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

In summary, someone tells you to use the curl-less-ness of electric field to get the induced surface-charge density. However, you disagree and think that the potential and charge distribution are not symmetric in "theta".
  • #1
bjnartowt
284
3

Homework Statement



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]

Homework Equations

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)?
 
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  • #2
bjnartowt said:
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]

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]

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)?

"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:
  • #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?
 

1. What is the purpose of determining the charge distribution on the surface of a metal sphere?

The purpose of determining the charge distribution on the surface of a metal sphere is to understand the behavior of electric charges in a particular system. This information can be useful in various applications such as designing electronic circuits, predicting the behavior of lightning strikes, and studying the properties of materials.

2. How is the charge distribution induced on the surface of a metal sphere?

The charge distribution on the surface of a metal sphere is induced by the presence of an external electric field. This electric field causes the free electrons in the metal to redistribute themselves on the surface of the sphere, creating a net charge on the surface.

3. What factors affect the charge distribution on the surface of a metal sphere?

The charge distribution on the surface of a metal sphere is affected by the shape and size of the sphere, the magnitude and direction of the external electric field, and the material properties of the metal. It can also be affected by the presence of other nearby objects or sources of charge.

4. How is the charge distribution on the surface of a metal sphere calculated from a known potential?

The charge distribution on the surface of a metal sphere can be calculated using the Laplace equation, which relates the potential to the charge distribution. This equation can be solved analytically for simple geometries, or numerically for more complex systems.

5. What are some real-world applications of studying charge distribution on the surface of a metal sphere?

Studying charge distribution on the surface of a metal sphere has many real-world applications, including designing lightning rods to protect buildings from lightning strikes, optimizing the performance of electronic circuits, and understanding the behavior of materials in high-voltage environments. It is also important in fields such as electrochemistry, where the distribution of charge on metal surfaces can affect chemical reactions.

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