0
$\begingroup$

This question is my attempt to find a solution to the question here

Electric potential at a point inside the charge distribution is:

$\displaystyle \psi (\mathbf{r})=\lim\limits_{\delta \to 0} \int_{V'-\delta} \dfrac{\rho (\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV'$

where:

$\delta$ is a small volume around point $\mathbf{r}=\mathbf{r'}$

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

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

$\rho (\mathbf{r'})$ is the density of charge distribution

If:

radius of $\delta=2h$, $\delta=\dfrac{4}{3}\pi (2h)^3=f(h)$

enter image description here

then are the following equations correct?

\begin{align} \dfrac{\partial \psi (\mathbf{r})}{\partial z} &= \lim\limits_{h \to 0} \dfrac{ \lim\limits_{\delta \to 0} \displaystyle\iiint_{(V'-\delta)}\rho' \dfrac{1}{|\mathbf{r}(x,y,z+h)-\mathbf{r}'|}dV' - \lim\limits_{\delta \to 0} \displaystyle\iiint_{(V'-\delta)}\rho' \dfrac{1}{|\mathbf{r}(x,y,z)-\mathbf{r}'|}dV'}{h} \tag1\\ &= \lim\limits_{h \to 0}\ \lim\limits_{\delta \to 0} \iiint_{(V'-\delta)} \dfrac{\rho'}{h} \left( \dfrac{1}{|\mathbf{r}(x,y,z+h)-\mathbf{r}'|}-\dfrac{1}{|\mathbf{r}(x,y,z)-\mathbf{r}'|} \right) dV' \tag2\\ &= \lim\limits_{\delta \to 0} \iiint_{(V'-\delta)} (\rho')\ \lim\limits_{h \to 0}\ \dfrac{1}{h} \left( \dfrac{1}{|\mathbf{r}(x,y,z+h)-\mathbf{r}'|}-\dfrac{1}{|\mathbf{r}(x,y,z)-\mathbf{r}'|} \right) dV' \tag3\\ &= \lim\limits_{\delta \to 0} \iiint_{(V'-\delta)} (\rho')\ \dfrac{\partial}{\partial z} \left( \dfrac{1}{|\mathbf{r}-\mathbf{r}'|} \right) dV'\tag4 \\ \end{align}

Equation $(1)$ is my definition for the derivative of an improper integral. Is it correct?

In equation $(3)$, I have taken one of the limits inside. Is it correct according to some theorem (like convergence theorem)? Here it must be noted that $\delta=f(h)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.