sadraj
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Hi guys. My question is related to proving Gauss's law by using Columb's law. Let start with a charge density \rho(\vec{r}) in R^3 . by Columb's law we have :
E(\vec{r})=\int{\frac{\rho(\vec{r'})d^3r'(\vec{r'}-\vec{r})}<br /> {|\vec{r'}-\vec{r}|^3}}
suppose that domain of function \rho(\vec{r}) is finite in R^3. Clearly it can be showed that above integral exists for any \vec{r} out of the domain. & it's divergence is zero out of the domain. So for any Gauss's surface out of domain we can use Divergence theoreom to prove that
\int{E(\vec{r}).\vec{dS}}=\int{\rho(\vec{r})d^3\vec{r}}=\frac{Q}{\epsilon 0}
But for a Gauss's surface that goes through domain of \rho(\vec{r}) we have two problems :
1. it is not clear that for all \vec{r} the integral of electric field exist and converges.
2. Suppose that E converges for any \vec{r}. But it is not easy to bring devergence operator in the integral. Divergence theoreom doesn't work here.
What is your idea? Please don't use delta dirac function. I don't understand it. Because I haven't studied distributions in mathematics.
Similar question can be asked when we have surface charge density on the Gauss's surface.
Sorry for bad english
E(\vec{r})=\int{\frac{\rho(\vec{r'})d^3r'(\vec{r'}-\vec{r})}<br /> {|\vec{r'}-\vec{r}|^3}}
suppose that domain of function \rho(\vec{r}) is finite in R^3. Clearly it can be showed that above integral exists for any \vec{r} out of the domain. & it's divergence is zero out of the domain. So for any Gauss's surface out of domain we can use Divergence theoreom to prove that
\int{E(\vec{r}).\vec{dS}}=\int{\rho(\vec{r})d^3\vec{r}}=\frac{Q}{\epsilon 0}
But for a Gauss's surface that goes through domain of \rho(\vec{r}) we have two problems :
1. it is not clear that for all \vec{r} the integral of electric field exist and converges.
2. Suppose that E converges for any \vec{r}. But it is not easy to bring devergence operator in the integral. Divergence theoreom doesn't work here.
What is your idea? Please don't use delta dirac function. I don't understand it. Because I haven't studied distributions in mathematics.
Similar question can be asked when we have surface charge density on the Gauss's surface.
Sorry for bad english
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