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\begin{document}
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\heading{Parallel capacitor}
\auname{I.D. 39900337}
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{\bf The problem:}
\Dn
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Consider a parallel capacitor made of two conductor disks of radius $R$ being charged up by a constant electric current.
\begin{enumerate}
\item Find the electric and magnetic fields in the gap between the disks, as functions of the distance $r$ from the axis and the time $t$ (assume the charge is zero at $t = 0$).
\item Find the energy density and the Poynting vector in the gap. Note especially the direction of the vector.
\item Determine the total energy in the gap, as a function of time. Calculate the total power flowing into the gap. Check that the power input is equal to the rate of increase of energy in the gap.
\end{enumerate}
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\Dn
{\bf The solution:}
\Dn
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1. The electric field in the gap is in the direction of $\hat z$.
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\begin{eqnarray}
V &=& \int {\vec E\cdot d\vec l = E d}\\
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Q &=& I t\\
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V &=& \frac{Q}{C}\\
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C &=& \frac{A}{{4\pi kd}} = \frac{{R^2 }}{{4kd}}
\end{eqnarray}
substituting (2),(3) and (4) into (1) we get
\begin{eqnarray}
\vec E = \frac{{4kIt}}{{R^2 }}\hat z
\end{eqnarray}
From the Maxwell's equation:
\begin{eqnarray}
\oint {\vec B\cdot d\vec l} = \mu _0 I + \mu _0 \varepsilon _0 \frac{d}{{dt}}\left(\int {\vec E} \cdot d\vec a\right)
\end{eqnarray}
(where $\mu _0 I = 0$ because there is no current in the gap) we get
\begin{eqnarray}
B_{(s)} 2\pi r = \mu _0 \varepsilon _0 \frac{d}{{dt}}\left(\frac{{4kIt}}{{R^2 }}\pi r^2 \right)
\end{eqnarray}
From the Ampere's law we conclude that the magnetic field has to be in the direction of $\hat \theta$ therefore
\begin{eqnarray}
\vec B = \frac{{2\mu _0 \varepsilon _0 kIr}}{{R^2 }}\hat \theta
\end{eqnarray}
\Dn
2. The energy density is given by the formulae:
\begin{eqnarray}
u_B = \frac{{B^2 }}{{2\mu _0 }}\\
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u_E = \frac{{\varepsilon _0 E^2 }}{2}
\end{eqnarray}
The Poynting vector is
\begin{eqnarray}
\vec S = \frac{{\vec E \times \vec B}}{{\mu _0 }} = \frac{1}{{\mu _0 }}\frac{{4kIt}}{{R^2 }}\hat z \times \frac{{2\mu _0 \varepsilon _0 kIr}}{{R^2 }}\hat \theta = \frac{{8\varepsilon _0 k^2 I^2 rt}}{{R^4 }}( - \hat r)
\end {eqnarray}
The minus sign means that the energy flows into the capacitor from the outside.
\Dn
3. The total energy in the gap is
\begin{eqnarray}
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U &=& \iiint\limits_V {(u_B + u_E )dV}
= \int\limits_0^{2\pi } {\int\limits_0^d {\int\limits_0^R
{\left(\frac{{B^2 }} {{2\mu _0 }} + \frac{{\varepsilon _0 E^2 }}
{2}\right)rd\theta dzds} } }= \\
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&=& \int\limits_0^{2\pi } {\int\limits_0^d {\int\limits_0^R
{\left(\frac{{(2\mu _0 \varepsilon _0 kI)^2 }}
{{2R^4 \mu _0 }}r^3 + \frac{{\varepsilon _0 (4kIt)^2 }}
{{2R4}}r\right)d\theta dzds} } } = \varepsilon _0 ^2 k^2 d\pi I^2 + \frac{{8\varepsilon _0 k^2 d\pi I^2 t^2 }}
{{R^2 }} \\
P_1 &=& VI = EdI = \frac{{4kdI^2 t}}{{R^2 }}\\
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P_2 &=& \frac{{dU}}{{dt}} = \frac{{16\varepsilon _0 k^2 d\pi I^2 t}}{{R^2 }};\varepsilon _0 = \frac{1}{{4\pi k}} \to P_2 = \frac{{4kdI^2 t}}{{R^2 }}
\end{eqnarray}
Notice that $P_{1}=P_{2}$
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