# How to take the derivative of potential?

The electric potential $$\psi$$ at a point inside the volume charge distribution is:

$$\displaystyle \psi = \lim\limits_{\delta \to 0} \int_{V'-\delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dV'.$$

Now how shall we take the derivative of the potential w.r.t. $$x$$ to get:

$$\displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-\delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dV' \right) = -\lim\limits_{\delta \to 0} \int_{V'-\delta} \rho'\ \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3} dV'.$$

In other words, how shall we take the derivative of potential at a point inside the volume charge distribution to get the negative of $$x$$−component of electric field at that point?

• What exactly is the problem? Are you asking about the conditions to bring the derivative "into" the limit and the integral or just about how to differentiate the function? – gented May 31 '19 at 9:55
• I am asking whether or not is it allowed to bring the derivative inside the limit. I already know that we can bring the derivative inside the integral (by Leibniz rule). – N.G.Tyson May 31 '19 at 9:59
• The electrostatic potential $\phi(\mathbf{r})$ at a point $\mathbf{r}$ due to a static volume charge of density $\varrho(\mathbf{r}')$ is $$\phi(\mathbf{r})\boldsymbol{=}\iiint\dfrac{\varrho(\mathbf{r}')}{\Vert\mathbf{r}\boldsymbol{-} \mathbf{r}'\Vert}\mathrm dV' \tag{1}\label{1}$$ I don't understand the limits and the $\delta$'s. The static electric field is $$\mathbf{E}(\mathbf{r})\boldsymbol{=-\nabla}\phi(\mathbf{r}) \tag{2}\label{2}$$ .... – Frobenius May 31 '19 at 14:01
• @Frobenius: If we are considering field at a point $P \in V'$ we have an integrand discontinuity at point $\mathbf{r}=\mathbf{r'}$. So we remove a small volume $\delta$ around the point $\mathbf{r}=\mathbf{r'}$ and then find the field at point $P$ in the limit as the volume $\delta$ approaches zero. – N.G.Tyson May 31 '19 at 14:12
• @Frobenius: If you do not mind, may I get your e-mail ID. Hope you can help in peer reviewing my notes on electromagnetism. – N.G.Tyson May 31 '19 at 15:11

Actually, the gradient of the potential is always $$-\times$$ the electric field. However, you are asking how to obtain it by calculating the derivative. According to the comment, I call $$\Delta$$ a sphere of radius $$\delta$$ around $$r$$. We write: $$\displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right) = \displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-S} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right) +\\ \displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{S-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right)$$ where $$S$$ is a sphere of radius $$s$$ centered around $$r$$. $$s$$ is chosen such that $$S$$ is inside the volume $$V$$. The first integral on the right-hand-side does not depend on $$\delta$$, hence we can write: $$\displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right) = \displaystyle \lim\limits_{\delta \to 0} \int_{V'-S} \rho'\ \dfrac{\partial}{\partial x} \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' +\\ \displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{S-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right)$$ The second integral on the right-hand-side can be usefully rewritten in polar coordinates, with origin in $$r$$: $$\displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right) = \displaystyle \lim\limits_{\delta \to 0} \int_{V'-S} \rho'\ \dfrac{\partial}{\partial x} \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' +\\ \displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{\delta}^s \rho'\ r'\sin(\theta) dr'd\theta d\phi \right)$$ For $$\rho$$ constant, the function has uniform convergence for $$\delta \to 0$$. Depending on $$\rho$$ as a function of the position, the convergence could be non uniform. Under the condition that $$\rho_{min} < \rho < \rho_{max}$$ the convergence is uniform. We can thus exchange the derivation with the limit.
$$\displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right) = \displaystyle \lim\limits_{\delta \to 0} \int_{V'-S} \rho'\ \dfrac{\partial}{\partial x} \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' +\\ \displaystyle \lim\limits_{\delta \to 0} \int_{S-\Delta} \rho'\ \dfrac{\partial}{\partial x} \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr'$$ Indeed, also the expression with the derivative inside the integral vanishes, due to simmetry reasons. (It is also possible to take the derivative inside the limit operation by writing in polar coordinates and noticing the uniform convergence!) Summing again: $$\displaystyle \dfrac{\partial}{\partial x} \left( \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr' \right) = \displaystyle \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{\partial}{\partial x} \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} dr'$$ Writing explicitly: $$\dots = \displaystyle \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{\partial}{\partial x} \dfrac{1}{ \sqrt{\left(x-x'\right)^2+\left(y-y'\right)^2+\left(z-z'\right)^2} } dr'$$ Calculating the derivative: $$\dots = - \displaystyle \lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{x-x'}{ \sqrt{\left(x-x'\right)^2+\left(y-y'\right)^2+\left(z-z'\right)^2}^3 } dr'$$ And finally: $$\dots = -\lim\limits_{\delta \to 0} \int_{V'-\Delta} \rho'\ \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3} dr'.$$
• I guess that the $\delta$ means that some part of the volume is excluded for some reason. So, maybe, there is some difficulty that I do not see. I suggest you to clarify why you need the $\delta$. Sorry if I answer with an answer this time! – Doriano Brogioli May 31 '19 at 10:39
• $\delta$ was introduced to get rid of integrand singularity at $\mathbf{r}= \mathbf{r'}$ – N.G.Tyson May 31 '19 at 11:36
• Fine, this was the kind of "regularization" I was considering: cutting the region just around $r'$. Now the question is: can we exchange the $\partial / \partial x$ with the limit? We know that we can exchange with the integral. If so, the calculation is straightforward. Are you asking about the validity of this exchange, or the answer is already detailed enough? – Doriano Brogioli May 31 '19 at 11:43