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Charge Distribution from Known Potential - Induced on Surface of Metal Sphere

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Using the method of images: discuss the problem of a point-charge “q” inside a hollow, grounded, conducting-sphere-shell of inner radius “a”. That charge "q" is located at [tex]{\left| {{{{\bf{\vec r}}}_0}} \right|}[/tex]. We don't care about its angular position. Oh yes: the problem is two dimensional: as you will see, potential is termed as r, theta dependent.

    (b) Find the induced surface-charge density. The potential is known to be:
    [tex]\Phi (r,\theta ) = \frac{q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{{\sqrt {{r^2} + {{\left| {{{{\bf{\vec r}}}_0}} \right|}^2} - 2r\left| {{{{\bf{\vec r}}}_0}} \right|\cos \theta } }} - \frac{a}{{\sqrt {{r^2}{{\left| {{{{\bf{\vec r}}}_0}} \right|}^2} + {a^4} - 2r\left| {{{{\bf{\vec r}}}_0}} \right|{a^2}\cos \theta } }}} \right)[/tex]

    2. Relevant equations 3. The attempt at a solution

    Surface charge density: it’s derived from integral Gauss’s Law:
    [tex]\int {{\bf{\vec E}} \bullet d{\bf{\vec A}}} = \frac{Q}{{{\varepsilon _0}}}{\rm{ }} \to {\rm{ }}{\bf{\vec E}}(r,\theta ) = \frac{Q}{{{\varepsilon _0}\int {d{\bf{\vec A}}} }} = \frac{{\sigma (r,\theta )}}{{{\varepsilon _0}}}[/tex] [I.2]

    And: curl-less-ness of electric field:
    [tex]{\bf{\vec E}} = - \vec \nabla \Phi [/tex] [I.3]

    Then: [I.2] and [I.3] give the surface-charge density as a function of the potential we derived as:
    [tex]\sigma = - {\varepsilon _0}\left| {\vec \nabla \Phi } \right|[/tex]

    The spherical gradient operator reduced to polar coordinates:
    [tex]\vec \nabla = {\bf{\hat r}}\frac{\partial }{{\partial r}} + \hat \theta \frac{1}{r}\frac{\partial }{{\partial \theta }}[/tex]

    Umm…I feel like I’m doing this really wrong. Never have I needed to use the "theta-hat" in computation…only in conceptually grasping the field of a dipole.

    Then: someone tells me to use:
    [tex]\sigma = {\left. {{\varepsilon _0}\frac{{\partial \Phi (r,\theta )}}{{\partial r}}} \right|_{r = a}}[/tex]

    But I disagree: neither the charge distribution nor the potential are symmetric in "theta". Surely: as you "tilt through theta" away from the charge hovering outside the metal sphere: the charge will go from positive to negative when you've gone from theta = 0 (right under the charge) to theta = pi (on the opposite pole of the metal-sphere, away from the charge)?
     
  2. jcsd
  3. Aug 30, 2010 #2

    gabbagabbahey

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    No, [itex]\int\textbf{E}\cdot d\textbf{A} = \textbf{E}\cdot\int d\textbf{A}[/itex] not [itex]\textbf{E}\int d\textbf{A}[/itex], and only in the case where [itex]\textbf{E}[/itex] is uniform/constant over the surface. And, [itex]\textbf{a}\cdot\textbf{b}= c[/itex] does not mean that [tex]\textbf{a}=\frac{c}{\textbf{b}}[/tex]

    "someone" is correct. The equation comes from the boundary conditions on the [itex]\textbf{E}[/itex]-field as it crosses a surface. The component of the normal [itex]\textbf{E}[/itex]-field that is perpendicular to the surface will be discontinuous by an amount [itex]\frac{\sigma}{\epsilon_{0}}[/itex] (this is likely derived in your textbook directly from Gauss' Law, and I strongly recommend you read that section of your text).
     
    Last edited: Aug 30, 2010
  4. Aug 30, 2010 #3
    I think I see it. The field discontinuity across a charged surface is:
    [itex]{{\bf{\vec E}}_{above}} - {{\bf{\vec E}}_{below}} = \sigma /{\varepsilon _0}[/itex]

    ...in which you use the field/potential relation:
    [itex]{\bf{\vec E}} = - {\mathop{\rm grad}\nolimits} (\Phi )[/itex]

    ...to get what "someone" told me:
    [itex]\sigma = - {\varepsilon _0}\frac{{\partial \Phi }}{{\partial r}}[/itex]

    ...and I'm just the short distance of "plug-n'-chug" away from a possibly-correct charge density. No?
     
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