If i place some charge on a conductor then it will distribute itself in such a way that electric field everywhere inside is zero. My text book says that only one kind of such charge distribution is possible. Or if i place charges outside the conductor then charges will be induced on the surface of a conductor to make field inside zero and again such charge distribution is unique. Intuitively it all seems correct but i am trying to figure out right arguments for this.

One such argument is that suppose i place some charge on conductor then i can solve poisson's equation to find potential at the surface(at boundary). Charge distribution outside the conductor is known. Then uniqueness theorem says that only one such potential function is possible which satisfies the given/known boundary conditions and poisson's equation outside. Since a particular potential function corresponds to a particular charge distribution the charge distribution must also be unique. This argument seems correct to me but at the same time something seems missing. Am i correct?

  • 2
    $\begingroup$ Your argument is correct, and since you don't state it even implicitly, one cannot know what "at the same time seems missing" to you... $\endgroup$
    – hyportnex
    Commented Sep 29, 2023 at 13:44

1 Answer 1


A valid argument for zero charge (and thus zero electric field) inside a conductor comes from a combination of the constitutive equation connecting the electric field $\mathbf{e}$ to the current density $\mathbf{j}$ through the resistivity $\rho_R$ or the conductivity $\sigma = \frac{1}{\rho_R}$ \begin{equation} \mathbf{e} = \rho_R \mathbf{j} \quad , \quad \mathbf{j} = \sigma \mathbf{e} \ , \end{equation} Gauss equation for the electric field \begin{equation} \dfrac{\rho}{\varepsilon} = \nabla \cdot \mathbf{e} = \rho_R \nabla \cdot \mathbf{j} \quad \rightarrow \nabla \cdot \mathbf{j} = \dfrac{1}{\varepsilon \rho_R} \rho \ , \end{equation} and the continuity equation for charge density \begin{equation} \partial_t \rho + \nabla \cdot \mathbf{j} = 0 \ , \end{equation} that can be recast as \begin{equation} \partial_t \rho + \dfrac{1}{\varepsilon \rho_R} \rho = 0 \ , \end{equation} whose solution reads \begin{equation} \rho(\mathbf{r},t) = \rho_0(\mathbf{r}) e^{-\frac{t}{\varepsilon \rho_R}} \ , \end{equation} telling us that charge density decays to zero exponentially fast, with a very small time constant $\tau = \varepsilon \rho_R$.

  • $\begingroup$ As always, LEL for anonymous downvoting $\endgroup$
    – basics
    Commented Sep 30, 2023 at 11:52

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