# Electrostatic Potential Energy

I have read many books on Mechanics and Electrodynamics and the one thing that has confused me about electrostatic potential energy is its derivation .One of the classical derivations is :

\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]\newln \Rightarrow \int^{r}_{\infty}k\cdot\frac{q.q_o}{r^2}d\mathrm{\mathbf{r}}=-U_r ~~~~~~~ [\textrm{Coulomb's Law}]\newln \Rightarrow kq\cdot q_o\int^{r}_{\infty}\frac{1}{r^2}d\mathbf{r}=-U_r\newln \Rightarrow kq\cdot q_o\left[\frac{-1}{r} \right]^r_\infty=-Ur\newln\Rightarrow \frac{-kq.q_o}{r}=-U_r\newln \Rightarrow U_r=\frac{kq.q_o}{r} \end{align}

My problem is in the fourth line of the derivation. shouldn't there be a $Cos\theta$ term there as it is a dot product.But using it would change the resulting equation to have an extra $Cos \theta$ term being multiplied to it.Why is it not used ? . The direction of the forces have to be defined right?

One more thing - In the derivation it is finally said that the result is sign sensitive but isn't the equation of Coulombic force just for the magnitude?Then how come the result is sign sensitive.

Could Some One give me or direct me to a definition and a derivation that caters to these querrys?

EDIT:

• [I saw the proof on Wikipedia it has some thing on my doubt but doesnt seen to use it.It uses cos theta but i cant seem to understand why they used only $\pi$ and not $2\pi$]

• [I know that $Cos\theta$ can be either +1 or -1 depending on the polarity of the charges and that it will be a constant but doing that will change the result altogether : How? :]

• If in the original result you put similar charges you get a positive result .
• But in this case if you have similar signs then the cos theta would be -1 as the displacement and the force of repulsion would be anti-paralell.
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Related: physics.stackexchange.com/q/17938/2451 and links therein. –  Qmechanic Apr 30 '12 at 14:54
how is it related 'cause there isn't an explicit explanation about the theta factor there. –  The-Ever-Kid Apr 30 '12 at 15:02
The potential energy depends on the charge. The electric potential (or "voltage") is the potential energy per unit charge, so you divide out by the charge, with sign, and you get the same answer whether you use the positive or negative charge to calculate the potential. The work changes sign, but the charge changes sign too, so the work per unit charge is unchanged. –  Ron Maimon May 1 '12 at 21:26

You don't need the $\cos\theta$ because you only need to use the potential function for $\vec F$ to prove the work done between two points is the difference in the potential function. Your derivation should be:

Let $\vec F$ be conservative so that $\vec F = \nabla\phi(\vec r)= \left(\frac {\partial\phi(\vec r)}{\partial x},\frac {\partial\phi(\vec r)}{\partial y},\frac {\partial\phi(\vec r)}{\partial z}\right)$

The applied force does work against $\vec F$ and is therefore $-\vec F$, giving the work done by $-\vec F$ between $a$ and $b$ as $$\int^b_a-\vec F\cdot\vec{dr} =\int^b_a-\nabla\phi(\vec r) \cdot\vec{dr}= \phi (\vec a) - \phi (\vec b)$$

For a Coulomb electric field $\vec E = k \frac {q_1\vec r} {r^3}$, the corresponding $\phi(\vec r) = -k \frac {q_1q_2} {r}+$ const

Defining $\phi(\infty) = 0$ finally gives $$\int^b_\infty-k \frac {q_1q_2\vec r} {r^3}\cdot\vec{dr} = k \frac {q_1q_2} {b}$$

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excellent answer . Love it!!! –  The-Ever-Kid May 1 '12 at 0:47

You are right...! There should be a $\cos\theta$ factor. If you choose a path $\mathcal{P}$ from $\infty$ to $r$, then along this path, you have to evaluate $$\mathbf{F} \cdot d\mathbf{r} = |\mathbf{F}| |d\mathbf{r}| \cos\theta$$ The angle $\theta$ will also depend on where you are on the path compared to the local direction of the force $\mathbf{F}$. True.

But when you learnt that in electrostatics one can define an electric potential, you must also have learnt that this potential is related to the electric field via $$\mathbf{E} = -\nabla U$$ The reason you can write this way is because the electric field is irrotational, i.e., $$\nabla \times \mathbf{E}=0$$ If this makes sense to you, then you must also know that via Stokes theorem the line-integral of $\mathbf{E}$ between two points is independent of the path between those points. Since $\mathbf{F} = q\mathbf{E}$, we have $$\int_A^B \mathbf{F} \cdot d\mathbf{r}$$ is independent of the path chosen for going from $A$ to $B$. So you can just choose a path such that this $\cos\theta$ factor is a constant. For a point charge, $\mathbf{E}$ will be radially directed from the charge. Choose the path along a radial direction from $\infty$ to $r$. Then $\cos\theta = \pm 1$ depending on the sign of the charge and comes out of the integral.

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But you see when i do that the equation becomes $(kCos\theta q q_o)/(r)$ now when the charges are the same then the force and the displacement are opposite to each other so $Cos\pi = -1$ this changes the result itself BTW the angle theta is the angle between the force on the charge and the direction of incremental displacement right? BTW you're Indian too? –  The-Ever-Kid Apr 30 '12 at 15:38
I am confused. Why does the result change? –  Vijay Murthy Apr 30 '12 at 15:53
okay tell me if both the charges are the same then what will cos theta be? –  The-Ever-Kid Apr 30 '12 at 15:54
Are you concerned that there is an ambiguity in the sign of $U(r)$ i.e. that it's positive if the charges are the same but negative if they're different? –  John Rennie Apr 30 '12 at 16:09
no im concerned that the previous results dont hold true when we add the cos theta term to it –  The-Ever-Kid Apr 30 '12 at 16:58

You're measuring the force between two point charges, and the force always acts along the line joining the charges. That means $cos(\theta)$ is always zero or $\pi$ i.e. $cos(\theta)$ is 1 or -1.

The equation for the force is not just the magnitude. The forces between two identical charges acts in a different direction to two opposite charges. You need to put $q$ into the equation with the correct sign and it will give you the force with the correct sign.

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the coulombs law doesn't use the signs of the charges at all .It only tells you the magnitude. The direction is found by physical analysis.And i know the forces act along the line joining them i know cos theta is 1 or -1 for either conditions depending upon the unlike and like condition . i want to know why we dont include it in the integration? –  The-Ever-Kid Apr 30 '12 at 15:12
See en.wikipedia.org/wiki/Coulomb's_law#The_law. Coulomb's law does use the sign of the charge and does give you the direction of the force. Re your second point, why would you include a constant factor of 1 or -1 in the integration? You'd take it outside the integral, or in the case of 1 just ignore it. Remember F and r always point along the same line so $cos(\theta)$ is a constant and doesn't depend on r. –  John Rennie Apr 30 '12 at 15:17
Incorrect the coulombs law states : (read the article carefully) "The magnitude of the Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them." to your second answer i know that the cos theta term would be extracted outside the integral but put that in there and see your final result it would be completely different. –  The-Ever-Kid Apr 30 '12 at 15:21
The magnitude is a scalar, so quite correct, the magnitude is proportional to the scalar product of the charges. However the force is not just a magnitude, it is a magnitude and direction so to calculate it you need the signs of the charges. I suspect we're disagreeing about exactly how Coulomb's law is defined, but I'm sure we both agree that including the signs gives you the direction of the force regardless of whether you call the resulting equation "Coulomb's Law" or not. –  John Rennie Apr 30 '12 at 15:28
We are going round in circles. If you choose a path that is radial then $cos(\theta)$ is a constant, $\pm 1$ and you don't need to include it in the integral, which is what I said right at the outset. –  John Rennie Apr 30 '12 at 15:54