Why is the electric force the negative of the position derivative of EPE?

[lys04]

Homework Statement

As title suggests

Relevant Equations

$$ F=-\frac{dU}{dx}$$

My interpretation of this equation is "how much does the electrical potential energy changes when I move the particle a little bit"? Suppose that I have some arrangement of charges, with Q at x and q at x'. Then I move the charge q a little bit in the direction of Q (dx), it will require some energy for me to do so, assuming they have the same sign (dU), which is precisely equal to the Coulomb force it experiences? As for the negative sign, if dU/dx is positive that means I had to do work to push the charge q a distance of dx closer to the charge Q, which means the Coulomb force is pushing it away, i.e in (-dx) direction, and hence I need the negative sign. Whereas if dU/dx is negative, then this means the electric field did work on the charge as opposed to me doing the work? Which means the force is attractive in this case and is in the same direction as (dx), so the extra negative sign cancels out and I get a force in the (dx) direction.

Now I want to extend this idea using position vectors which is where I get a bit confused.

Suppose I have $U(\vec{x}) = \frac{qq'}{4\pi \epsilon_0 r}$ where $r=|\vec{x}-\vec{x'}|$. Then if I compute $\nabla U(\vec{x}) = \frac{\partial U}{\partial x} \vec{x} + \frac{\partial U}{\partial y} \vec{y} + \frac{\partial U}{\partial z} \vec{z}$ , and compute this at some point (x, y, z), I'm measuring how much force the charge experiences at that point by moving in every direction a little bit? Or say I only want to know what the force in the x direction is at point (x,y,z), then I do $\nabla_x U$ or more generally $\nabla_i U$ for the ith direction?

However I don't understand what's happening when I do $\nabla ' U(\vec{x})$ instead. Can someone enlighten me?
I've drawn a diagram to illustrate what I mean.

 

[Gavran]

Suppose I have $U(\vec{x}) = \frac{qq'}{4\pi \epsilon_0 r}$ where $r=|\vec{x}-\vec{x'}|$. Then if I compute $\nabla U(\vec{x}) = \frac{\partial U}{\partial x} \vec{x} + \frac{\partial U}{\partial y} \vec{y} + \frac{\partial U}{\partial z} \vec{z}$ ,

Are you sure that $\vec x$ in $r=|\vec x-\vec{x'}|$ and $\nabla U(\vec x)$ is the same as $\vec x$ in $$ \frac{\partial U}{\partial x}\vec x+\frac{\partial U}{\partial y}\vec y +\frac{\partial U}{\partial z}\vec z $$?

However I don't understand what's happening when I do $\nabla ' U(\vec{x})$ instead. Can someone enlighten me?
I've drawn a diagram to illustrate what I mean.

What do you mean by $\nabla’$ in your expression $\nabla’ U(\vec x)$?

 

[PhDeezNutz]

Before considering it in 3-dimensions consider a general single variable function

$U\left( \left|x - x’\right| \right)$ and forget about the specific characteristics of what $U$ might be.

Can you see how $\frac{\partial U}{\partial x} = - \frac{\partial U}{\partial x’}$ because of the chain rule?

If we generalize to 3-dimensions $\nabla U = - \nabla’ U$

Where

$\nabla U = \frac{\partial U}{\partial x} \hat{x} + \frac{\partial U}{\partial y} \hat{y} + \frac{\partial U}{\partial z} \hat{z}$

And

$\nabla’ U = \frac{\partial U}{\partial x’} \hat{x} + \frac{\partial U}{\partial y’} \hat{y} + \frac{\partial U}{\partial z’} \hat{z}$

Notice how the basis vectors for both expressions are the non-primed coordinates.

I’m not sure $\nabla’ U$ has a physical interpretation. As far as I can remember the primed derivatives are just a mathematical trick to eventually arrive at the Reciprocity Theorem along with Greens Functions which you will encounter later. In graduate E&M 1 (statics) you will encounter some pretty weird/counterintuitive (but highly interesting and useful) approaches for calculating potentials. Most of which are not readily obvious.

 

[lys04]

Are you sure that x→ in r=|x→−x′→| and ∇U(x→) is the same as x→ in ∂U∂xx→+∂U∂yy→+∂U∂zz→?

I'm not sure what you mean here

What do you mean by ∇′ in your expression ∇′U(x→)?

Taking the gradient with respect to the primed coordinates.

 

[lys04]

Can you see how ∂U∂x=−∂U∂x′ because of the chain rule?

Yes.

I’m not sure ∇′U has a physical interpretation. As far as I can remember the primed derivatives are just a mathematical trick to eventually arrive at the Reciprocity Theorem along with Greens Functions which you will encounter later.

Alrighty. Thank you!

 

[Gavran]

I'm not sure what you mean here

Your notation is confusing. You use the same symbol $\vec x$ for two different things.
$\vec x$ in $r=|\vec x-\vec{x'}|$ and $\vec x$ in $$ \frac{\partial U}{\partial x}\vec x+\frac{\partial U}{\partial y}\vec y +\frac{\partial U}{\partial z}\vec z $$ have two different interpretations.

I would use the next approach for the second part of your explanation in the original post.
The potential energy of the system with two charges $q_1$ and $q_2$ is
$$ U_E=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{|\vec r_1-\vec r_2|} $$ where $\vec r_1$ and $\vec r_2$ are position vectors of charges $q_1$ and $q_2$, respectively. By using $\vec r_1-\vec r_2=(x_1-x_2)\hat x+(y_1-y_2)\hat y+(z_1-z_2)\hat z$ there will be $$ U_E=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}} $$ where $x_1$, $y_1$, $z_1$ and $x_2$, $y_2$, $z_2$ are coordinates of charges $q_1$ and $q_2$ in a three dimensional Cartesian coordinate system, respectively.

By moving the charge $q_1$ a little bit there will be

$\begin{align} dU_E&=\frac{\partial U_E}{\partial x_1}dx_1+\frac{\partial U_E}{\partial y_1}dy_1+\frac{\partial U_E}{\partial z_1}dz_1\nonumber\\ &=-\vec{F_1}\cdot(\hat xdx_1+\hat ydy_1+\hat zdz_1)\nonumber\\ \end{align}$

which means that the charge $q_1$ experiences the force $$ \vec{F_1}=-\frac{\partial U_E}{\partial x_1}\hat x-\frac{\partial U_E}{\partial y_1}\hat y-\frac{\partial U_E}{\partial z_1}\hat z=-\nabla_1U_E $$. The same holds for the charge $q_2$ and there will be $$ \vec{F_2}=-\frac{\partial U_E}{\partial x_2}\hat x-\frac{\partial U_E}{\partial y_2}\hat y-\frac{\partial U_E}{\partial z_2}\hat z=-\nabla_2U_E $$.
The forces $\vec{F_1}$ and $\vec{F_2}$ are equal in magnitude and opposite in direction because $\nabla_1U_E=-\nabla_2U_E$.

By the way, they must be equal in magnitude and opposite in direction because of The third Newton’s law of motion.