2.7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Potential energy of a system of charges \( q_{1} \) and \( q_{2} \) is directly proportional to the product of charges and inversely to the distance between them.
Consider first the simple case of two charges \(q_{1}\)and \(q_{2}\) with position vector \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) relative to some origin. Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges \(q_{1}\) and \(q_{2}\) initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge \(q_{1}\) is brought from infinity to the point \(\mathbf{r}_{1}\). There is no external field against which work needs to be done, so work done in bringing \(q_{1}\) from infinity to \(\mathbf{r}_{1}\) is zero. This charge produces a potential in space given by \[ V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1 \mathrm{P}}} \] where \(r_{1P}\) is the distance of a point \(P\) in space from the location of \(q_{1}\). From the definition of potential, work done in bringing charge \(q_{2}\) from infinity to the point \(\mathbf{r}_{2}\) is \(q_{2}\) times the potential at \(\mathbf{r}_{2}\) due to \(q_{1}\): \[ \text{work done on } q_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{12}} \] where \( r_{12} \) is the distance between points 1 and 2.
Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges \( q_{1} \) and \( q_{2} \) is
\[ U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{12}} \] (2.22)
Obviously, if \( q_{2} \) was brought first to its present location and \( q_{1} \) brought later, the potential energy \( U \) would be the same. More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force.
Equation (2.22) is true for any sign of \(q_{1}\) and \(q_{2}\). If \(q_{1}q_{2} > 0\), potential energy is positive. This is as expected, since for like charges (\(q_{1}q_{2} > 0\)), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges (\(q_{1} q_{2} < 0\)), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative.
Equation (2.22) is easily generalised for a system of any number of point charges. Let us calculate the potential energy of a system of three charges \(q_{1}\), \(q_{2}\) and \(q_{3}\) located at \(\mathbf{r}_{1}\), \(\mathbf{r}_{2}\), \(\mathbf{r}_{3}\), respectively.
Potential energy of a system of three charges is given by Eq. (2.26), with the notation given in the figure.
To bring \( q_{1} \) first from infinity to \(\mathbf{r}_{1}\), no work is required. Next we bring \(q _{2}\) from infinity to \(\mathbf{r}_{2}\). As before, work done in this step is
\[ q_{2} V_{1}\left(\mathbf{r}_{2}\right)=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{12}} \] (2.23)
The charges \( q_{1} \) and \( q_{2} \) produce a potential, which at any point \( P \) is given by
\[ V_{1,2}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r_{1 \mathrm{P}}}+\frac{q_{2}}{r_{2 \mathrm{P}}}\right) \] (2.24)
Work done next in bringing \(q_{3}\) from infinity to the point \(\mathbf{r}_{3}\) is \(q_{3}\) times \(V_{1,2}\) at \(\mathbf{r}_{3}\)
\[ q_{3} V_{1,2}\left(\mathbf{r}_{3}\right)=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right) \] (2.25)
The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)],
\[ U=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right) \] (2.26)
Again, because of the conservative nature of the electrostatic force (or equivalently, the path independence of work done), the final expression for \( U \), Eq. (2.26), is independent of the manner in which the configuration is assembled. The potential energy is characteristic of the present state of configuration, and not the way the state is achieved.
Example 2.4
Four charges are arranged at the corners of a square \( ABCD \) of side \( d \), as shown in Fig. 2.15.
(a) Find the work required to put together this arrangement.
(b) A charge \( q_{0} \) is brought to the centre \( E \) of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at \( A \), \( B \), \( C \) and \( D \). Suppose, first the charge \( +q \) is brought to \( A \), and then the charges \( -q \), \( +q \), and \( -q \) are brought to \( B \), \( C \) and \( D \), respectively. The total work needed can be calculated in steps:
- Work needed to bring charge \( +q \) to \( A \) when no charge is present elsewhere: this is zero.
-
Work needed to bring \( -q \) to \( B \) when \( +q \) is at \( A \). This is given by
(charge at \( B \)) × (electrostatic potential at \( B \) due to charge \( +q \) at \( A \)) \[ =-q \times\left(\frac{q}{4 \pi \varepsilon_{0} d}\right)=-\frac{q^{2}}{4 \pi \varepsilon_{0} d} \] - Work needed to bring charge \(+q\) to \(C\) when \(+q\) is at \(A\) and \(–q\) is at \(B\). This is given by (charge at \(C\)) × (potential at \(C\) due to charges at \(A\) and \(B\)) \[ \begin{array}{l} =+q\left(\frac{+q}{4 \pi \varepsilon_{0} d \sqrt{2}}+\frac{-q}{4 \pi \varepsilon_{0} d}\right) \\ =\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left(1-\frac{1}{\sqrt{2}}\right) \end{array} \]
- Work needed to bring charge \(-q\) to \(D\) when \(+q\) is at \(A\), \(–q\) is at \(B\) and \(+q\) is at \(C\). This is given by (charge at \(D\)) × (potential at \(D\) due to charges at \(A\),\(B\) and \(C\)) \[ \begin{array}{l} =-q\left(\frac{+q}{4 \pi \varepsilon_{0} d}+\frac{-q}{4 \pi \varepsilon_{0} d \sqrt{2}}+\frac{q}{4 \pi \varepsilon_{0} d}\right) \\ =\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left(2-\frac{1}{\sqrt{2}}\right) \end{array} \]
The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
(Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)
(b) The extra work necessary to bring a charge \(q_{0}\) to the point \(E\) when the four charges are at \(A\), \(B\), \(C\) and \(D\) is \(q_{0}\) × (electrostatic potential at \(E\) due to the charges at \(A\), \(B\), \(C\) and \(D\)). The electrostatic potential at \(E\) is clearly zero since potential due to \(A\) and \(C\) is cancelled by that due to \(B\) and \(D\). Hence, no work is required to bring any charge to point \(E\).