Understanding Electrical Potential Energy of a charge distribution

[iochoa2016]

I quite understand the fact the EPE (Electrical Potential Energy) of a system of two charges are U = k*qQ/r, Q is fix. however when it comes to three charges i get lost. because my reasoning is :
if q1 is fix then the EPE of the system when q2 is brought is U2 = k*q1*q2/r12, when q3 is brought near the previous system, the U3 = k*q1*q3/r13+k*q2*q3/r23. all good till here

but, I should change the U of the previous system , q1 is fix, but q2 now is affected by q3 so U2' = k*q1*q2/r12+k*q2*q3/r23 , then total energy:

Utotal = k * ( q1*q2/r12 + q2*q3/r23 + q1*q3/r13 + q2*q3/r23 )

but if it is compared to book's formula , this result is wrong. Why? what I am missing? I know U is for the system and not for a particular charge.

I have seem also, the derivation of the right formula using potentials (V), which follows the structure above but with more combination including q1, but they apply the term 1/2 which get rid of the factor 2 that appears when V1+V2+V3 is added up., books say that U = 1/2 (V1+V2) for a system of two changes so U is shared, BUT for three changers 1/2 is still use. why not 1/3 as sharing factor for 3 charges?

 

[vanhees71]

This is a bit subtle. Since you deal with point charges you have to leave out the self-energies of these charges, because they are undefined (divergent). For energy balances you don't need this constant but singular contribution anyway.

That said the argument is as follows: You want to calculate the mechanical potential energy needed to put the charges from infinity (by definition one chooses the arbitrary additive constant such that the energy is 0 when all charges are at infinity). To move the 1st charge from infinity to its place $\vec{r}_1$ you need no energy at all, because there are no forces acting on the charge. To move the 2nd charge from infinity to its place $\vec{r}_2$ (assuming $\vec{r}_2 \neq \vec{r}_1$, because otherwise you get infinity again!). For this you need an energy (I assume the Coulomb potential for the field of a point charge to be known)
$$\Delta E_1=\frac{Q_1 Q_2}{4 \pi \epsilon_0 |\vec{r}_1-\vec{r}_2|}.$$
Now take the 3rd charge and transport it from infinity to $\vec{r}_3$. For this you need work due to the interaction with both charges $Q_1$ and $Q_2$,
$$\Delta E_2=\frac{Q_1 Q_3}{4 \pi \epsilon_0 |\vec{r}_1-\vec{r}_3|} + \frac{Q_2 Q_3}{4 \pi \epsilon_0 |\vec{r}_2-\vec{r}_3|}.$$
Now you go on with this thought experiment until you have brought all $N$ charges to their positions. The final answer is
$$E_{\text{tot}}=\sum_{i<j} \frac{Q_i Q_j}{4 \pi \epsilon_0 |\vec{r}_i-\vec{r}_j|} =\frac{1}{2} \sum_{i \neq j}\frac{Q_i Q_j}{4 \pi \epsilon_0 |\vec{i} -\vec{r}_j|} .$$
You need the factor $1/2$ in the latter expression to compensate for double counting each pair of charges.

 

[etotheipi]

An easy way to think of it is that the potential energy of the system associated with certain conservative forces internal to the system is the negative of the total work done by those forces to configure the system from an arbitrarily defined $0$ potential energy state. Here we only deal with two body forces (like $\vec{F}_{ij} = -\vec{F}_{ji}$), so the potential energy goes like $$\begin{align*}U &= -(W_{12} + W_{21} + W_{13} + W_{31} + W_{23} + W_{32}) \\
&= -[W_{12} + W_{21}]- [W_{13} + W_{31}] - [W_{23} + W_{32}] \\
&= U_{12} + U_{13} + U_{23}
\end{align*}$$ To give some justification, we say any two particles interacting ($i, j$ s.t. $i\neq j$), in a system interacting via. only two body forces, contributes a term $U_{ij} = U(\vec{r}_i, \vec{r}_j)$ where $\vec{F}_{ij} = -\vec{F}_{ji} = -\nabla_{\vec{r}_i} U_{ij} = \nabla_{\vec{r}_j} U_{ij}$. You will notice that the total internal work done by the two forces in any given pair $$W_{ij} + W_{ji} = \int \vec{F}_{ij} \cdot d\vec{r}_i + \int \vec{F}_{ji} \cdot d\vec{r}_j = -\Delta U_{ij}$$because $$\begin{align*}
dU_{ij}(x_i, y_i, z_i, x_j, y_j, z_j) &= \left( \frac{\partial U_{ij}}{\partial x_i} dx_i + \frac{\partial U_{ij}}{\partial y_i} dy_i + \frac{\partial U_{ij}}{\partial z_i} dz_i \right) + \left( \frac{\partial U_{ij}}{\partial x_j} dx_j + \frac{\partial U_{ij}}{\partial y_j} dy_j + \frac{\partial U_{ij}}{\partial z_ij} dz_j \right) \\ \\

dU_{ij} &= \nabla_{\vec{r}_i}U_{ij} \cdot d\vec{r}_i + \nabla_{\vec{r}_j}U_{ij} \cdot d\vec{r}_j \\ \\

\Delta U_{ij} &= \int \nabla_{\vec{r}_i}U_{ij} \cdot d\vec{r}_i + \int \nabla_{\vec{r}_j}U_{ij} \cdot d\vec{r}_j = -\int \vec{F}_{ij} \cdot d\vec{r}_i - \int \vec{F}_{ji} \cdot d\vec{r}_j = -(W_{ij} + W_{ji})

\end{align*}$$ As @vanhees71 describes, the total potential energy is $$U = \frac{1}{2} \sum_{i\neq j} U(\vec{r}_i, \vec{r}_j)$$ and by extension, the total conservative force on any given particle is $\vec{F}_i = -\nabla_{\vec{r}_i} U$ which follows from the linearity of the $\nabla$ operator.

 

[iochoa2016]

Hi Thanks for your answer, great explanation but I still want to ask:

When charge 1 is brought I guess we take this as a reference, so we stick this charge on position $\vec{r}_1$.
when charge #2 is brought to position $\vec{r}_2$, how it is held in that position?, I visualize it that an external force is applied to cancel out the Electrical force, and set $\vec{v} = 0$, if I say a finger provides that external force then when charge #3 is brought close to the system, my finger will feel an extra push from the 3 er charge but charge # 2 never moves, however, the finger has to apply more force to keep $\vec{v} = 0$ , work=0 and Energy change =0. so then I can say that he energy of the ONLY two particle system remains constant, never change even with the presence of the 3er charge?.

 

[etotheipi]

Hi Thanks for your answer, great explanation but I still want to ask:

When charge 1 is brought I guess we take this as a reference, so we stick this charge on position r→1.
when charge #2 is brought to position r→2, how it is held in that position?, I visualize it that an external force is applied to cancel out the Electrical force, and set v→=0, if I say a finger provides that external force then when charge #3 is brought close to the system, my finger will feel an extra push from the 3 er charge but charge # 2 never moves, however, the finger has to apply more force to keep v→=0 , work=0 and Energy change =0. so then I can say that he energy of the ONLY two particle system remains constant, never change even with the presence of the 3er charge?.

You're correct that when we speak of "holding" something in place there must be an external force on the that static charge that exactly balances the net internal electric force on it.

However, we can ignore these details since the only thing that matters when computing the change in potential energy is the work done by the internal conservative force.

Likewise, when we "move" a charge, it doesn't matter what external force we apply to it since the work done by the electric force will be exactly the same so long as the start and end points of the path are the same.

 

[jbriggs444]

how it is held in that position?

The pithy answer is that "we do not care".

Maybe we nail it to a piece of [insulating] plywood. However we choose to hold it in place, no work is required to hold it motionless.