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Electric Potential, conducting sphere in a conducting shell

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data
    The electric potential at the center of a (5.00cm radius) metallic sphere is zero Volts. The sphere is surrounded by a concentric conducting shell of 10.0cm outer radius and a thickness of 2.0cm. The shell has a net charge of +20mC. a.)Find the charge on the sphere. b.)Give V(r) and then plot V(r). c.) What is the Voltage between the conductors.


    2. Relevant equations
    Outside: V = (1/4∏ε0)*(q/r)
    Inside: V = (1/4∏ε0)*(q/R) (R being the radius of the sphere)

    3. The attempt at a solution
    I'm honestly really lost with this one, and I was hoping to be pointed in the right direction. The thing that confuses me is that V at the center of the sphere is equal to V at the surface of the sphere. Working off of that, how can the sphere have a charge? Does this mean the net charge is located completely on the conducting shell? I feel like I'm missing something and I have this completely wrong. Thanks for your time reading this.
     
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  3. Mar 21, 2012 #2

    ehild

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    The static electric field is zero inside a metal. What can you say about the potential then? How are the electric field and te potential related?

    The charge of a metal can exist on the surface only. The sphere has some charge on its surface. That charge will induce equal and opposite charge on the inner surface of the shell and there is charge also on the outer surface of the shell. Apply Gauss Law two find the electric field in the different domains. The charge of the sphere is obtained from the condition that its potential is zero.


    ehild
     
  4. Mar 21, 2012 #3
    Re: Electric Potential, Charges, Conducting Sphere

    Assuming the charge on the outer shell is distributed evenly, the outer shell gives no contribution to the E-field between the shell and sphere. If there is a voltage difference between the shell and sphere, then the sphere must be charged. However, not knowing the charge on the inner sphere, or the E-field inside or outside, or the voltage drop, I can't imagine a way you'd go about finding those things. I'd love to hear the actual solution to this problem when you find it out, or anyone else's ideas.
     
    Last edited: Mar 21, 2012
  5. Mar 21, 2012 #4

    ehild

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    Some more hints:
    The potential is a continuous function.
    Outside the shell (r≥0.12 m) U(r)=(1/4∏ε0)*(q/r) where q is the charge confined inside the sphere of radius r. That contains both the charge of the shell, 20 mC, and the unknown charge Q of the sphere inside the shell.
    Inside the metal shell, the potential is constant, so U(0.1)=(1/4∏ε0)*(q/0.12)
    The potential between the metal sphere and the shell (0.05 m<r<0.1 m) depends on the charge of the sphere, and it must contain an additive term A (U(r)=(1/4∏ε0)*(Q/r)+A) so it matches the potential U(0.1)=(1/4∏ε0)*(q/0.12). The potential is constant inside and on the surface of the metal sphere: It is zero at r=0.05 m.


    ehild
     
  6. Mar 21, 2012 #5
    Re: Electric Potential, Charges, Conducting Sphere

    So I talked to my teacher today, and he told me that the electric potential of the sphere being 0 was a result of electric fields cancelling out. So going off of that, would it be safe for me to assume that the charge on the inner radius of the sphere would be -20mC? and going off of that, I could assume that the charge on the shpere would then be +20mC, cancelling the charge on the inner surface of the shell. Looking at it that way, it just seems to simple, I feel like I'm missing something.

    New thought, since I know that v=0 at the surface of the sphere, could I use that in conjunction with VB-VA = A(integral)B E*dl and use that to find the voltage potential at the surface of the inside of the shell, and then maybe find the charge required there to counteract the v at the surface of the sphere...any guidance would be appreciated.
     
    Last edited: Mar 21, 2012
  7. Mar 22, 2012 #6
    Re: Electric Potential, Charges, Conducting Sphere

    Er....I assumed the metallic sphere was conducting. The potential inside any conductor is constant, and it being 0 or not is just a matter of reference...
     
  8. Mar 22, 2012 #7

    ehild

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    Re: Electric Potential, Charges, Conducting Sphere

    Fornax, you have the same problem in the other thread at Introductory Physics. See it.

    ehild
     
  9. Mar 22, 2012 #8
    I'll give Gauss's law a try and get back to you. Thanks for the hints, and the help!
     
    Last edited: Mar 22, 2012
  10. Mar 22, 2012 #9

    ehild

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    The electric field is zero in the metallic sphere.

    ehild
     
  11. Mar 22, 2012 #10
    ok so outside of the sphere I got the E = 20 +Q/4∏r2(epsilon)0
    Inside I got that E = Q/4∏r2(epsilon)0

    Ok so going off of that, since the v=0 on the sphere, that means that E is strongest at the shpere, and so, the E field must be flowing towards it. That being said, the sphere must then have a negative charge, so the inside of the shell has a positive charge. So the outside of the shell woud have a charge of -20mC, the inside would have a charge of +40mC, and the sphere would have a charge of +20mC.

    E=20+20/4∏r20 = 3x10^13
    E = Q/4∏r20 = 1.5x10^13

    ok, good so far?
     
  12. Mar 22, 2012 #11

    ehild

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    Use parentheses, and write Q in coulombs. Outside the shell E = (0.020 +Q)/(4∏r2epsilon0)

    Why do you think so?

    Using the expressions for the electric field, set up equations for the potential, and use the information given.

    ehild
     
    Last edited: Mar 22, 2012
  13. Mar 22, 2012 #12
    Well I figure the sphere is -20mC because electric fields flow towards negative charges. Since Electric potential's strength decreases as the electric field strength increases, and the electric potential at the surface of the sphere is 0, the sphere must have a negative charge. Based off of that, the charge on the inside of the shell must be positive, and twice the outside negative charge, since the shell's net charge is +20mC. I may be assuming too much here, let me know.
     
  14. Mar 22, 2012 #13

    ehild

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    You are assuming too much. Why do you think that the charge of the sphere is -20 mC?

    How are the potential and electric field related? How do you get the potential of a charged sphere?

    ehild
     
  15. Mar 22, 2012 #14
    Ok, so I think I may have it.
    after using ΔV = -∫Edl
    Outside shell : ΔV = (20+Q)/4∏ε0r
    Inside shell : ΔV = Q/4∏ε0r

    then

    ΔV = (20+Q)/4∏ε0r

    ΔV = 20/4∏ε0r + Q/4∏ε0r

    -20/4∏ε0r = Q/4∏ε0r ----> multiply by 4∏ε0r

    -20mC = Q
     
  16. Mar 22, 2012 #15

    ehild

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    ΔV is the potential difference between two points. Once you use it for two points outside the shell, the other time it is inside the shell. Can they be identical?

    If they are identical it means that 20/4∏ε0r + Q/4∏ε0r = Q/4∏ε0r

    This equation is equivalent to 20=0.

    You got Q=-20 mC from the condition ΔV = 0. Potential difference between what points?


    ehild
     
  17. Mar 22, 2012 #16

    ehild

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    If the charge of the sphere is -20 mC, the net charge inside a Gaussian surface outside the shell is zero. The electric field would be zero, and the potential constant, the same as at infinity, that is zero. If the potential is zero at the outside of the shell, it is the same at the inner surface, as the potential is constant in a metal. If the surface of the inner shell is at zero potential and the sphere is also at zero potential, the electric field is zero in the cavity between the sphere and the shell. Zero electric field would mean zero charge on the sphere, which is contradiction.


    ehild
     
  18. Mar 22, 2012 #17
    Ok, hopefully the last stab at this,

    using Vb - Va =- Q / (4∏ε0) *[1/b-1/a]

    20 - 0 = -Q / (4∏ε0) * [1/.08 - 1/.05]
    20 = -Q / (4∏ε0*7.5) ---->multiply by -(4∏ε0*7.5)

    -2.4x10^10 = Q
     
  19. Mar 22, 2012 #18

    ehild

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    It is almost correct. Recalculate the potential on the left-hand side, using appropriate units and not forgetting 1/4∏ε0, which will cancel with the same factor on the right hand side. Also remember that the outer sphere contains not only the 20 mC charge but also the unknown Q. On the right-hand side, 1/0.08 -1/0.05=-7.5. But remember that the integral of 1/r^2 is -1/r, so the sign of Q will be really negative.

    ehild
     
    Last edited: Mar 22, 2012
  20. Mar 22, 2012 #19

    vela

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    Re: Electric Potential, Charges, Conducting Sphere

    I merged the two threads.

    Fornax, you had it right the first time. This problem belongs in the introductory physics forum. The advanced physics forum is for problems from upper-division classes and graduate classes.
     
  21. Mar 22, 2012 #20
    Thanks for merging the threads, I hope I didn't confuse anbody :/ . As for another attempt, and yet again I think I have it....

    (20 + Q)/(4∏ε0*7.5) - 0 = -Q / (4∏ε0) * -[1/.08 - 1/.05]
    20 + Q / (4∏ε0*7.5) = -Q / (4∏ε0*7.5) ---->multiply by -(4∏ε0*7.5)

    20+Q = -Q

    20 = -2Q
    Q = -10mC
     
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