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A conducting shell kept in a uniform electric field

  1. Apr 13, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    Find the force that tries to rip apart a conducting shell into two halves about its equator when kept in a uniform electric field of magnitude ##E##.
    Untitled.png

    2. Relevant equations
    Gauss Law: $$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$
    I think Laplace equation might also be helpful.

    3. The attempt at a solution
    Firstly, why should the two halves repel each other :oldconfused:?

    Can you just help me to figure out that part? Then I will attempt the question once more.
     
  2. jcsd
  3. Apr 13, 2016 #2
    I think there will be an induced polarisation on the shell, won't it?
    If the field lines go from left to right, the electric field should attract positive charges to the right side of the shell and negative ones to the left. In this case, the two sides repell eachother. Or am I wrong?
     
  4. Apr 13, 2016 #3
    If you wish to calculate the force that tries to rip apart the hemispheres- it means that the two halves are being attracyed to each other,
    why?
    the electric field must have done some displacement of the charges on the conducting shell and if the hemisphers have got charges then it must have some potential developed at the surface..
    let us try to model the system by imaginary cutting the spherical conductor by a plane.
     
  5. Apr 13, 2016 #4

    Titan97

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    @drvrm rip apart means to repel. If the two halves attract each other, they will stick to each other.
     
  6. Apr 13, 2016 #5

    haruspex

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    You are right about the induced charges being positive one side and negative the other, but I believe opposite charges attract. What other reason might there be a force pulling them apart?
     
  7. Apr 14, 2016 #6

    Titan97

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  8. Apr 14, 2016 #7
    actually there can be two types of force - the external electric field acting on the hemispherical shells as well as due tp the effective displacement of the charges leading to a dipole field equivalent to the two hemispherical charge distribution and the net force can be repulsive .
     
  9. Apr 14, 2016 #8

    Titan97

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    @drvrm can you explain in simple language?
     
  10. Apr 14, 2016 #9
    the external electric field will contribute to a potential at any point outside the spherical conductor say at P(r. theta) ; however the displacement of charges of the conductor (induced charges) can be looked as equivalent to a dipole placed at the centre of the conductor and its potential contribution at P can be estimated. these potentials can have different nature and
    the electric field intensity due to these two may have a net repulsion between the hemispheres
     
  11. Apr 14, 2016 #10

    Titan97

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    :oldconfused::oldfrown::oldcry: ok. I just saw a problem in purcell and morin's asking me to prove that it forms a dipole. I am gonna go through it.
     
  12. Apr 14, 2016 #11

    haruspex

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    I'm not so sure the dipole is relevant. The dipole results from the charges on both hemispheres, but the force on one hemisphere does not involve the field its own charges generate. It comes from its own charge in relation to the fields from the other hemisphere and the background field.
    However, I do not yet have any ideas on how you would quantify it.
     
  13. Apr 14, 2016 #12

    Titan97

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    @haruspex what force is responsible for repulsion?
     
  14. Apr 14, 2016 #13

    haruspex

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    Don't think of it specifically as repulsion. Just think what forces act on each hemisphere.
     
  15. Apr 14, 2016 #14
  16. Apr 14, 2016 #15

    Titan97

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    Why are they letting ##Q\to\infty##?
     
  17. Apr 14, 2016 #16
    i think they are trying to portray a field as Q tends to infinity R also tends to infinity giving Q/R.^2.. finite actually this field is responsible for induced charges in conductor later they will relate it to field intensity E(0)

    as finally they have to calculate E-field on the surface of the conductor and derive the charge density then can finally calculate force between two hemispheres.
     
  18. Apr 14, 2016 #17
  19. Apr 16, 2016 #18

    rude man

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    As posts 16 and 17 allude, the problem can be set up as a dipole of charge -q and +q centered at the center of the shell and collinear with the external E field.

    The external E field can in turn be modeled by two charges Q and -Q placed distances -d and +d away from the shell's center. The dipole charges are then immediately determined in magnitude, polarity and position inside the shell.

    The ensuing picture is that of 4 charges: the +Q charge pulls harder on the -q charge than it pushes on the +q charge. Similarly, the -Q charge pulls more heavily on the +q charge than it repels the -q charge. The net effect is to exert a force repelling the two shell halves from each other.
     
  20. Apr 16, 2016 #19

    Titan97

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    @haruspex due to the electric field, the charge distribution will be like this right?
    Untitled.png
    red represents negative charge and blue represents positive charge.
    Hence the left part tries to move left and the right part tries to move right. (repulsion)

    The field inside the shell will be cancelled out the external field (clarification needed).

    I can find the surface charge density by using the solution of Laplace equation in spherical coordinates.

    If this method is correct, I will reply the surface charge density appearing on the shell as a function of ##\theta##.
     
  21. Apr 16, 2016 #20

    Titan97

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