1
$\begingroup$

For conductors, we propose that the free current density is proportional to the applied Electric field and the constant of proportionality is defined as conductivity.

\begin{equation} \textbf{J}_\textbf{f} = \sigma\,\textbf{E} \end{equation}

Maxwell's equation in the conducting material (assuming linear media) take the form, \begin{equation} \vec{\nabla} \cdot \textbf{E} = \frac{\rho_{f}}{\epsilon} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vec{\nabla} \cdot \textbf{B} = 0 \end{equation}

\begin{equation} \vec{\nabla} \times \textbf{E} = -\partial_t \,\textbf{B} \;\;\;\;\;\;\;\;\;\;\ \vec{\nabla} \times \textbf{B} = \mu\sigma \,\textbf{E} \,+ \mu\epsilon \, \partial_t \,\textbf{E} \end{equation}

Taking the divergence of the $\vec{\nabla} \times \textbf{B}$ equation and substituting divergence of the electric field with charge density gives,

\begin{equation} \frac{\partial\rho_f}{\partial t} = - \frac{\sigma}{\epsilon}\rho_f \implies \rho_f =\rho_f(0) \exp(-\frac{\sigma}{\epsilon} \;t) \end{equation}

I don't understand what this equation is supposed to mean. If I take some conducting wire and connect it to two ends of a battery the wire still retains its free charges. So how can $\rho_f$ be an exponentially decreasing function of time?

$\endgroup$

1 Answer 1

1
$\begingroup$

You are considering differential, i.e., local form of the Maxwell equations, and all your quantities are local, i.e., referring to a specific point in space. There is nothing wrong with the charge at a certain point decreasing - it flows away from this point, as the continuity equation tells us: $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\mathbf{J}$$ Let me also point out that $\rho_f$ in your equations is not the total charge, but the net charge at this point, i.e., the difference between the amounts of positive and negative charge.

Now, if we take a wire or a whole circuit, we are assuming a global picture, which is better described by the integral form of the Maxwell equations. Thus, if we consider a surface enclosing the circuit, the amount of charge inside this surface is not going to change, even though the shape of the distribution of the charge may vary.

A more subtle point is that talking about circuits one implies lumped element model, where all the electromagnetic phenomena a reduced to basic parameters, such as resistance, current, voltage bias and capacitor's charge. Note also that simple circuit "battery+wire" does not really sustain charge oscillations.

$\endgroup$
2
  • $\begingroup$ Thank you, this sort of clears some doubts I had. However, I don't really get how the net charge at a point flows from one point to another when $\rho_f$ decreases everywhere. Does it get distributed in all directions on average? $\endgroup$ Commented Mar 16, 2021 at 9:27
  • 1
    $\begingroup$ @a_point_particle your material equation (Ohms law) is valid only within the wire. The solution says that, if the wire were not electrically neutral, the extra charge will flow away from it. $\endgroup$
    – Roger V.
    Commented Mar 16, 2021 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.