Energy as work to assemble charges
Consider two charges $q_1$ and $q_2$ initially at infinity. Bring $q_1$ to its position first. Since no field is present yet, the work done is zero. The charge $q_1$ now sets up a potential throughout space. Bringing $q_2$ in from infinity therefore requires work against the field of $q_1$, equal to $q_2$ times the potential at $q_2$'s location due to $q_1$. Because the electrostatic force is conservative, this work is stored as the potential energy of the configuration.
Two consequences follow directly, and both are tested. First, the energy is reckoned relative to the charges being infinitely far apart, where $U = 0$. Second, the result is independent of the order in which the charges are brought in and of the path taken — it depends only on the final arrangement. This is what NCERT means when it says the energy "is characteristic of the present state of configuration, and not the way the state is achieved."
Step 1 brings $q_1$ in for free; step 2 does work $q_2 V_1$ against the field of $q_1$, which is stored as $U$.
Potential energy of two charges
The potential at $q_2$'s location due to $q_1$ is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{12}}$, so the stored energy is $q_2 V$. NCERT Eq. (2.22) gives the standard result:
$$U = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r_{12}}$$
where $r_{12}$ is the separation between the two charges. The charges are inserted with their signs. The expression is valid for any combination of signs, and the constant $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2\,\text{C}^{-2}$.
Three or more charges: sum over pairs
Generalising is mechanical. To assemble three charges, bring $q_1$ (no work), then $q_2$ against $q_1$, then $q_3$ against both $q_1$ and $q_2$. Adding the steps gives NCERT Eq. (2.26):
$$U = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$$
Each term is the interaction energy of one distinct pair. A three-charge system has three pairs; a four-charge system has six. In general a system of $n$ charges has $\dfrac{n(n-1)}{2}$ pairs, and the total energy is the sum over all of them.
Each side of the triangle contributes one pair term to the sum.
Sum over PAIRS, not over charges
A frequent error is writing one term per charge. The energy lives in the interactions, so you count pairs. Three charges give three terms ($r_{12}, r_{13}, r_{23}$); four charges give six terms, not four.
Number of terms $= \dfrac{n(n-1)}{2}$. For $n=3$: three terms. For $n=4$: six terms.
Four charges $+q,\ -q,\ +q,\ -q$ sit at the corners A, B, C, D of a square of side $d$ (NCERT Example 2.4). Find the energy to assemble them.
There are six pairs. The four sides (AB, BC, CD, DA) are unlike-charge pairs at distance $d$, each contributing $-\dfrac{kq^2}{d}$. The two diagonals (AC, BD) are like-charge pairs at distance $d\sqrt{2}$, each contributing $+\dfrac{kq^2}{d\sqrt{2}}$.
$$U = \frac{kq^2}{d}\left(-4 + \frac{2}{\sqrt{2}}\right) = -\frac{kq^2}{d}\left(4 - \sqrt{2}\right) = -\frac{q^2}{4\pi\varepsilon_0\,d}\left(4 - \sqrt{2}\right)$$ The negative sign reflects that the attractive (unlike) pairs dominate.
Every $U$ here is a charge times a potential. If $V$ feels shaky, revisit electrostatic potential first.
A single charge in an external field
Section 2.7 dealt with charges that produce their own field. Section 2.8 asks a distinct question: what is the energy of a charge in an external field $E$ (with potential $V$) produced by some other sources we are not accounting for? Bringing $q$ from infinity to a point where the external potential is $V(\mathbf r)$ does work $qV(\mathbf r)$, which is stored as energy. NCERT Eq. (2.27):
$$U = q\,V(\mathbf r)$$
There is only one charge here, so there is no $1/r$ mutual term — just the charge times the local external potential. This is the same relation that defines $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$, the work done on an electron accelerated through one volt.
Single charge in a field is qV, not k q₁q₂/r
The $\dfrac{kq_1q_2}{r}$ formula is for two charges interacting with each other. For one charge sitting in an external field, the energy is simply $U = qV$, where $V$ is the externally supplied potential at that point. Mixing the two is a classic blunder.
One charge in an external field: $U = qV$. Two charges interacting: $U = \dfrac{kq_1q_2}{r_{12}}$.
Two charges in an external field
When two charges $q_1$ and $q_2$ are both placed in an external field, the total energy has three contributions: each charge's energy in the external field, plus the mutual interaction between them. NCERT Eq. (2.29):
$$U = q_1 V(\mathbf r_1) + q_2 V(\mathbf r_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$
The first two terms are the external-field energies (Eq. 2.27 applied to each charge); the last term is the mutual energy (Eq. 2.22), which is the same whether or not an external field is present. In NCERT Example 2.5, the mutual term of a $7\ \mu\text{C}$ and $-2\ \mu\text{C}$ pair stays at $-0.7\ \text{J}$ even after the external field is switched on; only the external-field terms are added.
| Situation | Potential energy | NCERT Eq. |
|---|---|---|
| Two charges (no external field) | U = k q₁q₂ / r₁₂ | 2.22 |
| Three charges (no external field) | U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃) | 2.26 |
| Single charge in external field | U = q V(r) | 2.27 |
| Two charges in external field | U = q₁V(r₁) + q₂V(r₂) + k q₁q₂/r₁₂ | 2.29 |
| Dipole in external field | U = −pE cosθ = −p·E | 2.32 |
A dipole in an external field
A dipole in a uniform field $E$ feels no net force but a torque $\boldsymbol\tau = \mathbf p \times \mathbf E$. Rotating it from $\theta_0$ to $\theta$ against this torque does work $pE(\cos\theta_0 - \cos\theta)$, stored as potential energy. Choosing the zero at $\theta_0 = 90^\circ$ gives NCERT Eq. (2.32):
$$U(\theta) = -pE\cos\theta = -\mathbf p \cdot \mathbf E$$
The energy is minimum ($U = -pE$) at $\theta = 0$, where $\mathbf p$ is aligned with $\mathbf E$ — the stable equilibrium. It is maximum ($U = +pE$) at $\theta = 180^\circ$ — the unstable equilibrium. At $\theta = 90^\circ$ the energy is zero by our choice of reference.
Aligned ($\theta = 0$): lowest energy, stable. Perpendicular: $U = 0$. Anti-aligned ($\theta = 180^\circ$): highest energy, unstable.
Dipole energy is MINIMUM at θ = 0, not at θ = 90°
The reference $U = 0$ is at $\theta = 90^\circ$, but that is not the minimum. Because $U = -pE\cos\theta$, the lowest energy is at $\theta = 0$ where $\cos\theta = 1$, giving $U = -pE$. Aligned with the field = stable = minimum energy.
Minimum $U = -pE$ at $\theta = 0$ (stable); maximum $U = +pE$ at $\theta = 180^\circ$ (unstable).
Reading the sign of U
The sign of $U$ carries physical meaning and is a favourite trap. For a pair, $U \propto q_1 q_2$. If $q_1 q_2 > 0$ (like charges), $U$ is positive: the force is repulsive, so positive external work is needed to push them together. If $q_1 q_2 < 0$ (unlike charges), $U$ is negative: the force is attractive, and the system is more bound the closer they sit.
| Charges | Force | Sign of U | Interpretation |
|---|---|---|---|
| Like ($q_1 q_2 > 0$) | Repulsive | $U > 0$ | Work done against force to assemble |
| Unlike ($q_1 q_2 < 0$) | Attractive | $U < 0$ | Bound system; energy needed to separate |
The work to separate two charges infinitely is $W = U_\infty - U = 0 - U = -U$. For the $7\ \mu\text{C}$, $-2\ \mu\text{C}$ pair of Example 2.5, $U = -0.7\ \text{J}$, so separating them requires $+0.7\ \text{J}$ of external work — exactly the energy that was released when they came together.
Potential Energy of a System of Charges in one screen
- Definition: $U$ = work to assemble charges from infinity; path- and order-independent because the force is conservative.
- Two charges: $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$, charges inserted with sign.
- Three+ charges: sum over all distinct pairs; $n$ charges give $\dfrac{n(n-1)}{2}$ terms — count pairs, not charges.
- Single charge in external field: $U = qV(\mathbf r)$ (no $1/r$ term).
- Two charges in external field: $U = q_1V(\mathbf r_1) + q_2V(\mathbf r_2) + \dfrac{kq_1q_2}{r_{12}}$.
- Dipole in external field: $U = -pE\cos\theta = -\mathbf p\cdot\mathbf E$; min at $\theta=0$, max at $\theta=180^\circ$.
- Sign: $U>0$ for like charges, $U<0$ for unlike; work to separate $= -U$.