- #1
bjnartowt
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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)?