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How can we prove that surface integral of the electric fieldfor a point charge that is outside a gaussian surface, $$\int\mathbf E\cdot\,\mathrm d\mathbf S,$$ without actually using the concept of flux?

Kyle Kanos
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lowkeyy
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  • You can prove this with a fairly difficult intigration – Sourabh Mar 03 '19 at 14:20
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    From what assumptions do you want to prove this claim? – ACuriousMind Mar 03 '19 at 14:24
  • I can give you a quick answer. IntigrationE.ds is total number of electric field lines passing through the surface but we know that gaussian surface is closed and the charge is outside the the gausssian surface .So total number of field line passing through is is equal to total number of lines exiting from it hence net E.ds is 0 – Sourabh Mar 03 '19 at 14:31
  • @Sourabh isn't it related to flux – lowkeyy Mar 03 '19 at 14:33
  • Yes it is related to flux – Sourabh Mar 03 '19 at 14:33
  • @ACuriousMind i wanted to prove it using basic electrostatics – lowkeyy Mar 03 '19 at 14:34
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    @Sourabh Please use answers, not comments, for answers. – ACuriousMind Mar 03 '19 at 14:39
  • Can you specify the case in which you want to prove it – Sourabh Mar 03 '19 at 14:40
  • Any random gaussian surface – lowkeyy Mar 03 '19 at 14:41
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    $\int E\cdot dS$ is the definition of flux. How can you prove it is zero without using its definition? – garyp Mar 03 '19 at 16:37
  • @garyp i was wondering if there exists a proof using basic electrostatics and coulombs law – lowkeyy Mar 03 '19 at 16:38
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    But how can you even talk about flux at all without referencing its definition? I can talk about an automobile because I have a definition of "automobile" implicit in my memory. How can I talk about a "Xxfghold" without defining what a "Xxfghold" is? How can I prove flux is zero without knowing what flux is? – garyp Mar 03 '19 at 16:41

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Let us start from Coulomb's law: $$\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}$$ where we take $\vec{\mathbf{r}}$ to be our coordinate, and $\varepsilon_0$ to be the vacuum permittivity.

Of course, you should be able to see how we can express this same law in terms of the charge density, instead of the charge itself, through integrating over space: $$\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{\mathbf{r}}_0)}{||\vec{\mathbf{r}}_0-\vec{\mathbf{r}}||^2}\hat{\mathbf{r}}~\mathrm{d}^3\vec{\mathbf{r}}$$

where $\rho$ is the aforementioned charge density. Here, we can take as a theorem the divergence of an inverse-square vector field to be $$\vec{\nabla}\cdot\frac{1}{r^2}\vec{\mathbf{r}} = 4\pi\delta(\vec{\mathbf{r}})$$

Using this, we now can take the divergence of our charge density-defined electric field equation: $$\vec{\nabla}\cdot\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{1}{\varepsilon_0}\int\rho(\vec{\mathbf{r}})\delta(\vec{\mathbf{r}}_0-\vec{\mathbf{r}})~\mathrm{d}^3\vec{\mathbf{r}}$$

which is obviously $$\vec{\nabla}\cdot\vec{\mathbf{E}}(\vec{\mathbf{r}}) = \frac{\rho(\vec{\mathbf{r}})}{\varepsilon_0}$$

By the divergence theorem, we can see that this is equivalent to $$\oint \vec{\mathbf{E}}\cdot\mathrm{d}\vec{\mathbf{S}} = \frac{Q}{\varepsilon_0}$$

which is your desired result.

Riley Scott Jacob
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