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Spherical conductor in electric field

  1. Feb 1, 2009 #1
    An uncharged solid spherical conductor is placed in a constant electric field [itex]\mathbf{E_0}[/itex]. Find:
    a) the reulting potential and electric field
    b) the induced surface charge density and sketch the field lines
    c) the force on the sphere

    I don't have a scooby where to start here so any help would be greatly appreciated. Cheers
     
  2. jcsd
  3. Feb 1, 2009 #2

    gabbagabbahey

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    Re: Electrostatics

    Have you learned separation of variables yet? Do you know he general solution to Laplace's equation for spherical symmetry charge distributions? Can you reason that the charge induced on the sphere will possess such symmetry?

    If so, then finding the potential is simply a matter of choosing appropriate boundary conditions and applying them.
     
  4. Feb 1, 2009 #3
    Re: Electrostatics

    lol probably should be able to do all of the above.
    why would the induced charge possess spherical symmetry though?
    not sure on the solution of laplaces equation - would that be just to say let [itex]\varphi=R(r) \Theta(\theta) [/itex] and solving by seperation of variables??

    and then how do i know whether a neumann or dirichlet boundary condition is more suitbale?
     
  5. Feb 1, 2009 #4

    gabbagabbahey

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    Re: Electrostatics

    You are free to orient your coordinate system so that E0 points in the z-direction. When you do this, is there any reason why the induced charge distribution wouldn't be azimuthally symmetric? Is there any reason why [itex]V(\phi=0)[/itex] would differ from [itex]V(\phi=\pi/4)[/itex]?

    Yes, that's exactly what you'd do....Although most intermediate level E&M texts go over the general solution. Have you not studied this yet?

    As for the boundary conditions, I'll help you through them after you obtain the correct general solution.
     
  6. Feb 1, 2009 #5
    Re: Electrostatics

    ok. so i can't find a reason why it won't be azimuthally symmetric. but surely [itex]V(\theta)[/itex] will vary because there will be more negative charges on the bottom of the sphere and positive charges at the top because of the way we've oriented [itex]E_0[/itex].. surely (even if it's azimuthally symmetric) this will affect the spherical symmetry???

    why is [itex]V=R(r) \Theta(\theta)[/itex] and not [itex]V=R(r) \Theta(\theta) \Phi(\phi)[/itex]?

    working through teh seperation of variables I get

    [itex]R(r)=Ae^{kr}+Be^{-kr}[/itex] and [itex]\Theta(\theta)=C \sin{k \theta} + d \cos{k \theta}[/itex] where k is the seperation constant.

    How am I doing so far?
     
  7. Feb 1, 2009 #6

    gabbagabbahey

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    Re: Electrostatics

    Correct, the problem will not b symmetric in theta....when I said 'spherically symmetric' in my first post, I meant azimuthally symmetric.

    Because the problem is azimuthally symmetric, [tex]\Phi(\phi)[/tex] must be a constant (otherwise the potential would have some phi dependence and the azimuathal symmetry would be broken), so for simplicity this constant is absorbed into [tex]R(r)[/tex] and/or [tex]\Theta(\theta)[/tex]....

    Poorly:frown:....Why not start by showing me the differential equation you get when you tave the Laplacian of [tex]V=R(r) \Theta(\theta)[/tex] in spherical coords?:smile:
     
  8. Feb 1, 2009 #7
    Re: Electrostatics

    lol. i did rush it in the hope you'd still be online haha!

    [itex]\nabla^2 V = 0 \Rightarrow \frac{\partial{R \Theta}}{\partial r^2} + \frac{\partial{R \Theta}}{\partial \theta^2}=0[/itex]

    giving [itex] R'' \Theta + R \ddot{\Theta}=0 [/itex]

    wrong i take it?
     
  9. Feb 1, 2009 #8

    gabbagabbahey

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    Re: Electrostatics

    Very wrong; the Laplacian in spherical coordinates is not [tex]\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{\partial^2}{\partial \theta^2} +\frac{\partial^2}{\partial \phi^2}[/tex]

    Look up the correct formula in your text or online:smile:
     
  10. Feb 1, 2009 #9
    Re: Electrostatics

    ah yes completely forgot we were doing this in spherical polars. i am a tool.
    how can we justify not doing the phi part of it though?

    i now get

    [itex]\frac{1}{r^2} \frac{\partial}{\partial{r}} (r[R' \Theta]) + \frac{1}{r^2 \sin{\theta}} \frac{\partial}{\partial{\theta^2}} (\sin{\theta} R \dot{\Theta})[/itex]

    which simplifies to
    [itex]\frac{\Theta}{r^2}(2rR' + r^2 R'') + \frac{R}{r^2 \sin{\theta}}(\cos{\theta} \dot{\Theta} + \sin{\theta} \ddot{\Theta})[/itex]

    and then,

    [itex]\frac{2}{r} R' \Theta + R'' \Theta + \frac{R \dot{\Theta}}{r^2 \tan{\theta}} + \frac{R \ddot{\theta}}{r^2}[/itex]
     
  11. Feb 1, 2009 #10
    Re: Electrostatics

    ah the phi part of the laplacian will just be 0 as our V isn't a function of phi, yes?

    also, this may seem like a stupid question to ask after so many posts but why are we doing the laplacian of V to find V. i mean, why did you decide to find V by solving Laplace's equation?
     
  12. Feb 1, 2009 #11

    gabbagabbahey

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    Re: Electrostatics

    Okay, so now you have:

    [tex]\frac{2}{r} R' \Theta + R''\Theta + \frac{R \Theta'}{r^2 \tan{\theta}} + \frac{R\Theta''}{r^2}=0[/tex]

    Multiply both sides of the equation by [tex]\frac{r^2}{R(r)\Theta(\theta)}[/tex] to get:

    [tex]\frac{2r R' + r^2R''}{R} + \frac{\Theta'}{\tan{\theta}\Theta} + \frac{\theta''}{\Theta}=0[/tex]

    Now, since [itex]r[/itex] and [itex]\theta[/itex] are independent variables, you must have:

    [tex]\frac{2 rR' + r^2R''}{R} =-\left(\frac{\Theta'}{\tan{\theta}\Theta} + \frac{\theta''}{\Theta}\right)=\text{some constant}[/tex]

    To make the solution for [tex]\Theta(\theta)[/tex] as simple as possible, choose that separation constant to be [tex]l(l+1)[/tex] so that:

    [tex]\frac{2 rR' + r^2R''}{R} =l(l+1)[/tex]

    And

    [tex]\frac{\Theta'}{\tan{\theta}\Theta} + \frac{\theta''}{\Theta}=-l(l+1)[/tex]

    The DE for [itex]R(r)[/itex] is easy to solve, and as a hint to solving the DE for [itex]\Theta(\theta)[/itex]; think Legendre Polynomials :wink:

    P.S.....Is this derivation really not in your textbook or course notes?
     
    Last edited: Feb 1, 2009
  13. Feb 1, 2009 #12

    gabbagabbahey

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    Re: Electrostatics

    Yes, that's right.

    Well, since you don't know what the induced charge density is, you have know way of applying Coulomb's law or Gauss's Law. That leaves you with two options, Method of images and Solving Poisson's equation. Since there is no charge anywhere except on the surface of the sphere Poisson's equation becomes Laplace's equation everywhere except at the surface of the sphere, and since the problem has azimuthal symmetry, Laplace's equation is relatively easy to solve.

    This is the type of reasoning you should try to apply when considering which method to use to solve a problem. First think of which methods you know, then rule out any methods which require more information than you are given, and finally choose the easiest method from those that remain.
     
  14. Feb 2, 2009 #13
    Re: Electrostatics

    hi again. having trouble with the R solution. something's going wrong somewhere.

    [itex]r^2 R'' + 2r R' - l(l+1)R=0[/itex]

    gives an auxiliary equation of [itex]r^2 \alpha^2 + 2r \alpha - l(l+1) \alpha = 0 [/itex]
    and how the heck do you slove that?
    am i meant to make a substitution somewhere?

    i'm also having difficulty rearranging the other one into

    [itex]\frac{\sin{\theta}}{\Theta} \frac{\partial}{\partial{\theta}}(\sin{\theta} \frac{d \Theta}{d \theta}) + l(l+1) \sin^2{\theta} = m^2 [/itex] as i reckon i can get a solution from that point

    sorry to bug you with this
     
    Last edited: Feb 2, 2009
  15. Feb 2, 2009 #14

    gabbagabbahey

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    Re: Electrostatics

    You can solve it using power series, or you can simply realize that [itex]Ar^l[/itex] and [itex]Br^{-l-1}[/itex] both satisfy the DE and therefor the general solution is [tex]R_l=A_lr^l+B_lr^{-l-1}[/tex] with no restrictions on the value of [itex]l[/itex].

    Try multiplying the equation

    [tex]
    \frac{\Theta'}{\tan{\theta}\Theta} + \frac{\Theta''}{\Theta}=-l(l+1)
    [/tex]

    By [itex]\sin^2\theta [/itex] and using the fact that

    [tex]\frac{\partial}{\partial{\theta}}(\sin{\theta} \frac{d \Theta}{d \theta})=\cos\theta\Theta'+\sin\theta\Theta''[/tex]
     
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