Finding the magnitude of an electric field at a distance from a surface charge

  1. 1. The problem statement, all variables and given/known data
    Find the magnitude of the electric field intensity at a distance 0.15m from the axis of a long, straight conductor, which has radius 0.03m, and a charge density of 4.0 C/m.


    2. Relevant equations
    Eflux = E.A
    Eflux = q/e0
    q = sigma*h


    3. The attempt at a solution

    Eflux = E*Area
    Thus, Eflux = E*2pi*r*h (Surface area of the gaussian cylinder we superimpose over the conductor)

    Also, Eflux = q/e0 -> where e0 is the permittivity of free space and q is the enclosed charge in the gaussian surface.

    q = sigma*h

    Therefore we substitute for the q, and equate the two Eflux equations to get:

    Eflux = E*2pi*r*h = sigma*h/e0

    Solve for E and we get:

    E = sigma/(2pi*r*e0)

    Now we substitute the given values:

    4/(2pi*(0.03)*(8.85*10^-12)) = 2.40*E12 or 2.40*10^12 N/C

    I have no idea how to find the electric field at the distance 0.15m away from the surface. Please help! I would need the value of q to do this, however this is not a point charge we're dealing with but a surface charge, so I'm a bit lost.
     
  2. jcsd
  3. If you look back over your logic, you should see that the r you plug in to the final formula is not the radius of the charged cylinder, it's the radius of the Gaussian surface, which is 0.15m. The radius of the cylinder is actually irrelevant.
     
  4. Wow...I can't believe I didn't notice that inflating the imaginary Gaussian surface would give the required 'r' of 0.15m.

    Though...why would they give the radius of the cylinder if it had no part in it?
     
  5. To see if you would make this mistake. :smile: One note: as long as the Gaussian cylinder is outside the wire, the radius of the wire doesn't matter. But another common problem you may encounter is to find the E field inside the wire some distance from the center -- then, the radius of the wire matters.
     
  6. Teachers do that alot. It's to confuse you and see that you really understand the problem.
     
  7. Hmm yea, I've noticed that kind of question appears to be common with coaxial wires.

    Thank you for your help!
     
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