- #1

Beelzedad

- 24

- 3

- TL;DR Summary
- Please help me understand three dimensional improper integral in electrostatics in the correct way.

In Physics/Electrostatics textbook, I am in a situation where we have to find the electric field at a point inside the volume charge distribution. In Cartesian coordinates, we can't do it the usual way because of the integrand singularity. So we use the three dimensional improper integral.

##\displaystyle \mathbf{E}=\lim\limits_{\epsilon\to 0} \int_{V'-\delta_{\epsilon}} \rho'\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag1##

where:

##\mathbf{r'}=(x',y',z')## is coordinates of source points

##\mathbf{r}=(x,y,z)## is coordinates of field points

##V'## is the volume occupied by the charge

##\delta_{\epsilon}## is an arbitrary volume contained in ##V'## around the singular point ##\mathbf{r}=\mathbf{r'}## with ##\epsilon## being its greatest chord.

##\rho'## is the charge density and is continuous throughout the volume ##V'-\delta##

From equation ##(1)##, we can get the ##x##-component of ##\mathbf{E}##:

##\displaystyle E_x=\lim\limits_{\epsilon\to 0} \int_{V'-\delta_{\epsilon}} \rho'\ \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag2##

_____________________________________________________________________________________________________________________

I view the steps of solving ##E_x## as follows:

1. Make a ##\delta_{\epsilon}## cavity (with ##\epsilon=a##) contained in ##V'## around the singular point ##\mathbf{r}=\mathbf{r'}##. Then take the Riemann integral over ##V'-\delta_{\epsilon}##

2. Repeat above procedure for all ##\epsilon## in the interval ##(0,a]## keeping the shape of ##\delta_{\epsilon}## unaltered.

3. Find the function which relates "Riemann integral over ##V'-\delta_{\epsilon}##" and "##\epsilon##". Then make a graph of "##\epsilon##" (##x##-axis) and "Riemann integral over ##V'-\delta_{\epsilon}##" (##y##-axis) over the interval ##(0,a]##

4. Find ##l## ##\ni##

##\forall \varepsilon > 0, \exists \delta \ni \text{when} |x-0|<\delta, |y-l|< \varepsilon##

Thus ##l## is the solution for ##E_x## in equation ##(2)##

_____________________________________________________________________________________________________________________

I know steps ##(1)## and ##(4)## can be done. But is there a way to execute step ##(2)## and ##(3)##? Please explain.

##\displaystyle \mathbf{E}=\lim\limits_{\epsilon\to 0} \int_{V'-\delta_{\epsilon}} \rho'\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag1##

where:

##\mathbf{r'}=(x',y',z')## is coordinates of source points

##\mathbf{r}=(x,y,z)## is coordinates of field points

##V'## is the volume occupied by the charge

##\delta_{\epsilon}## is an arbitrary volume contained in ##V'## around the singular point ##\mathbf{r}=\mathbf{r'}## with ##\epsilon## being its greatest chord.

##\rho'## is the charge density and is continuous throughout the volume ##V'-\delta##

**While taking the limit the shape of ##\delta_{\epsilon}## is kept unaltered**From equation ##(1)##, we can get the ##x##-component of ##\mathbf{E}##:

##\displaystyle E_x=\lim\limits_{\epsilon\to 0} \int_{V'-\delta_{\epsilon}} \rho'\ \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3} dV' \tag2##

_____________________________________________________________________________________________________________________

I view the steps of solving ##E_x## as follows:

1. Make a ##\delta_{\epsilon}## cavity (with ##\epsilon=a##) contained in ##V'## around the singular point ##\mathbf{r}=\mathbf{r'}##. Then take the Riemann integral over ##V'-\delta_{\epsilon}##

2. Repeat above procedure for all ##\epsilon## in the interval ##(0,a]## keeping the shape of ##\delta_{\epsilon}## unaltered.

3. Find the function which relates "Riemann integral over ##V'-\delta_{\epsilon}##" and "##\epsilon##". Then make a graph of "##\epsilon##" (##x##-axis) and "Riemann integral over ##V'-\delta_{\epsilon}##" (##y##-axis) over the interval ##(0,a]##

4. Find ##l## ##\ni##

##\forall \varepsilon > 0, \exists \delta \ni \text{when} |x-0|<\delta, |y-l|< \varepsilon##

Thus ##l## is the solution for ##E_x## in equation ##(2)##

**Am I correct? Is this the correct interpretation of equation ##(2)##?**_____________________________________________________________________________________________________________________

**If yes:**I know steps ##(1)## and ##(4)## can be done. But is there a way to execute step ##(2)## and ##(3)##? Please explain.