# How would you prove $E = -\vec{\nabla} V$ from the electric potential's line integral?

Basically, we were given an equation: $$V_1 - V_2 = \int_{r_1}^{r_2}\vec{E}\cdot d\vec{r}$$

where $\vec{E}$ is the electric field distribution and $d\vec{r}$ is the displacement vector of the charge.

Suppose that, in a 3D space, the electric field is $$\vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k}$$ and the displacement vector is $$d\vec{r} = d_x \hat{i} + d_y \hat{j} + d_z \hat{k}$$

Given $V = V(x, y, z)$, we have to prove that

$$E = -\vec{\nabla} V$$

Which make a whole lot of sense, considering that the equations for gradient theorem and line integral state so. However, assuming I have no clue what the gradient theorem is, how would I go about proving $E = -\vec{\nabla} V$ using only the given equations above?

I know that if $V$ and $E$ were merely functions of $x$: $$V(x) = \int_{x_o}^{x} E(x)dx$$ we can differentiate both sides with respect to the upper limit to get: $$\frac{\partial V}{\partial x} = -E(x)$$ But how would I go about doing that if it was a 3D space? Do I need to use some electrostatic concepts, or is this all just basic calculus that I've long forgotten? I think I'm really overthinking this because my brain says it's simple, yet I just can't think of a way.

## migrated from physics.stackexchange.comDec 22 '13 at 7:30

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• This appears to be off-topic as it is about mathematics and not physics. – Kyle Kanos Dec 22 '13 at 3:28
• Though, as a hint, you may want to investigate the fundamental theorem for line integrals (sometimes called the gradient theorem). – Kyle Kanos Dec 22 '13 at 3:30
• Yes, I am aware of the gradient theorem. But this is assuming that I have no clue what the gradient theorem is, and that I am only equipped with basic calculus (maybe from differential calculus to differential equations). And yeah, my initial thought was that we can do this simply with pure calculus, and it is just that it is interpreted in terms of electrostatics. But I doubted it, so I submitted it here. Sorry about that – user36160 Dec 22 '13 at 3:34
• Don't worry about posting on the wrong forum, I've flagged it to be migrated to math.SE. If you know of the gradient theorem, the you just need to apply its proof. – Kyle Kanos Dec 22 '13 at 3:40

It can be answered with plain mathematics. I'll paraphrase a proof given in Vector Analysis and Introduction to Tensor Analysis by Spiegel.

If $\mathbf{F}$ is a vector field whose path integral between two points is independent of the path taken, then denote $\phi(\mathbf{r})=\int_\mathbf{r_1}^\mathbf{r}\mathbf{F}\cdot d\mathbf{r}$, where $\mathbf{r_1}$ is an arbitrary but constant vector. If $s$ parameterizes position:

$$\phi(\mathbf{r})=\int_\mathbf{r_1}^\mathbf{r}\mathbf{F}\cdot \frac{d\mathbf{r}}{ds} ds$$

By differentiation with respect to $s$, (that is, as $r$ travels along a path):

$$\frac{d\phi}{ds}=\mathbf{F}\cdot \frac{d\mathbf{r}}{ds}$$

But we also know, with directional derivatives:

$$\frac{d\phi}{ds}=\nabla \phi \cdot \frac{d\mathbf{r}}{ds}$$

and since these two equalities hold irrespective of $\frac{d\mathbf{r}}{ds}$,

$$\mathbf{F}=\nabla \phi$$