How does one derive the electric potential from an electric field?

Question

I'm trying to work through the full derivation to find the equation $$ \underline{E}(\underline{r}) = -\underline{\nabla}V $$ where V is the electric potential and E is the electric field strength. So far I have the definition: $$ V(\underline{r})=\int_{\underline{r}}^{\infty}\underline{E}(\underline{r})\cdot d\underline{r} $$(Ignoring q for now), and then the result:

$$ V(\underline{r_1}) - V(\underline{r_2}) =\int_{\underline{r_1}}^{\infty}\underline{E}(\underline{r})\cdot d\underline{r} - \int_{\underline{r_2}}^{\infty}\underline{E}(\underline{r})\cdot d\underline{r} = \int_{\underline{r_1}}^{r_2}\underline{E}(\underline{r})\cdot d\underline{r}$$

From course notes, I have:

$$ dV=V(\underline{r}+d\underline{r})-V(\underline{r}) $$But I am struggling to understand how to go from get to the next line shown here:

$$V(\underline{r}+d\underline{r})-V(\underline{r}) = -\underline{E}(\underline{r}) \cdot d\underline{r} $$

I understand that you need to use the r1 r2 result, which gives: $$ V(\underline{r}+d\underline{r}) - V(\underline{r})=\int_{\underline{r}+d\underline{r}}^{\infty}\underline{E}(\underline{r})\cdot d\underline{r} - \int_{\underline{r}}^{\infty}\underline{E}(\underline{r})\cdot d\underline{r} = \int_{\underline{r}+d\underline{r}}^{\underline{r}}\underline{E}(\underline{r})\cdot d\underline{r} = -\int_{0}^{d\underline{r}}\underline{E}(\underline{r})\cdot d\underline{r}$$ But I do not understand how to get:

$$ -\int_{0}^{d\underline{r}}\underline{E}(\underline{r})\cdot d\underline{r} = -\underline{E}(\underline{r}) \cdot d\underline{r}$$ In the general case (without knowing E). I had the idea that in the infinitesimal dr, E(r) is constant with regards to r, and so the integral just integrates dr, which then gives dr again when the limits are applied. Is this correct? Any help is much appreciated!

P.S. For posterity, the rest of the derivation is:

Using the relation $$ dT=\underline{\nabla}T \cdot d\underline{r} $$

We find: $$ dV = \underline{\nabla}V \cdot d\underline{r}$$

And inserting into the dV equation above,

$$ \underline{\nabla}V \cdot d\underline{r} = -\underline{E}(\underline{r}) \cdot d\underline{r}$$

Hence

$$ \underline{E}(\underline{r})=-\underline{\nabla}V$$

Answer 1

You are trying to do this in a needlessly complicated and muddled way (work integral with a variable limit). It isn't wrong but it more complicated and confusing.

It is better to do it the simple easy way (see below) and then realize the relation between electric potential and the work integral with variables limit (hypothetical work of electrostatic field on unit positive charge that undergoes infinite motion). Also, I recommend deprecating that non-standard underscore notation for vectors, the formulae using it are hard to read.

So here is the simple derivation; first, it is important to realize that the relation

$$ \mathbf E = - \nabla V \tag{*} $$ between electric field and electric potential isn't a generally valid relation, but it is only valid for the Coulomb electric field $\mathbf E_C$ of all charges (in general, this is only a part of the total electric field). The Coulomb field is defined as integral over all space:

$$ \mathbf E_C(\mathbf x) = \int_{space} K\rho(\mathbf x') \frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|^3}\,d^3\mathbf x' $$ where $\rho$ is density of electric charge.

The electric potential in (*) is one of infinite number of possible potentials that are related to the Coulomb potential of all charges in space via arbitrary shift in its value $V_0$: $$ V(\mathbf x) = V_0 + \int_{space} K\rho(\mathbf x') \frac{1}{|\mathbf x - \mathbf x'|}\,d^3\mathbf x'. $$

Now, using these two definitions, it is easy to verify that the two quantities obey the relation (*). The nabla operator applied to $V$ means that derivatives of $V$ with respect to components of vector $\mathbf x$ are to be taken.

$$ -\nabla V (\mathbf x) = \left [ -\frac{\partial V}{\partial x_1}, -\frac{\partial V}{\partial x_2}, -\frac{\partial V}{\partial x_3} \right ] $$ You can easily show that $-\partial V/\partial x_1$ is the same as $\mathbf E_{C,1}$ and so on for components 2,3.