Electrostatic Potential Energy Derivation

Question

How is the boxed step , physically as well as mathematically justified and correct ?

Source:Wiki http://en.wikipedia.org/wiki/Electric_potential_energy

As work done = $- \Delta U $. for Conservative force and it shouldn't matter whether we take $ds$ or $-dr$ ?

And when $dr$ is just a notation to specify the variable and the real thing behind it , is a limit , why is it that $dr$ is so important here .

Image : http://www.artofproblemsolving.com/Forum/download/file.php?id=43358&mode=view

What is wrong here ?

$$\newcommand{\newln}{\\&\quad\quad{}} \begin{align}&\int^{r_b}_{r_a}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_a-U_b) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_r-U_\infty) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}} =-U_r ~~~~~~~ [U_\infty = 0], \cos\theta=-1 , \vec{A} \cdot \vec{B}=|A||B|\cos\theta\newln \Rightarrow -\int^{r}_{\infty}k\cdot\frac{q.q_o}{r^2}dr=-U_r ~~~~~~~ [\textrm{Coulomb's Law}]\newln \Rightarrow kq\cdot q_o\int^{r}_{\infty}\frac{1}{r^2}dr=U_r\newln \Rightarrow kq\cdot q_o\left[\frac{-1}{r} \right]^r_\infty=U_r\newln\Rightarrow \frac{-kq.q_o}{r}=U_r\newln \Rightarrow U_r=-\frac{kq.q_o}{r} \end{align} $$

Answer 1

When you calculate work, you do so along a given path. Here, that path has tangent vector $d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward.

Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate $r$ to parameterize the path, in that $r$ is larger at the start of the path and smaller at the end. This runs counter to what you would usually do when parameterizing such a path with an arbitrary parameter.

Let $\mathbf s_0$ and $\mathbf s_1$ be the starting and ending points of a path $\mathbf s(\lambda) = \mathbf s_0 + (\mathbf s_1 -\mathbf s_0)\lambda$. The work integral is then

$$W = \int_{\mathbf s_0}^{\mathbf s_1} \mathbf F(\mathbf s) \cdot d\mathbf s= \int_0^1 \mathbf F(\mathbf s(\lambda)) \cdot \frac{d\mathbf s}{d\lambda} \, d\lambda = \int_0^1 \mathbf F (\mathbf s(\lambda))\cdot (\mathbf s_1 - \mathbf s_0) \, d\lambda$$

For two finite points, the basic approach is sound, but it breaks down when you have a point at infinity involved. This is the reason that the problem of assembling a configuration is usually attacked with a different basic parameterization.

Instead, set $\mathbf s(\lambda) = \lambda \hat{\mathbf a}$ for some unit vector $\hat{\mathbf a}$ and set the bounds of the integral as being from $[\infty, R)$. This is the important point: even though the path is being traversed coming in from infinity, the parameterization means that $d\mathbf s/d\lambda = + \hat{\mathbf a}$, not minus as I originally thought. The path's still oriented outward; we're just traversing it backwards.

Here's how that integral looks:

$$W = \int_{\infty}^R \mathbf F(\lambda \hat{\mathbf a}) \cdot \hat{\mathbf a} \, d\lambda$$

Of course, we know the expression for the electric force:

$$\mathbf F(\mathbf r) = k\frac{qq_0 \mathbf r}{|r|^3}$$

Plug in $\mathbf r = \lambda \hat{\mathbf a}$ to get

$$\mathbf F(\lambda \hat{\mathbf a}) = k \frac{qq_0 \lambda \hat{\mathbf a}}{\lambda^3} = k \frac{q q_0 \hat{\mathbf a}}{\lambda^2}$$

We find that the integrand is then

$$W = \int_{\infty}^R k \frac{q q_0}{\lambda^2} \hat{\mathbf a} \cdot \hat{\mathbf a} \, d\lambda = \int_\infty^R k \frac{q q_0}{\lambda^2} \, d\lambda = - k \frac{q q_0}{R} < 0$$

The work is negative, so the change in potential energy $\Delta U = - W$ is positive as required.

So where is the problem then? As we've seen, there actually shouldn't be an extra negative sign coming in on line 4 (as posted in the OP's question). This is somewhat obscured because an explicit parameterization of the path is never written down in the first place--usually, you don't have to, but this problem is tricky enough that it helps immensely.

Answer 2

New version

The problem in your demonstration is when you write down $\vec{A}\cdot\vec{B} = ||\vec{A}||\,||\vec{B}||\,\cos\theta$. More exactly, in your case $||d\vec{r}||\neq dr$ because $dr<0$ when you go from $\infty$ to $r$ and a norm is positive by definition. So the sign error is introduced from 3rd to 4th line.

Old version

The demonstration on wikipedia is ill-defined on several places. In particular, when they note

$$\vec{F}\cdot d\vec{s} = |F|\,|ds|\,\cos(\pi) = -F\,ds$$

the writer is assuming that $ds>0$ and also $F=qQ/(4\pi\varepsilon_0r^2)>0$, the latter being wrong in the cases where $qQ<0$, that is the two charges have different signs... Whereas it needs a negative $dr$ in the next section because it's coming from the infinity which cause you so much trouble.

Anyway, let's do it right. The electrical force is written in spherical coordinates $(r,\theta,\varphi)$ in base $(\vec{e_r},\vec{e_\theta},\vec{e_\varphi})$ as $$ \vec{F} = \frac{Qq}{4\pi\varepsilon_0\,r^2}\,\vec{e_r} $$

In those coordinates, the elementary displacement would be $$ d\vec{s} = dr\,\vec{e_r} + rd\theta\,\vec{e_\theta} + r\sin\theta\,d\varphi\,\vec{e_\varphi} $$

where $dr$, $d\theta$ and $d\varphi$ could be positive or negative depending where you would like to go. Thus, the dot product gives $$ \vec{F}\cdot d\vec{s} = \frac{Qq}{4\pi\varepsilon_0\,r^2}\,\vec{e_r} \cdot (dr\,\vec{e_r} + rd\theta\,\vec{e_\theta} + r\sin\theta\,d\varphi\,\vec{e_\varphi}) = \frac{Qq}{4\pi\varepsilon_0\,r^2}\,dr $$ which was what you were looking for. You can then integrate as specified.

Answer 3

$\mathbf{r}$ is a position vector and $\mathbf{s}$ is a displacement vector between two points, let say A and B. In general case, they are not equal, but they can be if we properly choose the origin of the coordinate system: A={0,0,0} or B={0,0,0} The sign depends on at which point A or B the origin is placed.

Answer 4

Just to be clear, the potential energy of a particle of charge $q_2$ at a distance $r$ from a source of potential (supposidely at zero) of charge $q_1$ is the work that an external operator has to provide to bring the particle from infinity to $r$ at constant velocity. This reads then:

$\int_{\infty}^r \vec{F}_{op}\cdot \vec{ds}$

As people have said, the only part of $\vec{ds}$ that will contribute to the work is the radial one in our case and since we go from infinity to $r$ then $\vec{ds}\cdot \hat{u}_r = dr$ (this was my mistake before...the integral boundaries are already taking into account the fact that I am going backward, convention in integral calculations have been made this way).

Now, $\vec{F}_{op}$ has to exactly oppose the force exerted by the source charge on our particle otherwise the constant velocity constraint won't be satisfy. We thus have:

$\vec{F}_{op} = -\frac{q_1q_2}{4 \pi \epsilon_0 r^2}\hat{u}_r $

When I consider the scalar product of the force generated by the external operator and the working direction along the path, I end up with:

$U(r)-U(\infty) = -\int_{\infty}^{r} dr' \:\frac{q_1q_2}{4 \pi \epsilon_0 r'^2}= \frac{q_1q_2}{4 \pi \epsilon_0 r}$

Answer 5

I think I've understood it now .

$ds=dr$ . but $dr<0$ and $|dr|=-dr$ Because dr is a small position vector and position vector is directed along field .

Now why I can't use ds directly is because the limits in the integral , (the upper and lower limit in integral notation) are in terms of position vector and not the displacement .

Had they been in terms of displacement vector , the limits would have been $0$ to $\infty - R$ (not trying to enter into debate about $\infty$, one can as well imagine coming from $1000$ m to $1$m, your displacement is $0 $to $999$ , but in terms of integral in position vector, you will put limits as $1000$ to $1$ .

But they are from $\infty $ to $ r $ , now simple use the dot product definition and the correct answer will follow .

Hence we used $\vec{r}$ , since integration wasn't possible in terms of $ds$ as you can't put appropriate integral limits .

Suppose it was even the case of an infinite sheet ,where force is constant everywhere, you could never still use the displacement vector directly as what limits you will put in the integral then, so you've to express displacement in terms of position vectors here , it is must .

So in the end thr problem is you can't choose origin at infinity . And rest follows from the correct answer by JJ Fleck