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### Q:

Show that without the displacement current \( \PDi{t}{\BD} \), Maxwell’s equations will not satisfy conservation relations.

### A:

Without the displacement current, Maxwell’s equations are

\begin{equation}\label{eqn:continuityDisplacement:20}

\begin{aligned}

\spacegrad \cross \BE( \Br, t ) &= – \PD{t}{\BB}(\Br, t) \\

\spacegrad \cross \BH( \Br, t ) &= \BJ \\

\spacegrad \cdot \BD(\Br, t) &= \rho_{\mathrm{v}}(\Br, t) \\

\spacegrad \cdot \BB(\Br, t) &= 0.

\end{aligned}

\end{equation}

Assuming that the continuity equation must hold, we have

\begin{equation}\label{eqn:continuityDisplacement:40}

\begin{aligned}

0

&= \spacegrad \cdot \BJ + \PD{t}{\rho_\mathrm{v}} \\

&= \spacegrad \cdot \lr{ \spacegrad \cross \BH } + \PD{t}{} (\spacegrad \cdot \BD) \\

&= \PD{t}{} (\spacegrad \cdot \BD) \\

&\ne 0.

\end{aligned}

\end{equation}

This shows that the current in Ampere’s law must be transformed to

\begin{equation}\label{eqn:continuityDisplacement:60}

\BJ \rightarrow \BJ + \PD{t}{\BD},

\end{equation}

should we wish the continuity equation to be satisfied. With such an addition we have

\begin{equation}\label{eqn:continuityDisplacement:80}

\begin{aligned}

0

&= \spacegrad \cdot \BJ + \PD{t}{\rho_\mathrm{v}} \\

&= \spacegrad \cdot \lr{ \spacegrad \cross \BH – \PD{t}{\BD} } + \PD{t}{} (\spacegrad \cdot \BD) \\

&= \spacegrad \cdot \lr{ \spacegrad \cross \BH } – \spacegrad \cdot \PD{t}{\BD} + \PD{t}{} (\spacegrad \cdot \BD).

\end{aligned}

\end{equation}

The first term is zero (assuming sufficient continity of \(\BH\)) and the second two terms cancel when the space and time derivatives of one are commuted.

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