

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

EDIT: ap123 beat me to it!

 

[Colts]

So the electrical field is

E = [itex]\frac{β}{2πrε}[/itex]

 

[Colts]

Is the last question asking me to integrate E from a to b?

 

[SammyS]

Is the last question asking me to integrate E from a to b?

Integrate the force, F_Ext, 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.)

 

[Colts]

[itex]\int[/itex][itex]\frac{βxdx}{2πrε}[/itex]

x is the distance

does that look right? and the integral would be from a to b

 

[SammyS]

[itex]\int[/itex][itex]\frac{βxdx}{2πrε}[/itex]

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 .