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Could you please explain using just words why electric the field is potentially? I know the proof using integral: $$A = \int_{12}q\vec{E}\cdot{d}\vec{r} = qQ\int_{12}\frac{\vec{r}\cdot{d}\vec{r}}{r^3} = qQ\int_{r_1}^{r_2}\frac{dr}{r^2} = qQ\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$ (so the work depends only on the coordinates of the beginning and the end). I can hardly understand it and I want to know how to prove it using only words (maybe with simple school math).

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2 Answers 2

First you have to ask yourself, for what kind of force you can define a scalar potential. The short answer is a conservative force. Basically, a conservative force is a force for which the work done in moving a particle between any two given points in path independent. From this, it is obvious that for a particle traveling in a closed path, the total work will be zero. For such a conservative force, you can give some numerical value for the potential at any point. So when you move your particle from point A to point B, the force changes the potential energy by a given value independent of the path taken.

As a counterexample, take the force of friction. Being a dissipative force (i.e. not a conservative one) you cannot define for it a scalar potential. The potential differences would not be equal if you take two different paths connecting the same points A and B.

Now lets use a little math to see what conditions must a force $F$ satisfy to be a conservative one:

$$\nabla\times\vec{F}=0$$

$$W=-\oint_c\vec{F}\cdot d\vec{r}$$

$$\vec{F}-\nabla\Phi$$

If your force $F$ satisfies these three conditions, then $F$ is a conservative one. Now, from the gradient theorem it follows that

$$\int_{c[A,B]}\nabla\Phi(\vec{r})\cdot d\vec{r}=\Phi(B)-\Phi(A)$$

In words, the gradient theorem says that line integrals through a irrotational vector field is path independent. So, if you want to use just words you can simply say this: "for any conservative forces, the potential difference depends only on the end points because the work done to move a test particle between said end points is path independent". But focus on the word conservative.

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Ok, let's consider $\vec{r} \vec{dr}$, it is equal to $|\vec{r}||\vec{dr}|_\vec{r}$ where $|\vec{dr}|_\vec{r}$ is projection of $\vec{dr}$ on $\vec{r}$. If you draw $\vec{r}$ and a small (remember, you need infinitesimal!) $\vec{dr}$ you will notice that this projection is actual equal to $|\vec{dr}|_\vec{r} = d|\vec{r}|$, so $\vec{r} \vec{dr} = |\vec{r}|d|\vec{r}|$. This is why you change $\vec{r}\vec{dr} = rdr$. No matter what $\vec{dr}$ you choose, while $|\vec{dr}|_\vec{r}$ is the same you get the same value for $\vec{r} \vec{dr}$.

Electric field of a point charge in static case is $\vec{E} = f(r)\vec{r}$, where $f(r)=Q/r^3$, so you see that work $\vec{E} \vec{dr}$ depends only on the change in $|\vec{r}|$ and does not depend on the integration path.

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