Diagram for Second Uniqueness Theorem

What I don't understand is that we have a big Region with volume $V$ and there are two cavities inside the volume $V$ ,the cavities ,each of them hold $Q_1$ & $Q_2$ charges in them.The density of the charges in between the regions of the conductor is ρ .

Now Second Uniqueness Theorem postulates that the Electric Field of this setup is unique .

My question is if $E_1$ & $E_2$ are two Electric Fields, then due to Gauss Law. $$\nabla.E_1=\dfrac{ρ}{ε_0}$$.

Similarly $$\nabla.E_2=\dfrac{ρ}{ε_0}.$$

This equation states that Electric field due to the charge density at any point in between the cavities are $E_1$ & $E_2$.

Integral form states that the Electric Field due to charges on the surface on Cavities (ie) $Q_a$ & $Q_b$ are $$\oint\vec{E_1}.\vec{dA}=\dfrac{Q_a}{ε_0}$$ etc (The same for $E_2$).


  1. Is the electric fields $E_1$ & $E_2$ ,is it because of the charges on the surface of the cavities or is it because of the charge density in between the cavities?

  2. Are the charges $Q_a$ & $Q_b$ taken into account when estimating the charge density $ρ$?


1 Answer 1


Electric field $\mathbf{E}(\mathbf{r})$ is defined in the whole domain. You can write it as the contributions of different sets of electric charges, by superposition principle if the behavior of the system is linear. Gauss' theorem is "a unique" physical principle as well, there is not a Gauss' theorem for each electric field.

What you can do, is to apply the integral form of the Gauss' theorem to different regions of the space, i.e. evaluating the flux of the electromagnetic field through the boundary of two regions $V_1$, $V_2$ as the ratio of the total electric charge $Q^{in}$ contained in these regions,

$\Phi_{S_1}(\mathbf{E}) = \displaystyle \oint_{\partial V_1} \mathbf{E} \cdot \mathbf{\hat{n}} = \int_{V_1} \nabla \cdot \mathbf{E} = \int_{V_1} \dfrac{\rho}{\epsilon_0} = \dfrac{Q^{in}_1}{\epsilon_0}$

$\Phi_{S_2}(\mathbf{E}) = \displaystyle \oint_{\partial V_2} \mathbf{E} \cdot \mathbf{\hat{n}} = \int_{V_2} \nabla \cdot \mathbf{E} = \int_{V_2} \dfrac{\rho}{\epsilon_0} = \dfrac{Q^{in}_2}{\epsilon_0}$ .

The electric field in space of the problem depicted in the picture and the charge distribution in the volume $V$ really depend on the nature of the volume $V$ itself, i.e. if it's a conductor or a dielectric:

  • if it's a conductor, charges in $V$ are distributed on the surfaces of the holes, with a total charge opposed to the charges contained; this must be true, to have electric field equal to zero in the conductor and thus no current; the 'excess' of charge goes to the external surface;
  • if it's a dielectric, polarization occurs in the volume $V$, but you can't say that much about the electric field and the charge in $V$, with the information you have

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