Finding electric field of a wire within a cylinder

  • #1

Homework Statement


[/B]
A wire with linear charge density lambda is surrounded by a cylindrical conducting shell with radius r and surface charge density sigma. Find an expression for the electric field (magnitude and direction) as a function of distance d from the wire, on an axis perpendicular to the wire.

Homework Equations



Gauss's Law

The Attempt at a Solution


[/B]
I know the electric field of a wire is lambda/(2*pi*r*permittivity of free space) however, I don't know how to find the electric field of a cylinder with a wire within it rather than just a single point charge.
 

Answers and Replies

  • #2
collinsmark
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Homework Statement


[/B]
A wire with linear charge density lambda is surrounded by a cylindrical conducting shell with radius r and surface charge density sigma. Find an expression for the electric field (magnitude and direction) as a function of distance d from the wire, on an axis perpendicular to the wire.

Homework Equations



Gauss's Law

The Attempt at a Solution


[/B]
I know the electric field of a wire is lambda/(2*pi*r*permittivity of free space) however, I don't know how to find the electric field of a cylinder with a wire within it rather than just a single point charge.
First things first, you are going to have to rename one of your variables. The [itex] r [/itex] in the [itex] \vec E = \frac{\lambda}{2 \pi r \varepsilon_0} \hat {a_r} [/itex] is not the same [itex] r [/itex] as the radius of the cylindrical shell. You'll have to rename one of them to avoid confusion. [The problem statement tells you to change one of them to [itex] d [/itex].]

Second, can you use Gauss' Law to derive the [itex] \vec E = \frac{\lambda}{2 \pi r \varepsilon_0} \hat {a_r} [/itex] formula with only the wire (ignoring the cylindrical shell for the moment)? If you know how to do that, adding in the cylindrical shell is pretty straightforward.

Let's look at Gauss's Law:
[tex] \oint \vec E \cdot \vec {dA} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

The [itex] dA [/itex] is a differential area element on your chosen Gaussian surface. Choose your Gaussian surface wisely to take advantage of symmetry. If you choose it wisely in certain special cases like this one, the closed integral is easy to solve. The idea is that you want the dot product of the electric field and the differential area element to be constant over the entire surface [Edit: possibly ignoring things like cylindrical endcaps, if you can show that [itex] \vec E \cdot \vec{dA} = 0 [/itex] at the endcaps]. That way the left hand side simply becomes [itex] EA [/itex], where [itex] A [/itex] represents a formula for the Gaussian surface's area. (You can't do this for everything, but you can do this for a) things with spherical symmetry, b) infinitely long wires/cylinders [like this problem] or c) infinite planes.)

The right hand side only depends on the charge enclosed within the Gaussian surface. Anything outside the Gaussian surface doesn't matter.

Then solve for [itex] E [/itex]. You can add the unit direction vector in as a final step. :wink:
 
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  • #3
First things first, you are going to have to rename one of your variables. The [itex] r [/itex] in the [itex] \vec E = \frac{\lambda}{2 \pi r \varepsilon_0} \hat {a_r} [/itex] is not the same [itex] r [/itex] as the radius of the cylindrical shell. You'll have to rename one of them to avoid confusion. [The problem statement tells you to change one of them to [itex] d [/itex].]

Second, can you use Gauss' Law to derive the [itex] \vec E = \frac{\lambda}{2 \pi r \varepsilon_0} \hat {a_r} [/itex] formula with only the wire (ignoring the cylindrical shell for the moment)? If you know how to do that, adding in the cylindrical shell is pretty straightforward.

Let's look at Gauss's Law:
[tex] \oint \vec E \cdot \vec {dA} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

The [itex] dA [/itex] is a differential area element on your chosen Gaussian surface. Choose your Gaussian surface wisely to take advantage of symmetry. If you choose it wisely in certain special cases like this one, the closed integral is easy to solve. The idea is that you want the dot product of the electric field and the differential area element to be constant over the entire surface [Edit: possibly ignoring things like cylindrical endcaps, if you can show that [itex] \vec E \cdot \vec{dA} = 0 [/itex] at the endcaps]. That way the left hand side simply becomes [itex] EA [/itex], where [itex] A [/itex] represents a formula for the Gaussian surface's area. (You can't do this for everything, but you can do this for a) things with spherical symmetry, b) infinitely long wires/cylinders [like this problem] or c) infinite planes.)

The right hand side only depends on the charge enclosed within the Gaussian surface. Anything outside the Gaussian surface doesn't matter.

Then solve for [itex] E [/itex]. You can add the unit direction vector in as a final step. :wink:

So after I find the equation for the Electric field of the cylindrical shell, do I just simply add it to the other equation i.e. E-Field of wire+ E-field of Cylinder?
 
  • #4
collinsmark
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So after I find the equation for the Electric field of the cylindrical shell, do I just simply add it to the other equation i.e. E-Field of wire+ E-field of Cylinder?
You'll have to break it up into two parts: one part where [itex] r [/itex] is less than [itex] d [/itex] and the other part where [itex] r [/itex] is greater than [itex] d [/itex].

Remember the right hand side of Gauss' Law, particularly [itex] Q_{enc} [/itex]. The only thing that goes into the equation is the charge enclosed within the Gaussian surface. Anything outside doesn't count. When [itex] r [/itex] is less than [itex] d [/itex] is the charge on the cylindrical shell within the Gaussian surface? What about when [itex] r [/itex] is greater than [itex] d [/itex]?

[Edit: Since [itex] d [/itex] is the independent variable, perhaps I should have switched those around and rather asked, "When [itex] d [/itex] is less than [itex] r [/itex] is the charge on the cylindrical shell within the Gaussian surface? What about when [itex] d [/itex] is greater than [itex] r [/itex]?"]
 
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