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Magnitude of Electric Field in Insulating Cylinder

  • #1

Homework Statement


Consider a long, uniformly charged, cylindrical insulator of radius R and length l with charge density 1.8x10^-6 C/m^3. The value of the permittivity of space is 8.85419 × 10^−12.

What is the magnitude of the electric field inside the insulate at a distance 1.3cm from the axis (1.3cm < R)?


Homework Equations


[tex]\phi = \int{EdA} = q_{inside}/\epsilon_{0}[/tex]


The Attempt at a Solution


[tex]\int{EdA} = q_{inside} = [/tex] rho*V
[tex]\int{EdA} = q_{inside} = [/tex] rho*pi*r2*l

Due to symmetry, E must be constant and can be pulled outside the integral. Unfortunately, I get stuck at this step. I'm having trouble imagining the field and determining what the integral of dA actually is? I know it's not the complete surface area of the cylinder, but I'm unsure what it is.

Any help is greatly appreciated!
 
Last edited:

Answers and Replies

  • #2
collinsmark
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Guass' Law states,

[tex] \oint _S \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

Note the dot product on the left hand side. You're not multiplying two scalars together, but rather two vectors together. Specifically, you're taking the dot product of the two vectors.

And the dot product means that you are only multiplying the parts that are parallel. In other words, if the two vectors are perpendicular to each other, the dot product is zero. If the two vectors are completely parallel, the dot product is simply the multiplication of their magnitudes.

And the result of the dot product is not a vector. It's a scalar! You start by multiplying two vectors together (considering how parallel they are), and you end up with a scalar.

The direction of dA is the direction normal to the surface. So for the curved part of the cylinder, dA points radially away from the center-line.

The electric field E also points radially away from the center line.

What about the end caps of the cylinder? The dA direction of the end caps point in a direction parallel to the center line. That's perpendicular to the direction of the electric field. So what does that tell you about E · dA of the endcaps? :wink:
 
  • #3
I actually just figured out! The electric field is itself a cylinder coaxial with the given cylinder and radius 1.3cm. dA must therefore be equal to the surface area of the cylinder. Plugging everything in, I found the answer.

Your explanation is perfect, collinsmark. I understand in this case why you do not consider the endcaps of the cylinder. My only question is -- what exactly does the direction of an area mean? Does the direction come from the orientation of the area from the axis?
 
  • #4
collinsmark
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I actually just figured out! The electric field is itself a cylinder coaxial with the given cylinder and radius 1.3cm. dA must therefore be equal to the surface area of the cylinder. Plugging everything in, I found the answer.
It's good that you figured it out! :smile:

But before you go on, it might be a good idea to realize how you got the answer you did.

You were correct that E has a constant magnitude when using a Guassian surface with a fixed cylindrical radius. So yes, it can be pulled out of the integral. But by doing so, it also means that both E and dA become scalars.

Remember what I said in the last post, about two vectors being completely parallel? If two vectors are completely parallel, the dot product is simply their magnitudes multiplied. In our problem, E and dA are completely parallel at all places except for the endcaps where they are perpendicular. So ignoring the endcaps (the dot product is zero at the endcaps anyway),

[tex] \oint _S \vec E \cdot \vec{dA} [/tex]

reduces to (note how the vector notation and dot product go away. We can now use simple multiplication),

[tex] \int _S E \ dA = E \int _S dA = EA[/tex]

It's understood now, that the surface area A does not include the surface area of the end-caps (for reasons discussed above).

Do you understand how I got that?
 

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