Potential energy of a system of charges

Bring two positive charges close together and let go — they fly apart. Bring a positive and a negative charge close together and let go — they snap toward each other. In both cases something was stored while you held them in place, and it is released the moment you stop holding on. That "something" is electrostatic potential energy: the work you, as an external agent, had to invest to assemble a set of charges into a particular arrangement, starting from a state where they were all infinitely far apart and not interacting at all. This is not an abstract bookkeeping trick. It is the same energy that fires a spark between two charged plates, that a nucleus stores in packing many protons into a tiny volume, and that a capacitor releases when you connect it to a bulb. For CBSE Class 12 the idea is built with nothing beyond the formula for potential due to a point charge and simple algebra: pick up one charge at a time, ask how much work it takes to walk each new charge in from infinity to its final spot against the electric field of the charges already in place, and add up all those bits of work. Two charges give one term. Three charges give three terms — one for every pair. A hundred charges give thousands of terms, but the rule never changes: sum $\frac{1}{4\pi\varepsilon_0}\frac{q_iq_j}{r_{ij}}$ over every distinct pair $(i,j)$ . This article rebuilds that rule from scratch with a full step-by-step derivation for the three-charge case, works two numerical problems (a two-charge system and a three-charge equilateral triangle, extended further here), and flags the two mistakes — sign errors and pair double-counting — that quietly cost the most marks on this topic every single year.

Key takeaways

Potential energy of two charges: $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ — the work needed to bring $q_2$ from infinity to distance $r_{12}$ from the fixed charge $q_1$ .
Reference point: $U = 0$ when all charges are infinitely far apart (mutually non-interacting); $U$ measures the work done against the mutual field to assemble the given configuration from that zero.
Sign tells you the physics: like charges ( $q_1q_2 > 0$ ) give positive $U$ — you had to push against repulsion; unlike charges ( $q_1q_2 < 0$ ) give negative $U$ — the field itself pulled them together.
For three or more charges, sum over every pair: $U = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_{i<j}\dfrac{q_iq_j}{r_{ij}}$ — three charges give exactly three pair-terms, four charges give six, $n$ charges give $\binom{n}{2}$ .
Order of assembly does not matter. Bringing $q_3$ in before $q_2$ , or in any other order, gives the identical total $U$ — the final configuration alone fixes the energy.
$U$ is a single number for the whole system, unlike potential $V$ , which is a value defined at every point in space and depends on only one charge times the field of the rest, not on charge pairs.
Key idea: the potential energy of a system of point charges is the sum, over every distinct pair, of the pairwise interaction energy $\frac{1}{4\pi\varepsilon_0}\frac{q_iq_j}{r_{ij}}$ , equal to the total work needed to assemble the charges from infinite separation.

The core insight

Electrostatic potential energy of a system of point charges is the work needed to assemble it from infinite separation, and for any number of charges it equals the sum of the pairwise energies $\frac{1}{4\pi\varepsilon_0}\frac{q_iq_j}{r_{ij}}$ taken over every distinct pair, independent of the order of assembly.

Where it fits in the syllabus

This topic sits inside the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), directly after potential due to a point charge and a system of charges, and directly before potential energy of a dipole in an external field, $U = -\vec p\cdot\vec E$ (a related but separate idea covered in the next node). The chapter's prescribed scope names exactly the two-charge formula $U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r}$ as core content; the extension to three or more charges by pairwise summation is the standard board-level generalisation examiners expect a student to produce on their own, using only the two-charge result and simple addition — no new physics is introduced beyond what is already prescribed.

The physical picture

Picture assembling a system of point charges one at a time, each new charge walked in from a location infinitely far away where it does not yet feel any of the others. The work an external agent must do to bring in each new charge equals that charge times the potential already existing at its arrival point, due to every charge already placed. Summing this work over the whole assembly process gives the total potential energy stored in the final configuration. For two charges $q_1$ (fixed) and $q_2$ (brought in from infinity to a separation $r_{12}$ ), the work done equals $q_2$ times the potential of $q_1$ at that final location:

U = q_2\,V_1(r_{12}) = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r_{12}}

where: $U$ is the electrostatic potential energy of the pair (joules), $q_1$ and $q_2$ are the point charges (coulombs, with sign), $r_{12}$ is the distance between them (metres), and $\frac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}$ is Coulomb's constant. Read physically: if $q_1$ and $q_2$ have the same sign, the product $q_1q_2$ is positive and so is $U$ — an external agent had to push the charges together against their mutual repulsion, storing positive energy that would be released (accelerating the charges apart) if let go. If the signs are opposite, $q_1q_2$ and hence $U$ are negative — the mutual attraction did the work of pulling them together, so an external agent would instead have needed to hold them back (do negative work) to bring them in slowly without a sudden snap. For three charges, the same bookkeeping — bring $q_1$ in for free (nothing to interact with yet), then $q_2$ against the field of $q_1$ , then $q_3$ against the combined field of both $q_1$ and $q_2$ — produces three pairwise terms:

U = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)

where: each term is the interaction energy of one distinct pair, and $r_{12}, r_{13}, r_{23}$ are the three mutual separations. The general rule for $n$ charges is a sum over every distinct pair, $U = \frac{1}{4\pi\varepsilon_0}\sum_{i<j}\frac{q_iq_j}{r_{ij}}$ , with exactly $\binom{n}{2} = \frac{n(n-1)}{2}$ terms — $3$ terms for $3$ charges, $6$ for $4$ charges, $10$ for $5$ charges, and so on. Crucially, this total does not depend on the order in which the charges were notionally brought in from infinity; only the final geometry (the set of separations $r_{ij}$ ) and the charge values fix $U$ , which is why the formula can be written as a simple sum without reference to any assembly sequence at all.

Figure

Three point charges at the corners of a triangle with dashed lines joining every pair labelled with a distance, showing that potential energy has one term for each of the three pairs

This total $U$ must not be confused with the potential $V$ due to a system of charges at some external point. $U$ is a single number describing the energy of the whole configuration, built from products of pairs of the system's own charges; $V$ is a value that exists at every point of space and is a sum over single charges (each divided by its distance to the field point), evaluated at one chosen location that need not even hold a charge.

Derivation APotential energy of three point charges by pairwise assembly

The goal is to show, from the definition of work done against the electric field, that the potential energy of three point charges $q_1, q_2, q_3$ fixed at mutual separations $r_{12}, r_{13}, r_{23}$ equals exactly the sum of the three pairwise interaction energies $\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$ , and that this total is independent of the order in which the three charges are notionally brought in from infinity — establishing both the formula and the key property that makes it well defined for any configuration.

Step 1. Bring in $q_1$ first, to some chosen point in otherwise empty space. Since there are no other charges present yet, no field opposes this charge, so the work done is zero: $W_1 = 0$ .

Step 2. Bring in $q_2$ from infinity to its final position, a distance $r_{12}$ from $q_1$ . The only field present is that of $q_1$ , so the work done equals $q_2$ times the potential of $q_1$ at that point: $W_2 = q_2\,V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}$ .

Step 3. Bring in $q_3$ from infinity to its final position, at distances $r_{13}$ from $q_1$ and $r_{23}$ from $q_2$ . Both $q_1$ and $q_2$ are already in place and contribute to the potential at $q_3$ 's arrival point, so the work done is $q_3$ times the sum of their potentials there: $W_3 = q_3\left(V_1 + V_2\right) = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_3}{r_{13}} + \frac{1}{4\pi\varepsilon_0}\frac{q_2q_3}{r_{23}}$ .

Step 4. Add the three stages. The total work done by the external agent, which equals the potential energy stored in the final configuration, is $U = W_1 + W_2 + W_3 = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$ — exactly three terms, one per distinct pair, with no self-energy term and no pair counted twice.

Step 5. Check that the order does not matter. Suppose instead $q_3$ were brought in second and $q_2$ last. Repeating the same argument, $W_1' = 0$ for $q_1$ , then $W_2' = q_3(V_1) = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_3}{r_{13}}$ for $q_3$ against $q_1$ 's field alone, then $W_3' = q_2(V_1 + V_3) = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} + \frac{1}{4\pi\varepsilon_0}\frac{q_2q_3}{r_{23}}$ for $q_2$ against the combined field of $q_1$ and $q_3$ . Adding these gives the identical total, term for term, as in Step 4 — the reordering only changes which stage each pairwise term appears in, never the final sum. This confirms $U$ is a genuine property of the configuration, not of any particular assembly history, which is exactly what justifies writing it as the order-free sum $U = \frac{1}{4\pi\varepsilon_0}\sum_{i<j}\frac{q_iq_j}{r_{ij}}$ for any number of charges.

Worked examples

Board level - potential energy of two point charges

Two point charges, $q_1 = +3\ \mu\text{C}$ and $q_2 = -2\ \mu\text{C}$ , are separated by 15 cm. Find the electrostatic potential energy of this two-charge system.

Step 1. Write the formula: $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$ , using $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}$ .

Step 2. Substitute the values: $q_1q_2 = (3\times10^{-6})(-2\times10^{-6}) = -6\times10^{-12}\ \text{C}^2$ , and $r = 0.15\ \text{m}$ .

Step 3. Compute: $U = (9\times10^9)\dfrac{-6\times10^{-12}}{0.15} = (9\times10^9)(-4\times10^{-11}\ \text{C}^2/\text{m}) = -0.36\ \text{J}$ .

Answer: the potential energy of the system is $-0.36\ \text{J}$ , negative because the charges are unlike and attract — the field itself would do positive work pulling them together, so an external agent effectively did negative work to bring them in slowly.

Competitive level - three charges on an equilateral triangle, then breaking one bond

Three point charges, $q_1 = +1\ \mu\text{C}$ , $q_2 = +2\ \mu\text{C}$ , and $q_3 = +3\ \mu\text{C}$ , sit at the corners of an equilateral triangle of side 10 cm. (a) Find the total electrostatic potential energy of the system. (b) Find the work needed to move $q_3$ away to infinity, leaving $q_1$ and $q_2$ fixed in place.

Step 1. Since the triangle is equilateral, all three pairwise distances are equal: $r_{12} = r_{13} = r_{23} = 0.10\ \text{m}$ .

Step 2. Use $U = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_2}{r_{12}} + \dfrac{q_1q_3}{r_{13}} + \dfrac{q_2q_3}{r_{23}}\right)$ , and compute each pairwise product: $q_1q_2 = 2\times10^{-12}$ , $q_1q_3 = 3\times10^{-12}$ , $q_2q_3 = 6\times10^{-12}\ \text{C}^2$ ; their sum is $11\times10^{-12}\ \text{C}^2$ .

Step 3. Substitute: $U = (9\times10^9)\dfrac{11\times10^{-12}}{0.10} = (9\times10^9)(1.1\times10^{-10}) = 0.99\ \text{J}$ . Answer (a): the total potential energy of the three-charge system is $0.99\ \text{J}$ .

Step 4. For part (b), the work needed to remove $q_3$ to infinity equals the energy stored specifically in $q_3$ 's two bonds to $q_1$ and $q_2$ — the $q_1$ – $q_2$ term is untouched since that pair stays fixed. That removed energy is $\Delta U = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_3}{r_{13}} + \dfrac{q_2q_3}{r_{23}}\right) = (9\times10^9)\dfrac{(3+6)\times10^{-12}}{0.10} = (9\times10^9)(9\times10^{-11}) = 0.81\ \text{J}$ .

Answer (b): the external agent must supply $+0.81\ \text{J}$ of work to pull $q_3$ out to infinity against the repulsive field of $q_1$ and $q_2$ (all three charges being positive), leaving behind only the $q_1$ – $q_2$ pair with energy $0.99 - 0.81 = 0.18\ \text{J}$ , which matches $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}} = (9\times10^9)\dfrac{2\times10^{-12}}{0.10} = 0.18\ \text{J}$ computed directly — a useful self-check. Timing note: aim for under two minutes. The fast route is to compute the three pairwise products once, sum them for part (a), then for part (b) simply drop the one term that stays fixed (here $q_1q_2$ ) and evaluate the remaining piece directly — don't recompute from scratch. Cross-checking the leftover $q_1$ – $q_2$ term against $U_{\text{total}} - \Delta U$ is a fast way to catch arithmetic slips under exam pressure.

Common traps

Missing a pair is the single most common error for three-or-more-charge systems — students frequently write only two of the three terms for a triangle, or accidentally repeat one pair while skipping another; always count $\binom{n}{2}$ terms and cross them off a mental list of pairs as you go. A close second is a sign error: forgetting that $U$ takes the sign of the product $q_iq_j$ , so a system with one negative charge among several positives will have some negative and some positive terms in the sum — do not force every term to be positive. Students also frequently confuse $U$ (a single number for the whole assembly, built from products of pairs of charges) with $V$ (a value that exists at every point of space, built from single charges divided by distance to that one point) — these use similar-looking formulas but answer completely different questions. Finally, remember the reference convention: $U = 0$ corresponds to all charges mutually infinitely far apart, so a computed $U$ is always relative to that zero, and "moving a charge to infinity" removes exactly the pairwise terms involving that one charge, not the whole system's energy.

Common trapEvery pairwise term must be counted exactly once, and its sign is not a choice

The most damaging mistake on this topic is treating potential energy the way many students first (wrongly) treat potential — as if you could just add up "distances" the way you would sum simple magnitudes, ignoring pairs and signs. For $n$ charges there are exactly $\binom{n}{2} = \frac{n(n-1)}{2}$ distinct pairs, and every one contributes exactly one term $\frac{1}{4\pi\varepsilon_0}\frac{q_iq_j}{r_{ij}}$ with whatever sign the product $q_iq_j$ actually has — you do not get to decide the sign based on "energy should be positive." A system of three positive charges has all-positive terms and large positive $U$ ; swap one charge to negative and at least two terms flip sign, potentially making the total $U$ negative even though most of the charges repel each other. Always list every pair explicitly before summing, and always compute the sign of each $q_iq_j$ product rather than assuming.

You should be able to…

State and apply the formula for potential energy of a pair of point charges, $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ , including its sign.
Extend the two-charge formula to three or more point charges by summing over every distinct pair, without omission or double-counting.
Explain why the reference state $U = 0$ corresponds to charges infinitely far apart, and interpret positive versus negative $U$ physically.
Derive the three-charge formula from first principles using stepwise assembly against the field of charges already placed.
Compute the work needed to remove one charge from a multi-charge system to infinity, by isolating only the pairwise terms that involve it.
Distinguish clearly between $U$ (system self-energy, pairwise) and $V$ (potential at a point, single-charge sum).

Ask the tutor: want to work through a four-charge square configuration (six pairwise terms) or see how this connects to the energy stored in a charged capacitor — ask the AI tutor to walk through it with you.

Why does bringing the first charge into an empty region cost zero work, isn't that charge also "there"

Work against an electric field is only done when there is already a field present to push against. When you bring in the very first charge, there is no other charge anywhere yet, so there is no electric field at all in that empty region — nothing resists you, so the work done is exactly zero. This is not a special exception; it is the same rule as for every other charge, just applied to a case where "the field of everything already placed" happens to be zero because nothing has been placed yet. Once that first charge sits there, it creates a field, and every charge brought in afterward does have to push against something.

If order does not matter, why does the derivation bring charges in one at a time in a specific order?

The order-by-order assembly is just a method of calculation, not a claim about how the charges were physically placed in reality. The derivation proves, as its final step, that any order of assembly gives the same total $U$ — so you are free to pick whichever order makes the arithmetic easiest (usually the order you are given the charges in). The deeper reason this works is that $U$ is a property of the final geometry and charge values alone, similar to how the change in height of a ball depends only on its start and end position, not on the path it took to get there.

Is potential energy of a system of charges the same as the potential energy of a single charge in a field?

No, and mixing these up is a common Class 12 slip. "Potential energy of a charge $q$ in an external potential $V$ " is a single term, $U = qV$ , describing one charge sitting in a field created by something else entirely (for example, a charge in the field between capacitor plates). "Potential energy of a system of charges" is the self-energy of the whole group, built by summing pairwise terms among the charges themselves, with no external field involved at all. Both use the word "potential energy," but the first has one charge and an outside field, while the second has multiple charges and their own mutual field.

Does this formula still work if the charges are not point charges, like charged spheres?

For charged conducting spheres (or any spherically symmetric charge distribution) that do not overlap, you can treat each sphere as if all its charge were concentrated at its centre, and the pairwise formula $U = \frac{1}{4\pi\varepsilon_0}\frac{q_iq_j}{r_{ij}}$ still applies using centre-to-centre distances — this is a direct consequence of the shell theorem for electrostatics, which is why CBSE numericals often quietly use "spheres" and "point charges" interchangeably. It breaks down only if the charge distributions are not spherically symmetric or if they overlap, situations that are beyond the Class 12 syllabus.