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Geometry Problem in Context of Image-Charge Problem (Griffiths,p. 125, [3.15],[3.16])

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    You have charge inside metal-sphere-shell held at V = 0, and you know an image-charge goes outside the metal-sphere-shell to formulate the equivalent and unique potential. Oh, and the metal-sphere-shell is of radius "a". You get to the point where you superimpose the image and real charges' potentials like so:

    [tex]\Phi \left( {a{\bf{\hat r}}} \right) \equiv 0 = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{q}{{\left| {{{{\bf{\vec r}}}_0} - {\bf{\vec r}}} \right|}} + \frac{{{q_i}}}{{\left| {{{{\bf{\vec r}}}_i} - {\bf{\vec r}}} \right|}}} \right)[/tex]

    in which: r-arrow is position at which potential is being considered, r[0] is position vector of original charge, and r is position vector of image charge. also: q[0] and q are the charges of real and image charges, respectively.

    Prove that the charge of the image charge is:
    [itex]{q_i} = - \frac{a}{{\left| {{{{\bf{\vec r}}}_i}} \right|}}[/itex]

    ...and that the image charge is located a radial distance:
    [itex]\left| {{{{\bf{\vec r}}}_i}} \right| = \frac{{{a^2}}}{{\left| {{{{\bf{\vec r}}}_0}} \right|}}[/itex]


    2. Relevant equations
    uniqueness, and law of cosines. law of cosines seems key, as it is a Griffiths hint: consider it applied to one of the scalar-denominators of the superposition of potentials from (1)

    [tex]\left| {{{{\bf{\vec r}}}_0} - a{\bf{\hat r}}} \right| = \sqrt {({{{\bf{\vec r}}}_0} - a{\bf{\hat r}}) \bullet ({{{\bf{\vec r}}}_0} - a{\bf{\hat r}})} = \sqrt {{{\left| {{{{\bf{\vec r}}}_0}} \right|}^2} + {{\left| {a{\bf{\hat r}}} \right|}^2} - 2\left| {{{{\bf{\vec r}}}_0}} \right|\left| {a{\bf{\hat r}}} \right|\cos {\theta _{{{{\bf{\vec r}}}_0}{\bf{\hat r}}}}} [/tex]

    ...in which the large "dot" denotes the scalar/dot product. Similar law-of-cosine treatment for other scalar denominator (that of the image charge):
    [tex]\left| {{{{\bf{\vec r}}}_i} - a{\bf{\hat r}}} \right| = \sqrt {{{\left| {{{{\bf{\vec r}}}_i}} \right|}^2} + {{\left| {a{\bf{\hat r}}} \right|}^2} - 2\left| {{{{\bf{\vec r}}}_i}} \right|\left| {a{\bf{\hat r}}} \right|\cos {\theta _{{{{\bf{\vec r}}}_0}{\bf{\hat r}}}}} [/tex]

    Also: the image charge and real charge lie along the same line as the circle’s radius, so:

    [tex]{\theta _{{{{\bf{\vec r}}}_0}{\bf{\hat r}}}} = {\theta _{{{{\bf{\vec r}}}_i}{\bf{\hat r}}}} = \theta [/tex]

    3. The attempt at a solution

    Use law of cosines in the superposition:
    {Q_i}\left| {{{{\bf{\vec r}}}_0} - a{\bf{\hat r}}} \right| = - Q\left| {{{{\bf{\vec r}}}_i} - a{\bf{\hat r}}} \right| \\
    {Q_i}\sqrt {{{\left| {{{{\bf{\vec r}}}_0}} \right|}^2} + {a^2} - 2\left| {{{{\bf{\vec r}}}_0}} \right|a\cos \theta } = - {Q_0}\sqrt {{{\left| {{{{\bf{\vec r}}}_i}} \right|}^2} + {a^2} - 2\left| {{{{\bf{\vec r}}}_i}} \right|a\cos \theta } \\
    {Q_i} = - {Q_0}\frac{{\left| {{{{\bf{\vec r}}}_i}} \right|}}{{\left| {{{{\bf{\vec r}}}_0}} \right|}}\left( {\frac{{\left| {{{{\bf{\vec r}}}_i}} \right| - 2a\cos \theta }}{{\left| {{{{\bf{\vec r}}}_0}} \right| - 2a\cos \theta }}} \right) \\

    Crude approach: plug in various values of "theta", and mandate they give the same image charge potential. Let us take a walk to the line between the already-parallel r or r[0]. Now: r, our position-vector, is parallel to both r and r[0] , meaning theta -> 0 , which makes this relation just under (3) of "Attempt..." into:
    [tex]{Q_i} = - {Q_0}\frac{{\left| {{{{\bf{\vec r}}}_i}} \right|}}{{\left| {{{{\bf{\vec r}}}_0}} \right|}}\left( {\frac{{\left| {{{{\bf{\vec r}}}_i}} \right| - 2a\cos 0}}{{\left| {{{{\bf{\vec r}}}_0}} \right| - 2a\cos 0}}} \right) = - {Q_0}\frac{{\left| {{{{\bf{\vec r}}}_i}} \right|}}{{\left| {{{{\bf{\vec r}}}_0}} \right|}}\left( {\frac{{\left| {{{{\bf{\vec r}}}_i}} \right| - 2a}}{{\left| {{{{\bf{\vec r}}}_0}} \right| - 2a}}} \right)[/tex]

    What if theta -> pi/2?

    [tex]{Q_i} = - {Q_0}\frac{{\left| {{{{\bf{\vec r}}}_i}} \right|}}{{\left| {{{{\bf{\vec r}}}_0}} \right|}}\left( {\frac{{\left| {{{{\bf{\vec r}}}_i}} \right| - 2a\cos {\textstyle{\pi \over 2}}}}{{\left| {{{{\bf{\vec r}}}_0}} \right| - 2a\cos {\textstyle{\pi \over 2}}}}} \right) = - {Q_0}\frac{{{{\left| {{{{\bf{\vec r}}}_i}} \right|}^2}}}{{{{\left| {{{{\bf{\vec r}}}_0}} \right|}^2}}}[/tex]

    Stupid! I am shy about setting something equal to anther thing...I'll prolly generate an algebra-mess.

    Griffiths does say that picking image-charge magnitude and location is a bit of an art rather than a science. Sigh.
  2. jcsd
  3. Aug 27, 2010 #2
    Re: Geometry Problem in Context of Image-Charge Problem (Griffiths,p. 125, [3.15],[3.

    I reached a contradiction by setting the theta -> 0 and theta -> pi/2 results equal to one another:

    \frac{{\left| {{{{\bf{\vec r}}}_i}} \right|}}{{\left| {{{{\bf{\vec r}}}_0}} \right|}}\left( {\frac{{\left| {{{{\bf{\vec r}}}_i}} \right| - 2a}}{{\left| {{{{\bf{\vec r}}}_0}} \right| - 2a}}} \right) = - \frac{{{Q_i}}}{{{Q_0}}} = \frac{{{{\left| {{{{\bf{\vec r}}}_i}} \right|}^2}}}{{{{\left| {{{{\bf{\vec r}}}_0}} \right|}^2}}} \\
    \left( {\left| {{{{\bf{\vec r}}}_i}} \right| - 2a} \right)\left| {{{{\bf{\vec r}}}_0}} \right| = \left| {{{{\bf{\vec r}}}_i}} \right|\left( {\left| {{{{\bf{\vec r}}}_0}} \right| - 2a} \right) \\
    \left| {{{{\bf{\vec r}}}_i}} \right|\left| {{{{\bf{\vec r}}}_0}} \right| - 2a\left| {{{{\bf{\vec r}}}_0}} \right| = \left| {{{{\bf{\vec r}}}_i}} \right|\left| {{{{\bf{\vec r}}}_0}} \right| - 2a\left| {{{{\bf{\vec r}}}_i}} \right| \\
    \left| {{{{\bf{\vec r}}}_0}} \right| = \left| {{{{\bf{\vec r}}}_i}} \right| \\
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