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A capacitor made of two circular plates of radius $L$ separated by $d$, initially the plates carry $\pm$ Q charge.

Then a wire of resistance R is placed between them, how do I go about deriving a time-dependent expressiion

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2 Answers 2

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You have to first derive the time-dependent current $I(t)$ which runs through the wire.

The capacitor is described by the relation $$Q(t) = C V(t) \qquad(1)$$ with $C= \epsilon \pi L^2/d$ and $\epsilon$ the dielectric constant of the medium between the two plates.

The wire is described by $$V(t) = R I(t). \qquad(2)$$

Charge conservation yields $$ \dot Q(t) = I(t). \qquad(3)$$

Taking the time derivative of (1) and plugging in (3) and then (2), we obtain $$ RC \,\dot{I}(t) = I(t)$$ with the solution $$ I(t) = I_0 e^{-t/RC}, \qquad I_0 = \frac{Q}{CR} = \frac{Qd}{\epsilon\pi L^2 R}.$$

The electric field $E(t)$ in the capacitor is constant throughout the capacitor (in the limit of $L/d \gg 1$) and points along the axis of the capacitor. Is is given by $E(t) = V(t)/d =R I(t)/d$.

Similarly, the magnetic field $B(t)$ is always perpendicular to the electric field and curls around the axis of the capacitor. Its size is given by $B(t) = \mu_0 I(t)/2\pi r$ with $r$ the (radial) distance to the axis.

Together, the magnitude of the Poynting vector is given by $$|P(t)| = E(t) B(t)/\mu_0 = \frac{R I(t)^2}{2\pi r d}.$$

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I think you neglected the capacitor displacement current. See my answer. –  Art Brown Nov 14 '12 at 2:57

This question sounds like homework, but it's too sweet not to dive into...

The capacitance $ C = \epsilon \pi L^2 /d $ and resistance $R$ form a simple first-order circuit, with voltage $v(t)$ and current $i(t)$ given by:

$$ v(t)=v(0)e^{-t/\tau} \, , \, i(t) = \frac{v(0)}{R} e^{-t/\tau} = \frac{Q}{\tau} e^{-t/\tau} $$

with

$$ v(0)=Q/C \, , \, \tau=RC $$

Note $v=iR$ at all times (equivalent to neglecting wire inductance).

To calculate the Poynting vector $\boldsymbol{P}$ we need the $\boldsymbol{E}$ and $\boldsymbol{B}$ fields.

1) Neglecting fringing fields, the electric field $E$ between the capacitor plates is just $v(t)/d$ (independent of where you measure between the plates):

$$ \boldsymbol{E} = - \frac{v(t)}{d} \boldsymbol{\hat{z}} = - \frac{i(t)R}{d} \boldsymbol{\hat{z}} = - \frac{Q}{Cd} e^{-t/\tau} \, \boldsymbol{\hat{z}} = - \frac{1}{\epsilon} \frac{Q}{\pi L^2} e^{-t/\tau} \, \boldsymbol{\hat{z}}$$

where I've introduced cylindrical coordinates with z increasing from the -Q "bottom" plate (bottom) to the +Q "top" plate, so the electric field is pointing in the $-\boldsymbol{\hat{z}}$ direction.

(Note that, as expected, the electric field is just the surface charge density $/ \epsilon$.)

2) Now the B-field. I'm not sure where exactly you've positioned the wire, so I'm going to assume it's on the z-axis, which is the simplest (and to me the most interesting) case. With this arrangement, the wire current flows down the z-axis from the top to the bottom plate.

Because the current and the electric field are vertical, $\boldsymbol{B}$ is purely azimuthal: $\boldsymbol{B} = B_\theta \boldsymbol{\hat{\theta}}$.

Applying the generalized Ampere's law (don't forget the displacement current!) to a circular surface (within the capacitor) of radius $r, 0<r<L$, centered on the z-axis, you get:

$$ 2 \pi r \frac{B_\theta}{\mu} = -i + \pi r^2 \epsilon \frac{d\boldsymbol{E}}{dt} = -i + \pi r^2 \epsilon \frac{Q}{\tau Cd} e^{-t/\tau} = -i\left[ 1 - \left(\frac{r}{L}\right)^2 \right] $$

( $-i$ because the current is flowing down, and since $\boldsymbol{E}$ points downward but is decreasing in magnitude, $d\boldsymbol{E}/dt$ points up.)

The displacement current contribution tends to cancel that of the electron current, completely canceling at $r=L$ (expected, since the total displacement current equals the electron current).

Completing the calculation,

$$ \boldsymbol{B} = - \frac{\mu i}{2 \pi r} \left[ 1 - \left(\frac{r}{L}\right)^2 \right] \boldsymbol{\hat{\theta}} $$

$$ \boldsymbol{P} = \frac{1}{\mu} \boldsymbol{E \times B} = - \frac{1}{2 \pi r d} i^2(t) R \left[1-\left(\frac{r}{L}\right)^2 \right] \boldsymbol{\hat{r}} \, , \, 0<r<L $$

As the capacitor discharges, its stored energy flows inward, into the wire (hence the $-\boldsymbol{\hat{r}}$ direction of $P$), where it is dissipated. The capacitor energy is contained inside its radius $L$, and $\boldsymbol{P}$ correspondingly falls to 0 at that boundary, since no energy is flowing in from outside the capacitor structure.

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