Deriving $-\vec{\nabla}V$ from $V_b-V_a=-\int\vec{E}\cdot d\vec{l}$

Question

I'm trying to understand how to derive $-\vec{\nabla}V$ from $V_b-V_a=-\int\vec{E}\cdot d\vec{l}$. I'm not really familiar with the gradient operator, I know how to compute it and I know that $\vec{\nabla}=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$.

From that formula, I guess that writing $\vec{\nabla}f$ would give as result a vector whose direction depends on the increase/decrease of $f$ in the $x$, $y$ and $z$ directions (it would point towards the positive $x$ direction, positive $y$ direction and negative $z$ direction if $f$ is increasing in the $x$ and $y$ direction, and $f$ is decreasing in the $z$ direction), and the length of the vector in those directions depends on how much $f$ it's increasing and decreasing in those directions. I don't know if this is correct, I've never read anything about the $\vec{\nabla}$ operator, and the formula above is all I was given.

The book I'm reading derives that equation on the following page as follows.

The footnote is as follows.

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What I don't understand is how is $-\vec{E}\cdot d \vec{l}=-E_ldl$? The dot product $\vec{E}\cdot d \vec{l}$ is equivalent to $E_xdx+E_ydy_+E_zdz$ or equivalent to $(E)(dl)cos(\theta)$ but I don't get how these last two are equivalent to $-E_ldl$. And I don't get how equation (8) is equal equation (9), which is equal to $-\vec{\nabla}V$.

Could you help me out please?

Answer 1

It seems to be that this is just a question of semantics. The statement $$\mathbf{E}\cdot\text{d}\ell = E_\ell\, \text{d}\ell$$ is not telling you anything new: it's just writing the dot product in a different way. Just as $E_x$ represents the "amount" of Electric Field in the $\mathbf{\hat{x}}$ direction, $E_\ell$ represents the amount of the field in the $\text{d}\mathbf{\hat{\ell}}$ direction, whatever that direction might be. For example, it should be clear to you that:

$$\mathbf{E}\cdot \text{d} \mathbf{x} = E_x \text{d}x.$$

So by the very definition of the dot product, $\mathbf{E}\cdot\text{d}\mathbf{\ell} = E_\ell \text{d}\ell$, since that's what the dot product is doing: it's picking out the component of the electric field along the direction given by the $\text{d}\hat{\ell}$ vector.

If you wish to use your standard definition of $$\mathbf{E}\cdot \text{d} \mathbf{\ell} = E \,\cos\theta\,\, \text{d}\ell,$$ where $\theta$ is the angle between $\mathbf{E}$ and $\text{d}\mathbf{\ell}$, then the quantity $E\cos\theta$ is precisely what one would call $E_\ell$. i.e. $$E_\ell = E \cos{\theta}.$$

(Of course, keep in mind that as you integrate over a curve, the angle between $\mathbf{E}$ and $\text{d}\ell$ could change, and so in general, $\theta$ is a function of $\ell$ and not a constant.)

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As to how Equations (8) and (9) are related, the author simply chooses the (arbitrary) direction $\ell$ respectively in Equation (8) to be $x$, $y$, and $z$. This leads to Equation (9). The partial derivatives appear since you're only varying one coordinate at a time, keeping the others fixed.

Answer 2

You are right that the gradient does give the direction of increase.

If we know that any component of electric field $E_l$ is equal to the negative of the rate of change of electric potential with respect to distance in that direction. One can simply rewrite the vector field $\vec{E}$ as the vector sum of all its orthogonal components in terms of orthonormal basis vectors $\vec{E}=E_xe_x+E_ye_y+E_ze_z$. Then of course since the negative of your electric field can be expressed as a gradient of some potential function $V$, you get $\vec{E}=-\nabla V$ as a result.