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What is the integral form for the Gauss Law for Electric Fields?





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up vote 4 down vote accepted

In CGS, we have that $$\nabla\cdot\mathbf{E}=4\pi\rho_{T}$$ Integrating in a volume $V$, we have: $$\iiint_{V}\nabla\cdot\mathbf{E}dV=4\pi\iiint_{V}\rho_{T}dV=4\pi Q_{T}$$ And with the Gauss-Ostrogradsky theorem:

$$\iiint_{V}\nabla\cdot\mathbf{E}dV=\iint_{\partial V}\mathbf{E}\cdot\mathbf{\hat{n}}dS$$ And finally, we get $$\mathbf{\iint_{\partial V}E}\cdot\mathbf{\hat{n}}dS=4\pi Q_{T}$$ This is the correct forms of the Gauss law in cgs. In MKS we have: $$\nabla\cdot\mathbf{E}=\rho/\epsilon_{0}$$ And with the same procedure, we arrive to the result $$\mathbf{\iint_{\partial V}E}\cdot\mathbf{\hat{n}}dS=Q_{T}/\epsilon_{0}$$ Where $Q_{T}$ is the total charge inside the volume $V$.

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good explanation =) – arthurg Nov 20 '12 at 3:34

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