I was learning about Electrostatic Potential.

First they introduced,

$\overrightarrow E(\overrightarrow r) = -\nabla \phi(\overrightarrow r) \quad\quad-(eq \;i)$

I understood the above equation because both are equal to zero.

But as I was reading further, I came across,

$-\int^{\overrightarrow r}_{\overrightarrow r_{ref}}\overrightarrow E(\overrightarrow {r'}) \;.\;\overrightarrow {dr'}=\int^{\overrightarrow r}_{\overrightarrow r_{ref} }\nabla\phi(\overrightarrow {r'})\cdot \overrightarrow{ dr'}=\int ^{\overrightarrow r}_{\overrightarrow r_{ref}}\frac{d\phi(\overrightarrow {r'})}{dr'} \;.\;\overrightarrow{dr'}=\phi(\overrightarrow{r'})-\phi(\overrightarrow{r_{ref}})\quad\quad-(eq \;ii)$

Let me assume eq(ii) this way,


Now, it is pretty much obvious why $I_1=I_2$ (from eq i).

Now, what I am facing difficulty is in figuring out how $I_2=I_3$.

If we look closely at $I_2$, we find that function of the integral consists of $\nabla\phi$, which is nothing but gradient. But as far as I know, $\nabla$ consists of partial fraction symbols $\partial$. Now, how did $\partial$ convert into $d$ in $I_2=I_3$?

If we closely compare $I_2$ and $I_3$, we get,

$\nabla\phi(\overrightarrow{r'})=\frac{d\phi(\overrightarrow{r'})}{dr'}\;.\;\overrightarrow {dr'}$. How is this part derived? This is my confusion, Can anyone please clarify it (with derivation if possible) ?

See this(pg1) And this (pg2)


2 Answers 2


\begin{align} \nabla \phi ({\overrightarrow{r'}})\cdot {\overrightarrow{dr'}} &= (\frac{\partial \phi}{\partial x'} \hat{i} + \frac{\partial \phi}{\partial y'} \hat{j} + \frac{\partial \phi}{\partial z'} \hat{k}) \cdot (\vec{dx'} \hat{i} + \vec{dy'} \hat{i} + \vec{dz'} \hat{k}) \\ &= \frac{\partial \phi}{\partial x'} dx'+ \frac{\partial \phi}{\partial y'}dy' + \frac{\partial \phi}{\partial z'} dz' \\ &= d\phi ({\overrightarrow{r'}}) \end{align}

Hope this helps you!


All your integrals are path integrals, meaning that, in more formal terms, you would have to pick a curve $\gamma$ joining the two ends and compute

$$\int_0^1\mathbf E(\gamma(t))\cdot\dot\gamma(t)\text dt$$

In general, this integral will depend on the choice of the path $\gamma$, but if $\mathbf E$ is a conservative field, that is, there exists a scalar field $\phi$ such that $\mathbf E=\nabla\phi$, then the integral only depends on the ends of the curve $\gamma(0)$ and $\gamma(1)$.


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