Potential due to a system of charges

Stand a metre away from three charged pins stuck into a corkboard — one positive, one negative, one positive again — and ask a simple question: what is the electric potential at your fingertip? For the electric field this would mean carefully adding three arrows, each pointing a different way, each of a different length. For the potential it means something almost embarrassingly easier: add three plain numbers, with their signs, and you are done. That contrast is the whole story of this topic. Electric potential is a scalar — it has a size but no direction — so when several charges act together, their potentials at any point simply add like ordinary arithmetic, $V = V_1 + V_2 + V_3 + \cdots$ , with negative charges subtracting and positive charges adding. This is the superposition principle for potential, and it is the single biggest reason potential is the workhorse quantity of electrostatics: no components, no angles, no resolving along axes — just distances, charges, and a sum. This article builds that idea from the one-charge formula you already know, shows exactly how and why it generalises to any number of charges (and, in spirit, to continuous charge distributions too), and then puts it to work on the same two- and three-charge problems that show up again and again in board papers and competitive exams — including the classic trap where the potential at a point is zero while the electric field there is very much not.

Key takeaways

Potential is a scalar. For a system of point charges, the net potential at any point is the plain algebraic sum of the potentials each charge would produce alone: $V = \dfrac{1}{4\pi\varepsilon_0}\sum_i \dfrac{q_i}{r_i}$ .
Signs matter, directions do not. A negative charge's term is subtracted, not just ignored; but there are no angles or components to resolve, unlike for the electric field.
Each $r_i$ is measured to the same point $P$ . Different charges generally sit at different distances from $P$ — never use one shared distance unless the geometry actually forces it.
Contrast with the field. Electric field superposition, $\vec E = \sum_i \vec E_i$ , is a vector sum requiring components; potential superposition is a scalar sum, which is why potential problems are usually the easier route to a hard field problem.
Zero potential does not mean zero field. A point equidistant from an equal-and-opposite charge pair has $V = 0$ there, yet the field at that same point is generally non-zero — the two quantities answer different questions.
It extends to continuous charge. The same idea — add up $\dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r}$ over every bit of charge — underlies the potential of rods, discs and shells, though the full integral machinery is beyond this chapter's numerical core.
Key idea: because potential is a scalar, the net potential of a system of charges at any point is the ordinary signed sum of each charge's individual contribution, $V=\frac{1}{4\pi\varepsilon_0}\sum_i q_i/r_i$ , which is why potential problems skip the vector algebra that field problems demand.

The core insight

Potential due to a system of point charges is the algebraic (scalar) sum of the potentials each charge produces on its own, $V = \dfrac{1}{4\pi\varepsilon_0}\sum_i \dfrac{q_i}{r_i}$ , with signed charges and each distance measured from that charge to the same point.

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 single point charge and potential due to a dipole, both of which this article assumes as known building blocks. It is squarely inside the 2025-26 prescribed scope — the syllabus explicitly lists "potential due to a point charge, a dipole, and a system of charges" as core content, with no rationalisation cuts applying here. The board expects a student to state and apply the superposition principle for potential, distinguish it clearly from vector superposition of the field, and handle two- and three-charge numerical problems confidently — exactly the skill this article drills.

The physical picture

Begin with what is already known: the potential due to a single point charge $q$ at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

where: $V$ is measured in volts, $q$ in coulombs (with its sign), $r$ in metres, and $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is the familiar Coulomb constant. Physically, $V$ here is the work done per unit positive charge in bringing a small test charge from infinity (where $V$ is taken as zero) up to the point at distance $r$ , so a positive source charge gives a positive potential and a negative source charge gives a negative one. Now place several point charges $q_1, q_2, \ldots, q_n$ at fixed positions, and pick any point $P$ in space at respective distances $r_1, r_2, \ldots, r_n$ from each charge. Because potential is defined as work done per unit charge — a scalar quantity with no direction attached — the net potential at $P$ due to the whole system is simply the sum of what each charge would produce there on its own, sign included:

V = V_1 + V_2 + \cdots + V_n = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots + \frac{q_n}{r_n}\right) = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i}

where: each $q_i$ carries its own sign (positive charges add a positive term, negative charges add a negative term), and each $r_i$ is the straight-line distance from that particular charge to the one point $P$ where the potential is being evaluated — not to each other, and not some shared or average distance unless the geometry of the problem happens to make all the $r_i$ equal. If $P$ is moved to a different location, every single $r_i$ must be recalculated from scratch, since the charges' positions do not change but their distances to the new $P$ generally do. Compare this with the corresponding statement for the electric field, $\vec E = \vec E_1 + \vec E_2 + \cdots + \vec E_n$ , which is a vector sum: each $\vec E_i$ has both a magnitude and a direction, and combining them requires resolving components along chosen axes and adding those components separately before recombining into a resultant. Potential superposition sidesteps all of that — it is why, in a great many hard field problems, the smart route is to compute $V$ first (an easy scalar sum) and then, if the field is genuinely needed, recover it from how $V$ changes with position, a relationship covered elsewhere in this chapter.

Figure

Three point charges at different distances from a common point P, with three straight dashed lines from each charge to P labelled with distance r one r two and r three and their signed potential terms adding into one scalar total V at P

Derivation ABuilding the superposition rule for potential from work done

The goal is to show, from the definition of potential as work done per unit charge in bringing a test charge from infinity, why the potentials of several source charges simply add for a system of point charges — rather than combine like vectors the way electric fields do — using nothing beyond the single-charge potential formula and the physical meaning of work.

Step 1. Recall what $V$ at a point means for one charge. The potential at a point $P$ due to a single charge $q$ is defined as the work done, per unit positive test charge, in bringing that test charge from infinity to $P$ against the electric force of $q$ , without acceleration. For a point charge this works out to $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ , where $r$ is the distance from $q$ to $P$ , and $V=0$ is the reference at infinity.

Step 2. Extend to many source charges by using the work-energy meaning directly. Suppose charges $q_1, q_2, \ldots, q_n$ are all fixed in place, and a small test charge $q_0$ is brought from infinity to point $P$ . The total work done against the combined electric force of all the source charges is, by the general property that work is additive for a sum of forces acting along the same path, exactly the sum of the work that would be needed against each source charge acting alone: $W_{\text{total}} = W_1 + W_2 + \cdots + W_n$ .

Step 3. Divide through by the test charge $q_0$ to convert work into potential. Since $V = W/q_0$ by definition, dividing the Step 2 equation by $q_0$ gives $V = \dfrac{W_1}{q_0} + \dfrac{W_2}{q_0} + \cdots + \dfrac{W_n}{q_0} = V_1 + V_2 + \cdots + V_n$ , where each $V_i = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_i}$ is exactly the potential that source charge $q_i$ alone would create at $P$ , with $r_i$ its own distance from $q_i$ to $P$ .

Step 4. Write the final result and note why no vector algebra was needed. Substituting each single-charge formula gives $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_{i=1}^{n}\dfrac{q_i}{r_i}$ . Because work and potential are both scalars — plain numbers, with sign but no direction — the additivity of work in Step 2 translates directly into additivity of potential, with no components or angles anywhere in the argument. This is the structural reason potential superposition is arithmetic while field superposition, built from the vector force $\vec F = q_0\vec E$ , must be done vector-first. Verified: setting all $q_i$ equal for $n$ identical charges each at the same distance $r$ from $P$ reduces the sum to $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{nq}{r}$ , matching the expected symmetric-arrangement result, and setting $n=1$ recovers the single point-charge formula exactly.

Worked examples

Board level - net potential from three point charges

Three point charges are placed near a point $P$ : $q_1 = +2\ \mu\text{C}$ at $r_1 = 10\ \text{cm}$ from $P$ , $q_2 = -3\ \mu\text{C}$ at $r_2 = 15\ \text{cm}$ from $P$ , and $q_3 = +4\ \mu\text{C}$ at $r_3 = 20\ \text{cm}$ from $P$ . Find the net potential at $P$ .

Step 1. Write the superposition formula: $V = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}+\dfrac{q_3}{r_3}\right)$ , using $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ .

Step 2. Convert every distance to metres, since the formula is built on SI units throughout: $r_1 = 0.10\ \text{m}$ , $r_2 = 0.15\ \text{m}$ , $r_3 = 0.20\ \text{m}$ .

Step 3. Compute each signed term separately, carrying the sign of each charge through the division: $\dfrac{q_1}{r_1} = \dfrac{2\times10^{-6}}{0.10} = 2\times10^{-5}\ \text{C/m}$ , $\dfrac{q_2}{r_2} = \dfrac{-3\times10^{-6}}{0.15} = -2\times10^{-5}\ \text{C/m}$ , $\dfrac{q_3}{r_3} = \dfrac{4\times10^{-6}}{0.20} = 2\times10^{-5}\ \text{C/m}$ .

Step 4. Add the three terms algebraically: $2\times10^{-5} - 2\times10^{-5} + 2\times10^{-5} = 2\times10^{-5}\ \text{C/m}$ . Notice the middle negative term genuinely cancels one of the positive ones — this is the scalar addition doing its work.

Step 5. Multiply by the Coulomb constant: $V = (9\times10^9)(2\times10^{-5}) = 1.8\times10^5\ \text{V}$ .

Answer: the net potential at $P$ is $1.8\times10^5\ \text{V}$ , that is, $180\ \text{kV}$ .

Competitive level - locating a point of zero potential and comparing it with the field there

Two charges $+5\ \mu\text{C}$ and $-5\ \mu\text{C}$ are placed $50\ \text{cm}$ apart. Consider the point $Q$ on the perpendicular bisector of the line joining them, so that $Q$ is equidistant from both charges at $r = 25\ \text{cm}$ from each. Find the potential at $Q$ , and state (without full calculation) whether the electric field at $Q$ is also zero.

Step 1. Apply superposition with the two equal, opposite-sign, equal-distance charges: $V = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}\right) = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{5\times10^{-6}}{0.25}+\dfrac{-5\times10^{-6}}{0.25}\right)$ .

Step 2. Since $r_1 = r_2 = 0.25\ \text{m}$ exactly and the charges are equal in magnitude but opposite in sign, the two terms are equal and opposite numbers, so they cancel exactly: $V = 0\ \text{V}$ , regardless of what the common distance actually is.

Step 3. Now consider the field at the same point $Q$ . Unlike potential, the field is a vector: the field of $+5\ \mu\text{C}$ points away from it (toward $Q$ , roughly along the bisector direction) and the field of $-5\ \mu\text{C}$ points toward it (also with a component along the same general direction, since it pulls a test charge back toward itself). Because both fields have equal magnitude here (equal charge magnitude, equal distance) but their directions are not opposite along the bisector — they add a net component instead of cancelling — the resultant field at $Q$ is non-zero, in fact directed parallel to the line joining the charges. This is exactly the equatorial-point result for a dipole, since a $+q,-q$ pair separated by a fixed distance behaves as a dipole, and the equatorial field of a dipole is known to be non-zero even though the equatorial potential is always zero.

Answer: $V_Q = 0\ \text{V}$ , but the electric field at $Q$ is not zero — it is a finite vector directed antiparallel to the dipole moment. Zero potential at a point never guarantees zero field there. Timing note: under ninety seconds. The moment you spot "equal magnitude, equal distance, opposite sign," write $V=0$ immediately without computing individual terms — the real exam skill being tested is recognising the equidistant-equal-opposite pattern instantly and not confusing the zero-potential conclusion with a (false) zero-field conclusion, which is exactly the kind of one-mark trap competitive papers set.

Common traps

Trying to resolve "components" of potential the way one resolves field vectors — potential addition is purely algebraic; there are no directions or angles to worry about, ever.
Dropping the sign of a negative charge and adding its magnitude instead of subtracting it — the sign of $q_i$ must be carried all the way through the calculation.
Using one shared distance for every charge when the charges actually sit at different distances from the point of interest — each $r_i$ belongs to its own charge and must be recalculated if the point of interest changes.
Concluding that a point of zero net potential must also be a point of zero net electric field — the second worked example shows this is false in general; potential and field answer different physical questions and can vanish at completely different points.
Forgetting to convert all distances to a single consistent unit (usually metres) before substituting into the formula, which silently introduces a factor-of-100 or factor-of-1000 error.

Common trapZero potential does not mean zero field, and potential superposition is never a vector sum

The single biggest scoring trap in this topic is conflating potential and field. Because both involve "adding contributions from several charges," it is tempting to treat them the same way — but potential superposition is a scalar sum of signed numbers with no directions involved, while field superposition is a genuine vector sum requiring components and angles. A direct consequence, seen in the second worked example, is that a point can have exactly zero net potential (because equal-and-opposite contributions cancel numerically) while still having a strong, non-zero electric field there (because the field vectors from those same charges do not point in opposite directions and so do not cancel). Never infer one quantity's value from the other without doing the correct — scalar or vector — calculation for that specific quantity.

You should be able to…

State the superposition principle for electric potential due to a system of point charges, and justify it from the fact that potential is a scalar.
Compute the net potential at a specified point due to several given point charges and their distances, carrying signs correctly throughout.
Explain, with a concrete reason, why potential superposition is scalar addition while electric field superposition requires vector addition.
Recognise the pattern of equal-magnitude, opposite-sign, equidistant charges and immediately conclude the potential there is zero, without lengthy calculation.
Correctly judge whether the electric field at a point of zero potential is also zero, and justify the answer physically rather than by assumption.

Ask the tutor: Want to try a four- or five-charge arrangement, or see why the point of zero potential on a dipole's equatorial line is never a point of zero field — ask the AI tutor for a guided problem set.

Why does potential just add up like plain numbers, but field needs vector addition?

Potential at a point is defined as work done per unit charge to bring a test charge there from infinity, and work is a scalar — it has a size but no direction, so combining contributions from several source charges is just ordinary addition, with sign. Electric field, on the other hand, is defined as force per unit charge, and force is inherently a vector with both magnitude and direction, so combining several fields at a point genuinely means adding arrows head-to-tail (or adding their components). The two definitions look similar in words but differ in this one crucial mathematical property.

If two charges are equal and opposite and equidistant from a point, is the potential there always exactly zero?

Yes, as long as the magnitudes truly match and the distances truly match — the two terms in the sum are then equal in size but opposite in sign, so they cancel completely regardless of what that common distance actually is. This is exactly the situation on the perpendicular bisector of a dipole (the equatorial line), which is why the equatorial potential of a dipole is always zero, at every point along that line, not just at one special spot.

Does the superposition principle for potential also work for charges that are not point charges, like a charged sphere or rod?

Yes — the underlying idea is identical, only the sum becomes an integral. A continuous charge distribution is imagined as an enormous number of tiny point charges $dq$ , each contributing $\dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r}$ to the potential at a given point, and the total potential is the sum (integral) of all these tiny contributions. The board-level numerical work in this chapter sticks to discrete point charges, but knowing that the principle generalises this way is useful for recognising why potential calculations for extended objects are set up the way they are.

In the worked example with three charges, why did the negative charge not make the total potential negative?

Because superposition adds signed numbers, not magnitudes — the negative charge's term did subtract from the total, it just did not subtract enough to overcome the two positive contributions. If the negative charge had been larger in magnitude, or closer to $P$ (making its term's magnitude bigger), the total could easily have come out negative instead. Always compute every term with its correct sign and let the arithmetic decide the final sign; never guess it from "there are more positive charges" or similar shortcuts.