Fall 2005 13 Em

Two halves of a spherical metallic shell of radius $R$ are separated by a small insulating gap. The alternating voltage $V \cos \omega t$ is applied to the top half and the alternating voltage $-V \cos \omega t$ to the bottom half. Suppose that $\omega \ll c/R$ where $c$ is the speed of light. Compute the amplitude of the oscillating dipole moment of the system and the time-average power it radiates

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From the relation given to us $\omega \ll c/R$ at any point in time we can approximate the solution of potential everywhere in space (and electric field) as an electrostatic problem. The relation is saying that period for oscillations is much longer than the time it takes for the changes in the electric field to propagate to the rest of the sphere.

This problem is both a boundary value problem and a radiation problem. We want to calculate the power it radiates through electric dipole radiation, but to do this we much solve a boundary value problem.

Since we know the potential at the surface we can calculate the potential everywhere. The potential must satisfy the equation

(1)
\begin{align} \nabla^2 \phi = 0 \end{align}

for $r \not= R$. The solution in spherical coordinates for the potential with azimuthal symmetry is

(2)
\begin{align} \phi = \sum_{\ell} \left( A_{\ell} r^{\ell} + \frac{B_{\ell}}{r^{\ell+1}} \right) P_{\ell} (cos \theta) \end{align}

The potential cannot blow up at $r = \infty$ or $r=0$. Therefore we can write the inside and outside solutions as

(3)
\begin{eqnarray} \phi_{in}(r,\theta) = \sum_{\ell} A_{\ell} r^{\ell} P_{\ell} (cos \theta) \\ \phi_{out}(r,\theta) = \sum_{\ell} \frac{B_{\ell}}{r^{\ell + 1}} P_{\ell} (cos \theta) \end{eqnarray}

Our constraint on the potential is

(4)
\begin{align} \phi_{in}(R,\theta) = \phi_{out}(R,\theta) = \phi(\theta) \end{align}

where $\phi(\theta)$ is the potential given to us on the spherical shell. What we know want to do is find some other constraint that will allow us to calculate only one or two coefficients. To do this we can to plug in our general solution when calculating the dipole moment and power radiated.

The power radiated due to an electric dipole moment is

(5)
\begin{align} P = \frac{2}{3} \frac{ \left| \ddot{\vec{p}} \right|^2}{c^3} \end{align}

where $\vec{p}$ is the dipole moment. Since the potential on the spherical shell is oscillating we can assume the dipole moment is also oscillating.

(6)
\begin{align} \vec{p} = \vec{p_0} \cos \omega t \end{align}

We can also assume from azimuthal symmetry that the dipole is pointing in the $\hat{z}$ direction. Therefore

(7)
\begin{align} \vec{p_0} = p_0 \hat{z} \end{align}

We therefore get for the time-average power radiated as

(8)
\begin{align} P = \frac{1}{3} \frac{\omega^4 p_0^2}{c^3} \end{align}

where the $1/2$ comes from time-averaging. Calculating the $z$ component of the dipole moment is

(9)
\begin{eqnarray} p_0 &=& \int d\Omega z \sigma(\theta) \\ &=& 2 \pi R^3 \int_0^{\pi} d\theta \sin \theta \cos \theta \sigma(\theta) \end{eqnarray}

where $\sigma$ is the surface charge. We can calculate the surface charge from the potential. From one of Maxwell's equations, we know that

(10)
\begin{align} \left( \vec{E}_{out} - \vec{E}_{in} \right) \cdot \hat{r} \bigg|_{r=R} = 4 \pi \sigma \end{align}

Using our solutions for the potential we get

(11)
\begin{align} \sigma = \frac{1}{4 \pi} \sum_{\ell} \left( \ell A_{\ell} R^{\ell-1} + (\ell + 1) \frac{B_{\ell}}{R^{\ell+2}} \right) P_{\ell} (cos \theta) \end{align}

We know that the orthogonality relation for Legendre polynomials is

(12)
\begin{align} \int_0^{\pi} d\theta sin \theta P_m (\theta) P_n (\theta) = \frac{2}{2n+1} \delta_{mn} \end{align}

Since $P_1 (cos \theta) = cos \theta$, our integral for the dipole moment picks out just the $\ell = 1$ coefficients of the potential expansion.

(13)
\begin{align} p_0 = \frac{2}{3} B_1 + \frac{1}{3} R^3 A_1 \end{align}

Now all we have to do is calculate these two coefficients

(14)
\begin{eqnarray} \int_0^{\pi} d\theta sin \theta P_1 (cos \theta) \phi_{out}(R,\theta) = \frac{2}{3} \frac{B_1}{R^2} \\ \int_0^{\pi} d\theta sin \theta P_1 (cos \theta) \phi_{in}(R,\theta) = \frac{2}{3} R A_1 \\ \int_0^{\pi} d\theta sin \theta P_1 (cos \theta) \phi (\theta) = V \end{eqnarray}

Plugging in these solutions to our equations 4, 13, 8 we get

(15)
\begin{align} P = \frac{3}{4} \frac{\omega^4 V^2 R^4}{c^3} \end{align}