Question

Determine the electric potential energy for a system of three-point charges.

Answer 1

Electric Potential Energy of a System of Charges
Electric potential energy of a system of charges is equal to the amount of work done in forming the system of charges by bringing them at their particular positions from infinity without any acceleration and against the electrostatic force. It is denoted by U.$U=W=qV\left(r\right)$
(a) Electric potential energy of system of two charges:
Suppose two charges $+q_1$ and $+q_2$ are situated at a distance $r$. We have to find out the electrical potential energy of this system.
When the charge $q_1$ is brought from infinity to in its position, no work is done because there is no other charge to repel or attract it. The electric potential due to this charge $q_1$ at a distance i.e., in the position of charge $q_2$
$V_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r}$
Therefore the work done in bringing the charge $q_2$ from infinity to its own position i.e., the electric potential energy of the system.
$W=U=V_1{q}_2$
or $\cup =\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}\dots$(1)
If both the charges are of same nature, the potential energy will be positive and for unlike charges it will be negative. Therefore during the process of evaluation of potential energy, the value of charge should be used with its sign.
(b) Electric potential energy of a system of three charges:
Let us first of all consider a system of three point charges $q_1,q_2$ and $q_3$. Such that $q_2$ and $q_3$ are initially at infinite distance from the charge $\left(q_1\right)\text{[In figure}3.18\left(a\right)]$, Work done to bringing the charge $q_2$ from infinity to point $B$
$W_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{12}}\dots$(2)
Work done bringing charge $q_3$ (In the presence of charge $q_1$ and $q_2$ ) from infinity to point $C$
$\begin{array}{cc}{W}_{123}& =V_1{q}_3+V_2{q}_3\\ & =\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_3}{r_{31}}+\frac{1}{4\pi {\epsilon }_0}\frac{q_2{q}_3}{r_{23}}\dots \left(3\right)\end{array}$
Now the total work done to bringing all the three charges from infinity to their respective position is given by
$\begin{array}{cc}W& =W_{12}+W_{123}\\ & =\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{12}}+\frac{1}{4\pi {\epsilon }_0}+\frac{q_1{q}_3}{r_{31}}+\frac{1}{4\pi {\epsilon }_0}\frac{q_2{q}_3}{r_{23}}\\ & =\frac{1}{4\pi {\epsilon }_0}[\frac{q_1{q}_2}{r_{12}}+\frac{q_1{q}_3}{r_{31}}+\frac{q_2{q}_3}{r_{23}}]\end{array}$
This work done is stored in the form of potential energy
$\therefore U=W=$ potential energy of three system of
$=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1{q}_2}{r_{12}}+\frac{q_1{q}_3}{r_{31}}+\frac{q_2{q}_3}{r_{23}}]$
or $U=\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^3\sum _{j=1,i\ne j}^3\frac{q_i{q}_j}{r_{ij}}\right]$
(c) Electric potential energy due to four system of charges:
Suppose there are four charges in a system of charges, situated as shown in figure $(3.18$ (b)).
The total energy of this system can be obtained by adding the potential energies of each possible pair.
Hence $U=U_{12}+U_{23}+U_{34}^{\prime }+U_{41}+U_{13}+U_{24}$
$=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_3{q}_4}{r_{34}}+\frac{q_4{q}_1}{r_{41}}+\frac{q_1{q}_3}{r_{13}}+\frac{q_2{q}_4}{r_{24}}]$
Similarly, the electrical potential energy of the system of n charges can be obtained. We can write theequivalent expression as under,
$U=\frac{1}{4\pi {\epsilon }_0}\sum _{\text{All pairs}}\frac{q_i{q}_j}{r_{ij}}\dots$(4)
To obtain this sum, we have to use each pair of charge only once. Therefore the general formula forelectrical potential energy will be as under
$U=\frac{1}{2}\sum _{i=1}^n\sum _{j=1,i\ne j}^n\frac{1}{4\pi {\epsilon }_0}\frac{q_i{q}_j}{r_{ij}}$
Here we have to multiply the expression by $\frac{1}{2}$ because in this expression each pair comes two times.