1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss's law in action

  1. Mar 20, 2004 #1
    First I set about to calculate the electric field:

    Solve for Q with the capacitance of two coaxial cylinders:

    substituting:

    Where 'r' is a distance measured from the center. Trying several distances (r = 5cm -> E = 69.2 Volts / meter, r = 2cm [on the internal cylinder] -> 432.8 Volts / meter) I discovered that the electric field varies with distance, therefore its not uniform.

    The equation:

    Cannot be applied to this situation.

    This is (guess) because less charge from the power source settles on the inner cylinder because of its relative radius compared to the outer cylinder.

    Am I correct?

    Peter
     
  2. jcsd
  3. Mar 20, 2004 #2
    You are correct that the field is not uniform but your reason is wrong. Both cylinders have the same quanta of charge but with opposite signs, i.e if the inner cylinder is positively charged by +q the outer cylinder is negatively charged by -q. The reason the electric field is not constant is because only the inner cylinder contributes to it, and the cylinder can be treated as a long line in which the charges are uniformly distributed. Therefore the electrid field near it is given by the formula:

    [tex]E = 2Ke\frac{\lambda}{r}[/tex]

    Where [tex]\lambda = \frac{dq}{dl}[/tex], i.e the amount of charge for a length unit. We assume it is constant here, even thought that closer to the ends of the cylinder it is not. But the height of the cylinders are so big compared to their radii that this doesn't matter.
     
    Last edited: Mar 20, 2004
  4. Mar 20, 2004 #3
    Quanta? I'll assume it means amount. So my intuition was correct about the maximum available charge is spread between the cylinders.

    Calculating the electric field at the outer cylinder:

    So you are correct, the electric field does extend beyond the outer cylinder and some distance beyond where it becomes negligible. The only way this could be if the electric field produced by the inner cylinder was stronger than the outer cylinder, or that there is no electric field internal to a cylinder?

    Which would mean that the forces internal to the cylinder, the charges on the surface, exactly balance each other perpendicular to the surface. Which makes sense given the symmetry of a slice of the cylinder, a circle.

    So given that line of reasoning, there is no electric field internal to the outer cylinder and only the inner cylinder contributes to the field inside the cylindrical shell. Which would explain why the field dissipates with distance as if it were alone.

    Again, is there a flaw in my logic?

    Peter
     
  5. Mar 21, 2004 #4
    Before doing anything here you need to understand exactly how the charge is distributed. Do you know (and understand why) the charge of the outer cylinder is concentrated in its inner surface only, and there is no charge on the outer surface?

    That's correct. When you are in the space between the cylinders, only the inner cylinder contributes to the field (you can easily prove this and calculate the field with Gauss's law [tex]\oint E\cdot dA = \frac{q}{\epsilon _0}[/tex]). The field between the cylinders is anything but constant, and it weakens as you move away from the inner cylinder and towards the outer one.

    I never said that (I hope). As I said above, there is no charge on the outer surface of the outer cylinder.* But more importantly, the net charge of the whole system is 0 (it was not charged to begin with, the cylinders were just 'polarized'). So if you look at it from the outside, both cylinders contribute the exact same field, but in opposite directions, making the net electric field outside the system 0.


    *The reason for this is that the electric field inside the shell is 0, since it's conductive. So if you define a Gaussian space that includes the inside of the shell, the inner surface of it and the inner cylinder, the net charge of them should be zero. This means that the charge of the inner surface of the outer shell is equal (and opposite) to the charge of the inner cylinder, leaving no more charge to reside on the outer surface of the shell.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gauss's law in action
  1. Gauss's law in action (Replies: 1)

  2. Gauss' law (Replies: 2)

Loading...