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A spherical capacitor with inner radius $r_1$ and outer radius $r_2$ is filled with dielectric material with permittivity $\epsilon=\epsilon_0+\epsilon_1\cos^2\theta $.
$\theta$ is the polar angle. Furthermore there is a charge $Q$ on the inner surface at $r_1$ and a charge $-Q$ on the surface at $r_2$.

Give reasons why the potential $\Phi(r,\theta,\varphi)$ is for all $r$ independent of $\theta$ and $\varphi$.

The independence of $\varphi$ is clear as our problem is invariant under $\varphi$-rotations. I'm not able to find a valid argument why the potential is independent of $\theta$.

Thanks in advance for help !

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  • $\begingroup$ What are your thoughts about the problem? $\endgroup$
    – ProfRob
    Jul 18, 2014 at 17:03

1 Answer 1

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In the region between two spherical plates, we have, $$\begin{array}{l} \nabla \times E = 0\,\,\, (1)\\ \nabla .\left( {\varepsilon \left( \theta \right)E} \right) = 0\,\,\, (2) \end{array}$$ The first equation leads to the definition of scalar potential $\Phi$, i.e. $E = - \nabla \Phi $. Therefore, from Eq. (2) we have, $$\nabla .\left( {\varepsilon \left( \theta \right)\nabla \Phi } \right) = \nabla \varepsilon \left( \theta \right).\nabla \Phi + \varepsilon \left( \theta \right){\nabla ^2}\Phi = 0 \,\,\, (3)$$ Since the problem is invariant under $\varphi$-rotation, $\Phi=\Phi(r,\theta)$. Using the separation of variable method, $\Phi(r,\theta)=R(r)\Theta(\theta)$. By substituting this function into the Eq. (3), after some straightforward calculations (in spherical coordinate system), one can easily obtain the following equation,

$$-\frac{1}{{R(r)}}\frac{d}{{dr}}\left( {{r^2}\frac{dR(r)}{dr}} \right) = {\frac{1}{{\varepsilon \left( \theta \right)\Theta \left( \theta \right)}}\frac{{d\varepsilon \left( \theta \right)}}{{d\theta }}\frac{{d\Theta \left( \theta \right)}}{{d\theta }} + \frac{1}{{\sin \left( \theta \right)\Theta \left( \theta \right)}}\frac{d}{{d\theta }}\left( {\sin \left( \theta \right)\frac{{d\Theta \left( \theta \right)}}{{d\theta }}} \right)} $$

One side of this equation is r-dependent and the other side is $\theta$-dependent. Therefore both side must be a constant value, namely $\lambda$, $$\frac{d}{{dr}}\left( {{r^2}\frac{dR(r)}{dr}} \right) + \lambda R(r) = 0\,\,\, (4)$$ $$\lambda = \frac{1}{{\varepsilon \left( \theta \right)\Theta \left( \theta \right)}}\frac{{d\varepsilon \left( \theta \right)}}{{d\theta }}\frac{{d\Theta \left( \theta \right)}}{{d\theta }} + \frac{1}{{\sin \left( \theta \right)\Theta \left( \theta \right)}}\frac{d}{{d\theta }}\left( {\sin \left( \theta \right)\frac{{d\Theta \left( \theta \right)}}{{d\theta }}} \right)\,\,\, (5)$$

On the other hand, we know that a conductor has an equipotential surface , i.e. the functions $\Phi(r=r_1,\theta)$ and $\Phi(r=r_2,\theta)$ must be independent of $\theta$. Therefore, using $\Phi(r,\theta)=R(r)\Theta(\theta)$, we conclude that $\Theta(\theta)$ must be constant, namely one, and therefore $$\Phi(r,\theta)=R(r)\Theta(\theta)=R(r)$$ is independent of $\theta$ for all values of $r$. This is the desired result.

In addition, in this case, $\frac{d\Theta(\theta)}{d\theta}=0$, from Eq. (5), we find $\lambda=0$, and therefore by using Eq. (4), we have $$\frac{d}{{dr}}\left( {{r^2}\frac{{dR(r)}}{{dr}}} \right) = 0$$ with the solution, $$R(r) = {c_1} + \frac{{{c_2}}}{r} \,\,\, (6)$$ where, $c_1$ and $c_2$ can be determined from the boundary conditions. (Here the total charge at $r=r_1$ is $Q$ and the total charge at $r=r_2$ is $-Q$).

By setting the reference point at infinity, i.e. by assuming $\Phi(r=\infty)=0$, we find $c_1=0$. For calculation of $c_2$ we use this fact that the total charge at $r=r_1$ is $Q$ (or, equivalently, the total charge at $r=r_2$ is $-Q$), $$\int {\sigma \left( \theta \right)da} = \int {\sigma \left( \theta \right)r_1^2\sin } \left( \theta \right)d\theta d\varphi = Q \,\,\, (7)$$ Consider the general boundary condition $$\left( {{\varepsilon _2}{E_2} - {\varepsilon _1}{E_1}} \right).{\hat n_{12}} = \sigma$$ where $\hat n_{12}$ is a unit vector from region 1 to region 2. In this problem the region 1 is inside the spherical plate at $r=r_1$ and the region 2 is its outside surface. Since the electric field inside a conductor is zero, we have

$$\sigma(\theta) = {\left. {\varepsilon \left( \theta \right){E_r}} \right|_{r = {r_1}}} = - \varepsilon \left( \theta \right){\left. {\frac{{d\Phi \left( r \right)}}{{dr}}} \right|_{r = {r_1}}} = - \varepsilon \left( \theta \right){\left. {\frac{{dR\left( r \right)}}{{dr}}} \right|_{r = {r_1}}} = \frac{{{c_2}}}{{{r_1}^2}}\varepsilon \left( \theta \right) \,\,\, (8)$$ Substituting Eq. (8) in Eq. (7), gives, $${c_2} = \frac{Q}{{\int {\varepsilon \left( \theta \right)\sin } \left( \theta \right)d\theta d\varphi }} = \frac{Q}{{2\pi \int_0^\pi {\varepsilon \left( \theta \right)\sin \left( \theta \right)d\theta } }}\,\,\, (9)$$ which can be obtained easily.


Another Solution:

If, at beginning, we know that the electric potential is independent of $\theta$, we can use the Gauss's law, $$\int {\varepsilon \left( \theta \right)E.\hat nda = Q} \,\,\, (10)$$ In this case since $\Phi$ is independent of $\theta$, $E = - \nabla \Phi = - \frac{{d\Phi (r)}}{{dr}}\hat r = e(r)\hat r$ is also independent of $\theta$ and radial. Therefore, using Eq. (10), we have, $$e(r){r^2}\int {\varepsilon \left( \theta \right)\sin \left( \theta \right)d\theta d\varphi = Q} $$ and therefore (see Eq. (9)), $$e(r) = \frac{Q}{{{r^2}\int {\varepsilon \left( \theta \right)\sin \left( \theta \right)d\theta d\varphi } }} = \frac{Q}{{2\pi {r^2}\int_0^\pi {\varepsilon \left( \theta \right)\sin \left( \theta \right)d\theta } }} = \frac{{{c_2}}}{{{r^2}}}\,\,\, (11)$$

Integrating the differential equation $- \frac{{d\Phi (r)}}{{dr}} = e(r) = \frac{{{c_2}}}{{{r^2}}}$, one obtain, $$\Phi (r) = {c_1} + \frac{{{c_2}}}{r}$$ which is exactly the previous result ($c_1=0$ if we set the reference point at infinity).

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  • $\begingroup$ Super answer thanks ! I'm just wondering because we were asked in the task to give reasons for the $\theta$ independence so I suppose qualitative arguments. In your (great !) answer though you go into some calculations. Is is possible to resume your answer a little bit ? $\endgroup$ Jul 19, 2014 at 9:16
  • $\begingroup$ In brief, the main reason is that a conductor has an equipotential surface. Therefore, the potential $\Phi$ must be independent of $\theta$ at both $r=r_1$ and $r=r_2$. By considering the fact that we can separate the $r$ and $\theta$ dependent parts of the $\Phi$, we conclude that $\Phi$ must be independent of $\theta$ not only for $r=r_1$ and $r=r_2$, but also for all other values of $r$. $\endgroup$ Jul 19, 2014 at 9:34
  • $\begingroup$ One last question: How exactly do you compute $c_1$ and $c_2$ ? $\endgroup$ Jul 19, 2014 at 9:38
  • $\begingroup$ See the edited version of the answer. $\endgroup$ Jul 19, 2014 at 10:39
  • $\begingroup$ @MojtabaGolshani : would you care to have a stab at tinyurl.com/jrbz6fn ? $\endgroup$ Feb 13, 2017 at 13:39

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