Question

= 0 ( [this point is away from 0] A conducting sphere of radius a is hollow and grounded (V = 0). A particle of charge q is a placed inside at a distance b from the center. We wish to find the potential anywhere inside the sphere. This problem can be solved with the image charge method. The image charge d' should be placed a distance b = from the center, so that the center 0, q and q are on one line (see picture). (a) Take any point P on the surface of) the sphere. Let R and R' be the distances from P to q and q. Show that = constant independent of where P is on the surface. Calculate the constant. Hint: In Cartesian coordinates, with q at (0,0,6), q at (0,0,), and Pat (2, y, z), use x² + y2 + 2 = a’ to express R2 and R' in terms of only. (b) Use the constant from part (a) to determine what the charge d' should be in terms of q, a,b). (c) Write the potential V(r, 0,) inside the sphere. Use spherical coordinates with the 0 = 0 axis along the ray from 0 to q. (d) Use the result of part (c) to write down the Green's function G7,7") for the Dirichlet problem with V given by the interior of the sphere or radius b. You may express r and r in either cartesian or spherical coordinates, whichever you prefer. (e) Will your answer to part (c) change if the the sphere is not grounded, but is kept neutral (i.e., with total charge 0)? If so, how?

Only need help with (d)

Answer 1

d) If the charge q is at a point $\vec b$, then the potential at a point $\vec r\,\,$ inside the sphere of radius is given by

$V(\vec r)=\frac{q}{4\pi \epsilon_0}\left(\frac{1}{\sqrt{r^2+b^2-2rb\cos\gamma}} -\frac{1}{\sqrt{a^2+\frac{r^2b^2}{a^2}-2rb\cos\gamma} }\right )$

Here the angle gamma is the angle between $\vec b$ and $\vec r\,\,$

The potenital can also be written as

$V(\vec r)= \int_v \rho(\vec r\,')G_D(r,r')d^3 r'=q\int\delta^3(\vec r\,'-\vec b)G_D(r,r')d^3 r'$

Then the Dirichlet Green's function in spherical polar coordinates is simply

$G_D(r,r')=\left(\frac{1}{\sqrt{r^2+r'^2-2rr'\cos\gamma}} -\frac{1}{\sqrt{a^2+\frac{r^2r'^2}{a^2}-2rr'\cos\gamma} }\right )$

and

$\cos\gamma = \frac{\vec r.\vec r\,'}{r.r'}=\sin(\theta)\sin(\theta')\cos(\phi-\phi')+\cos(\theta)\cos(\theta')$

Here is the radius of the sphere.

Note that G remains the same if you interchange and r'. This is the property of Green's function.