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Gauss law is given by

$$\oint_{\partial S}\vec E\cdot d\vec {A}=\dfrac{q_\text{enclosed}}{ε_0}.$$

$$q_\text{enclosed}=\iiint \rho\ dV.$$

For a closed surface $$\oint_{\partial S}\vec E\cdot d\vec{A}=\iiint (\nabla\cdot \vec{E})\ dV$$.

$$\implies \iiint (\nabla\cdot \vec{E})\ dV =\dfrac{\iiint \rho\ dV}{ε_0}.$$

What's the correct and intuitive way to remove the integral and $dV$ (volume element ) and give us the result (I don't know to establish a proper reasoning for eliminating the integral and $dV$).

$$\nabla\cdot E=\dfrac{ρ}{ε_0}$$

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2 Answers 2

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The key realization is that the equality $$ \int_\mathcal{V} \vec{\nabla} \cdot \vec{E} \, dV = \int_\mathcal{V} \frac{\rho}{\epsilon_0} \, dV $$ must hold for all volumes $\mathcal{V}$. By this, I mean that it is true when we integrate both the left-hand side and the right-hand side over any given region of space $\mathcal{V}$: a sphere of radius 1 m around a point charge, my kitchen, a 50-light-year cube centered at Alpha Centauri, whatever. We must also assume continuity of all functions involved.

Under these assumptions, we can prove the contrapositive of the given statement:

Suppose there exists some point $P$ in space where $\vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0 = c \neq 0$. Suppose first that $c > 0$. By continuity, this means that there is some small volume $\mathcal{V}$ surrounding $P$ such that $\vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0 > 0$ for all points within $\mathcal{V}$ (since by continuity, the value of $\vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0$ must be "close" to $c > 0$ for points sufficiently close to $P$.) This means that we must have $$ \int \left( \vec{\nabla} \cdot \vec{E} - \rho/\epsilon_0 \right) \, dV > 0 \quad \Rightarrow \quad \int_\mathcal{V} \vec{\nabla} \cdot \vec{E} \, dV > \int_\mathcal{V} \frac{\rho}{\epsilon_0} \, dV$$ and so the integrals cannot be equal. A similar argument follows if $c < 0$.

Therefore, the existence of a point $P$ at which $\vec{\nabla} \cdot \vec{E} \neq \rho/\epsilon_0$ implies that a volume exists for which the integral form of Gauss's Law does not hold. This therefore means that if Gauss's Law holds for all volumes, it must be the case that $\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0$ at all points.

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  • $\begingroup$ I don't understand what you mean by the term "for all volumes " $\endgroup$
    – Harry Case
    Commented Aug 26, 2022 at 11:57
  • $\begingroup$ @HarryCase: Edited to elaborate on what this means. $\endgroup$ Commented Aug 26, 2022 at 12:00
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    $\begingroup$ Are we coming to the conclusion that this integral holds for every possible Volume ,so it has to be the case that the function over which we are integrating has to equal else this isn't possible ,this is the same case with every Maxwell law which can be written in two forms ,a differential and an integral form $\endgroup$
    – Harry Case
    Commented Aug 26, 2022 at 12:11
  • $\begingroup$ Yes, that's the conclusion we have reached; and yes, similar arguments hold for the other three of Maxwell's equations. $\endgroup$ Commented Aug 26, 2022 at 12:13
  • $\begingroup$ Thank you so much . $\endgroup$
    – Harry Case
    Commented Aug 26, 2022 at 12:14
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The integral equation $$\iiint_V (\nabla\cdot \vec{E})\ dV =\frac{\iiint_V \rho\ dV}{ε_0}$$ is true for every volume $V$ independent of its shape.

This can only be true if the integrand functions on the left and right side are equal at every point in space: $$\nabla\cdot \vec{E}=\dfrac{\rho}{ε_0}$$

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  • $\begingroup$ But how can we be sure that the volume we are talking about on the left is the same as the volume on the right ? $\endgroup$
    – Harry Case
    Commented Aug 26, 2022 at 11:55
  • $\begingroup$ @HarryCase: They have to be the same volume for Gauss's Law to work. Gauss's Law doesn't hold if you take the volume on the left-hand side to be my kitchen and the volume on the right-hand side to be a one-light-year sphere around Alpha Centauri. $\endgroup$ Commented Aug 26, 2022 at 11:56
  • $\begingroup$ Why not ? I know I am being stupid and missing something so obvious here ,but why can't there be a case where the divergence and $\dfrac{ρ}{ε_0}$ match the differences in Volume and make them equal $\endgroup$
    – Harry Case
    Commented Aug 26, 2022 at 11:59
  • $\begingroup$ @HarryCase: In your argument, you started with the flux through the surface $S$ enclosing some volume $V$. The enclosed charge inside that region is the integral of $\rho$ over that same volume $V$. And Gauss's Law says that you can relate the flux through $S$ to the integral of the divergence of the field over the volume enclosed by $S$, i.e., $V$. So the two integrals are over the same region of space. $\endgroup$ Commented Aug 26, 2022 at 12:03
  • $\begingroup$ Yeah sorry totally forgot about that ,we are in effect considering the same region . $\endgroup$
    – Harry Case
    Commented Aug 26, 2022 at 12:05

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