Proving the Image Charge in a Metal-Sphere-Shell

In summary, the problem involves finding the charge and location of an image charge in a metal-sphere-shell held at zero potential. Using the law of cosines and the uniqueness principle, it can be shown that the charge of the image charge is equal to -a divided by the distance of the image charge from the center of the sphere. The distance of the image charge from the center is also equal to the square of the radius of the sphere divided by the distance of the real charge from the center.
  • #1
bjnartowt
284
3

Homework Statement



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]

...away.


Homework 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]



The Attempt at a Solution



Use law of cosines in the superposition:
[tex]\begin{array}{c}
{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) \\
\end{array}[/tex]

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.
 
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  • #2


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

[tex]\begin{array}{c}
\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| \\
\end{array}[/tex]
 

1. What is the image charge in a metal-sphere-shell?

The image charge in a metal-sphere-shell is an imaginary charge that is placed inside the shell to mimic the behavior of a real charge outside the shell. It is used to simplify the calculations of electric fields and potentials in a system with a conductive sphere and a surrounding conducting shell.

2. How is the image charge determined in a metal-sphere-shell?

The image charge is determined by using the method of images, which involves reflecting the real charge across the surface of the sphere and shell. The distance and magnitude of the image charge can be calculated using the known properties of the system, such as the radius and potential of the sphere and shell.

3. What is the purpose of using an image charge in a metal-sphere-shell?

The purpose of using an image charge in a metal-sphere-shell is to simplify the calculations of electric fields and potentials in a complex system. It allows us to treat the system as if it only contains a single charge, making the analysis much easier and more manageable.

4. Are there any limitations to using an image charge in a metal-sphere-shell?

Yes, there are some limitations to using an image charge in a metal-sphere-shell. This method is only applicable in situations where the sphere and shell are perfectly conducting and there are no other charges or conductors present. Additionally, the image charge will only accurately mimic the behavior of a real charge at distances much larger than the size of the sphere and shell.

5. How is the image charge used to prove the behavior of the electric field in a metal-sphere-shell?

The image charge can be used to prove the behavior of the electric field in a metal-sphere-shell by comparing the electric field calculated using the image charge method with the electric field calculated using the more complex method of integrating the charge distribution over the surface of the sphere and shell. If the two results are the same, it provides evidence that the image charge accurately represents the behavior of the electric field in the system.

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