2
$\begingroup$

My question arised while solving Classical Electrodynamics(J. D. Jackson) Pro 4.9a

Find a potential at all points in space for a point charge $q$ located in a free space a distance $d$ from the center of a solid dielectric sphere of radius $a(a < d)$ and dielectric constant $\frac{\epsilon}{\epsilon_0}$.

My second trial after failure of using image charge method was

  1. to divide the space into two regions - inside(1) and outside(2) the sphere

  2. In (1) use a solution of Laplace equation($\sum_{l=0}^{\infty} A_l r^l P_l(\cos{\theta})$), in (2) superpose $\frac{q}{4\pi\epsilon_0r}$ and solution of Laplace equation($\sum_{l=0}^{\infty} B_l \frac{1}{r^{l+1}} P_l(\cos\theta)$).

  3. Get the expression of $A_l$ and $B_l$ using boundary condition at sphere shell$(r = a)$.

Resulting potential is

At (1) :$\quad \Phi(x) =\frac{q}{4\pi\epsilon_0d} + \sum_{l=1}^{\infty}\frac{q}{4\pi d^{l+1}}\frac{(2l+1)/l}{\epsilon+\epsilon_0\frac{l+1}{l}}r^lP_l(\cos\theta) \quad \cdots(*)$

At (2) :$\quad \Phi(x) = \frac{q}{4\pi\epsilon_0} \frac{1}{|x-d\hat{z}|} + \sum_{l=0}^{\infty}\frac{q}{4\pi d^{l+1}}(1-\frac{\epsilon}{\epsilon_0})\frac{a^{2l+1}}{\epsilon+\epsilon_0\frac{l+1}{l}}\frac{1}{r^{l+1}}P_l(\cos\theta)\quad \cdots(*)$

My first attempt was to locate image charge($q'$) at (2)($x'$) for potential at (1) and use different charge($q''$) at the same location($d\hat z$) for potential at (2), i.e.,

At (1) :$\quad\Phi(x) = \frac{1}{4\pi\epsilon}\frac{q''}{|x-d\hat z|}$

At (2) :$\quad\Phi(x) = \frac{1}{4\pi\epsilon_0} \left(\frac{q}{|x-d\hat z|} + \frac{q'}{|x-x'|}\right)$

and use boundary condition to determine $q',q''$, and $x'$ .

But this turned out to be not working - there was no solution to equations from boundary conditions.

And as a proof, the potential(*) at (2) doesn't seem to be decomposed into two point charges, and potential at (1) doesn't seem to be unified with rescaling of $q$ to $q''$.

This implied either that I used wrong setting for image charge method - number of image charges, location - or that image charge method is not applicable to this problem.

Here is my question.

Is it possible to use method of image in this specific example? I hope the answer to be considering the physical situation of this problem - polarization due to a nearby point charge.

And if some problems don't allow using this approach, which situation allows using this method?

Thank you in advance.

$\endgroup$
0
$\begingroup$

Well, accidentally see this, so I decided to register this cite XD.

The answer is yes. In case of dielectric spheres, there will be an image point charge, and an extra image line charge density distributed from the sphere center to the Kelvin image point.

For detailed formula, you may check out Eq.(10) in the paper:

https://journals.aps.org/pre/abstract/10.1103/PhysRevE.84.016705

The solution is not well known (actually it was re-derived by different people in history). Any way, you can first obtain the solution in terms of spherical harmonic expansion, then you may use certain way to transform the expansion into an integral form (for example using the finite Mellin transform), then you will find that, the integrand contains a Dirac delta function, which gives a Kelvin image, and an extra smooth line charge density.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.