# Total Electrical potential energy of two particle system

I recently have been studying Electro-statics and I couldn't understand properly how the potential energy of two particle system is found.

Suppose you have two particles with charges $Q_1$ and $Q_2$ respectively. The distance between them is $r$. What is the total electrical potential energy of the system comprising of the two particles? Well, I know the answer. I want to know the reasons behind the answer.

My book says "Fix one of the charges and then bring the other from infinity. Hence we get the answer. I am not satisfied with the answer, as I saw in Wikipedia they find the potential of each other with respect to the other and then add and divide by $2$. Why is this taking potential with respect to each other valid? Why do they divide by two? What does taking potential with respect to other mean? Do they fix one charge and bring another one?

what about three charges $Q_1, Q_2$ and $Q_3$ which are situated in the vertices of an equilateral triangle with side length $r$?

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Can you add a link to the Wikipedia article –  John Rennie Mar 30 '14 at 7:54

As far as I remember, that $\frac{1}{2}$ comes because $Q_1$ isn't just in a potential $\varphi_2(\mathbf{r})$ that just happens to be there, but is being created by another charge, $Q_2$. Taking potential with respect to the other means: let $\varphi_1(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q_1}{|\mathbf{r} - \mathbf{r}_1|}$ potential generated by the 1st charge and $\varphi_2(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q_2}{|\mathbf{r} - \mathbf{r}_2|}$ the potential of the 2nd charge. When you calculate the energy, you have to "put" each charge in the field of the other, $E = Q_2\varphi_1(\mathbf{r}_2) + Q_1\varphi_2(\mathbf{r}_1)$, but because you actually have to put them into eachother's field simultaneously you have to multiply this by $1/2$ -- because all the present charges are creating those fields simultaneously. So for 3 charges it's $$E = \frac{1}{2}\big[Q_2\varphi_1(\mathbf{r}_2) + Q_3\varphi_1(\mathbf{r}_3) + Q_1\varphi_2(\mathbf{r}_1) + Q_3\varphi_2(\mathbf{r_3}) + Q_1\varphi_3(\mathbf{r}_1) + Q_2\varphi_3(\mathbf{r}_2)\big]$$ Note: I used bold font to denote vectors, so $\mathbf{r} = (x,y,z)$.

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If a positive charge $Q_1$ is fixed at some point in space and another positive charge $Q_2$ is brought close to it, it will experience a repulsive force and will therefore have potential energy.

Now to find this potential energy, we assume that the potential energy with respect to that point at infinity is $0$. So if a test charge $q$ is brought from infinity to a distance $r$ from $Q_1$, the work done in doing so is said to be the potential energy of $Q_1$ which is equal to $$U=k\frac{qQ_1}{r}$$ If you want to find the Total Electric Potential Energy of a system with charges $Q_1$ and $Q_2$ with respect to a test charge $q$, you just add the individual potential energies, so it will be $$U_t=kq(\frac{Q_1}{r_1}+\frac{Q_2}{r_2})$$ where $r_1$ and $r_2$ are the respective distances. Remember, this is not the potential energy of the charges with respect to each other but with respect to a test charge, which when taken as a unit test charge will be the Electric Potential of the system.

I think you are mistaken, it is never divided by $2$ for any reason. That might be the magnitude of the test charge.

For charges in a triangle, just add the three individual potential energies of each side.

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The electric potential energy of a system of charges is defined as the energy stored in the electric field between the charges that make up the system.

Consider a positive charge $Q$ $C$ $fixed$ at a point. The electric field lines point away from the charge. Now see what happens when I bring a unit positive charge $q$ ( where $q$ $= +1$ $C$ ) $slowly$ from $\infty$ to a point which is at a distance $r$ from $Q$. At $\infty$, $\vec{E} = 0$ and thus the two charges do not affect each other.

Since both the charges are positive, $q$ will experience an electric force of repulsion( increasing in magnitude as a result of the inverse distance squared relationship ) as you keep moving it $slowly$ towards $Q$. This force tends to prevent $q$ from approaching $Q$( look at the intermediate position ).

In order to overcome the force of repulsion and move $q$ towards $Q$ until you reach the point, you need to do work on charge $q$. In this case, you need to spend energy and since $\vec{E}$ is a conservative field( also assuming that the non-conservative forces are absent ), any amount of energy that you spend gets stored in the system of these two charges in the form of electric potential energy which is present in the electric field between these two charges. The energy goes nowhere but remains in the system of two charges. The mechanical energy as delivered by you transforms into the electric potential energy. Now, you may ask me a question 'Since $q$ is moved towards $Q$, won't there be Kinetic energy( $K.E$ ) involved?'. If you have observed, I have used the word $slowly$ while describing the motion of $q$. That was used in order to ignore any changes in $K.E$ so that all the energy spent gets converted to Electric potential energy. So, $\Delta K.E \approx 0$. So in doing work, you have lost energy and this energy gets stored in the form of electric potential energy of the system of two charges and the system is said to have gained energy. So, energy is conserved.

You can have a system of many charges. Now, to the system of two charges, I bring another positive charge $q1$. Now, you have to do work in order to overcome the force of repulsion due to $Q$ and $q$ and move it $slowly$ towards both $Q$ and $q$. You need to invest more energy in order to assemble the charges and thus the electric potential energy of the system of three charges will also be more.

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The potential energy is the energy required (or work done) to pick up one of the charges from infinitely far away, and push it towards the other particle to the distance you want. Since there is a force acting on these particles (either positive or negative, depending on their relative charges) it takes energy to move them towards or away from each other.

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