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Electric Fields on a Hollowed Metal Sphere

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A hollow metal sphere has 6 cm and 10 cm inner and outer radii, respectively. The surface charge density on the inside surface is -100 nC/m^2. The surface charge density on the exterior surface is + 100 nC/m^2 .

    What is the strength of the electric field at point 4 cm from the center?

    What is the strength of the electric field at point 8 cm from the center?

    2. Relevant equations
    Inner Surface Charge = n = Q/A
    E = K*q/r^2

    3. The attempt at a solution
    Well I got the 1st one right:
    Convert all the nC and cm to C and M.
    E (at .04 m) =
    = [K * q]/r^2
    = [K * nA]/r^2
    = [K * n(4(pi)R^2)]/r^2 (note R = inner sphere radius and r = 4 cm from center)


    For the 2nd part I'm unsure. How do I found out the charge on the part of the sphere in between the hollow and exterior at 8 cm?

    Is it q = nA - nA_2?
    q = [1*10^-17 * 4pi(.1m)^2] - [-1*10^-17 * 4pi(.6m)^2]
    Then repeat previous problem with E (at .08m)?
     
    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 2, 2007 #2

    dynamicsolo

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    I'm a little unclear on how those opposing charges on the inner and outer surfaces of this conducting metal sphere are staying separated. Are you sure this isn't an insulating sphere? Alternatively, are the signs on those surface densities really opposite? Is there a charge at the center of this shell?

    As for working out the electric fields, have you had Gauss' Law yet?
     
  4. Oct 3, 2007 #3
    I don't have that information. The problem was copied and pasted in its entirely from the online homework program I have to use. There are also is no diagram.

    The only thing I know so far is that.
    The first question's answer is
    E = 2.54 * 10^24 N/C
    I don't understand why its positive since my answer was negative. But it gave a hint to check my signs and I did and that's how E was found.

    And I know that the direction of the electric field at point .04 m from the center is outward, which I also don't understand. I was under the impression that an Electric field goes from a positive charge to a negative charge.

    This suggests that (for reasons I don't understand) the center of the metal sphere must have a more positive charge than a spot .04m away? I'm gonna go out on a limb and assume the center is 0, at 6 cm its negative, and at 10 cm its positive. Sound logical?

    I'm looking at Gauss's law but I still don't get what to put in for Q_in. 8 cm is directly between 6 cm negative, and 10 cm positive. Q_in = 0?

    E*A_sphere = Q_in/e_0 where e = epsilon
     
  5. Oct 3, 2007 #4

    dynamicsolo

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    That cinches it then. There has to be a positive charge at the center of the sphere (which seems not to be mentioned). If there were no charge inside the shell, the electric field would be zero (Gauss' Law -- no enclosed charge). If the field at r = 0.04 m points outward, it has to be coming from a positive charge further inward. Also, the charge separation on the walls on the shell could only occur if something inside the shell is repelling the positive charges to the surface and attracting the negative charges to the inner wall. Since the magnitudes of the charge densities are equal on both surfaces, the conducting shell must also have an "excess" charge (it isn't neutral overall).

    You found the charges on the inner and outer surfaces of the shell from the surfaces densities and areas:

    Q_out: (+100 nC/m^2)(4pi)[(0.1 m)^2) = +12.57 nC and
    Q_inner: (-100 nC/m^2)(4pi)[(0.06 m)^2) = -4.52 nC .

    So the "excess charge" on the shell is +8.05 nC.

    Now, the field inside the conductor once equilibrium is reached is zero; otherwise, a current would still be flowing inside the shell, which means charges would still be moving. If we put a "Gaussian surface" at a radius inside the shell then (r = 0.08 m, for instance), the total enclosed charge would be zero (Gauss' Law). [So we have our answer for the second question.] But the enclosed charge there is the sum of the charge Q_inner plus the charge at the center, q. Thus q = +4.52 nC.

    The field at r = 0.04 m would then be found from Gauss' Law, giving the result from Coulomb's law for a point charge q at that distance.

    Does this sound plausible? (I don't know why the central charge isn't mentioned, unless part of the intent of the problem is to force the student to realize it would have to be present...) None of the fields will be as spectacular as E = 2.54 * 10^24 N/C (!!).
     
    Last edited: Oct 3, 2007
  6. Oct 3, 2007 #5
    There was a follow up multiple choice question to first question that I didn't originally mention, sorry:

    "What is the direction of the electric field at point 4 cm from the center?"
    Choices were: Outward, Inward, Field is 0

    The answer was outward, but I didn't understand why (i picked inward originall). I understand now but now I don't understand how I was suppose to figure that answer out.

    Using the surface charge density equation for question 1 to find Q_in yielded a negative number for me: -4.52 nC. Q_out I didn't bother solving for but I knew it would be a positive number (which you found to be 12.52 nC). And I'm 99% confident that a E-field always goes from positive to negative, in this case, outer to inner, which I would think is INWARD!

    And yes the answer turns out to be E = 0 at 8 cm. However, is this because r = 8 cm is directly between 10 cm and 6 cm? Would a place such as r = 7 cm, or r = 9.2 cm (which is still part of the "shell") yield zero electric field?

    Thanks a lot by the way.
     
  7. Oct 3, 2007 #6

    dynamicsolo

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    Your statement about the direction of an electric field is correct. The fact that the electric field at r = 4 cm. points outward then tells us that there has to be a positive charge somewhere for r < 4 cm. (Gauss' Law again: there is a net outward electric flux at r = 4 cm.)

    Yes, it would be E = 0 for 6 cm < r < 10 cm. If the shell had been electrically neutral overall, the surface charges would have been Q_in = -4.52 nC and Q_out = +4.52 nC. All of the 8.05 nC of positive excess charge is repelled to the outer surface and the field inside the shell remains zero. That occurs because the field within the shell is now the sum of the field between the inner and outer walls plus the field from the central charge (you can confirm that they cancel exactly).

    That's why I was mystified as to how the situation could exist without a central charge. Opposing charges would never stay in place on a conductor by themselves. If there were the positive excess charge present on the shell without the central charge there, Q_in and Q_out would both be positive.

    I'm still wondering if the statement of the problem is incomplete because of an omission or whether the omission was deliberate to get you to figure out how this charge configuration could possibly exist. Is that all of the questions related to this situation?
     
    Last edited: Oct 3, 2007
  8. Oct 3, 2007 #7
    You've been a big help, thank you.

    The problem statement I gave is word for word. There is more questions, I just omitted some. But what I provided was word for word. In its entirety and in order which they are presented:

    What is the strength of the electric field at point 4 cm from the center?
    E = 2.54*10^4 N/C
    What is the direction of the electric field at point 4 cm from the center?
    Outward
    What is the strength of the electric field at point 8 cm from the center?
    0 N/C
    What is the direction of the electric field at point 8 cm from the center?
    The field is Zero
    What is the strength of the electric field at point 12 cm from the center?
    E = 7860 N/C
    What is the direction of the electric field at point 12 cm from the center?
    Outward

    Once again, thanks.
     
  9. Oct 3, 2007 #8

    dynamicsolo

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    For the various distances, the enclosed charge is

    r = 4 cm: q = +4.52 nC
    r = 8 cm: q + Q_in = 0
    r = 12 cm: q + Q_in + Q_out = Q_out = q + "excess charge on shell" = +12.57 nC .

    These give the fields you list. I think we've got this covered. I'm glad I was of help.
     
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