Potential of Point Charges & Charge Systems

The previous module minted the currency — energy per coulomb — but never drew the map. This one draws it. Around every point charge, space carries a definite potential at every distance: a $1/r$ hill around positive charge, a $1/r$ well around negative. Learn that one shape and you can price any arrangement of charges at any point, because potentials combine by the friendliest rule in physics: plain addition of signed numbers. Feel the contrast with what chapter E1 made you do. Finding the electric field of four charges meant four vectors, four angles, components resolved and recombined — the Superposition module's honest, sweaty machinery. Finding the potential of the same four charges means four divisions and three additions. No angles exist in the calculation at all. That asymmetry — vector work versus scalar arithmetic — is why competitive papers are stuffed with potential questions that look hard and fall in under a minute, and why the strongest students reach for $V$ first and recover forces later only if asked. The map also carries this chapter's most planted trap. Because potentials add as signed numbers, they can cancel: points exist where $V = 0$ while the field is at full strength, and points exist where the field cancels while $V$ stands tall. Zero on one map never implies zero on the other — and by the end of this article you will see through that trap at a glance, because you will have computed both maps at the same point and watched them disagree.

Key takeaways

A point charge $Q$ builds the potential $V = kQ/r$ at distance $r$ : a hill for $Q > 0$ , a well for $Q < 0$ , zero at infinity. Note the first power of $r$ .
Many charges: add the numbers, signs included — $V = k\sum q_i/r_i$ . No components, no angles, ever.
$V$ is blind to direction: only the distance to each charge enters. Symmetric arrangements become one multiplication.
Cash the map into energy with $U = qV$ : work to bring $q$ from infinity to a point is $q$ times the potential there.
$V = 0$ does not mean $E = 0$ , and $E = 0$ does not mean $V = 0$ — the two maps record different things and cancel in different places.
Key idea: the potential map of any charge arrangement is a sum of signed $kq/r$ terms — scalar arithmetic that replaces the vector work of field superposition.

The core insight

Each point charge contributes $V = kq/r$ — sign included, distance only — and the potential of any arrangement is the plain sum of these contributions: one signed number per charge, no directions, with $U = qV$ converting the finished map into energy on demand.

Where this lives today

Open any modern chemistry or drug-design paper and you will meet this article's map wearing lab clothes: the molecular electrostatic potential surface, a coloured skin drawn over a molecule showing where the census of all its nuclear and electronic charges sums positive (blue) or negative (red). Medicinal chemists read those colours to predict where a drug candidate will dock onto a protein — negative patches greet positive ones — and the software underneath is computing $k\sum q_i/r_i$ over thousands of partial charges, exactly your equation (2) at industrial scale. The map also polices safety. When a live wire touches ground, the earth around the contact point becomes a steep potential hill; a person walking nearby spans a step potential — two feet, two different altitudes on the hill, a difference across the body. Substation engineers design grounding grids by computing potential maps around fault points so that no reachable pair of points offers a lethal difference — the pigeon principle from the previous module, engineered in reverse. Same census, same $1/r$ terms, lives depending on the arithmetic.

The physical picture

A positive point charge builds a hill that is infinitely steep at the summit and dies away as $1/r$ : double your distance and the altitude halves. A negative charge digs the mirror-image well. Every landscape in electrostatics — around molecules, capacitor plates, thunderclouds — is stitched together from copies of this one shape, raised or sunk, because arrangements are built from point charges and their landscapes simply add.

Figure

The two elementary landscapes — a 1/r hill and a 1/r well

Now the working picture for many charges. Stand at the point where you want the potential and conduct a census: ask every charge in the problem two questions only — what is your signed size? and how far are you from me? Each answers with one number, $kq_i/r_i$ , and the potential where you stand is the total. Near charges shout, far charges whisper, negative charges subtract. Nobody is asked for a direction, because altitude has no direction.

Figure

The scalar census — every charge reports one signed number, distance only

Hold on to the contrast in the caption. In Superposition: Forces & Fields of Many Charges you resolved components because fields fight about direction. Potentials cannot fight — they are altitudes, and altitudes just add. That is the entire strategic content of this module: when a problem permits the energy view, take it, because the arithmetic is an order of magnitude cheaper.

The point-charge potential

The formula falls straight out of the two modules behind you — no new physics, just a division. Electrostatic Potential Energy priced the pair: a test charge $q$ parked at distance $r$ from a source $Q$ holds $U = kQq/r$ . Electric Potential & Potential Difference then defined the site's per-coulomb price as $V = U/q$ . Divide, and the traveller cancels:

V = \frac{kQ}{r} \qquad (1)

where: $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ (all boards; $8.99\times10^9$ if the options run tight); $Q$ — the source charge, sign included; $r$ — the plain distance from it, never squared. Read the anatomy twice. The sign of $V$ is the sign of $Q$ : positive sources build hills, negative dig wells, with no extra decision from you. And the single power of $r$ is the energy fingerprint you met in the pair formula — fields fall as $1/r^2$ , potentials as $1/r$ , so the potential map dies more slowly and reaches further than the field it will later regenerate. At $r = 1.0$ m from a $+1.0\ \mu$ C charge, $V = 9000$ V; at 2.0 m, 4500 V; at 3.0 m, 3000 V — a hill you can tabulate in your head.

Derivation AThe point charge potential in three moves

We want equation (1) from ingredients already banked, making explicit that nothing new is being smuggled in. The strategy: price the trip of a test charge from infinity by the pair-energy formula, then strip off the traveller by the per-coulomb definition. The one thing to watch is which charge cancels — the test charge must vanish from the answer, and the source must survive.

Step 1. Park a small test charge $q$ at distance $r$ from a pinned source $Q$ . The previous-but-one module's pair formula prices the arrangement: $U = \dfrac{kQq}{r}$ , with the sign riding on the product $Qq$ .

Step 2. The previous module defines the potential at the parking spot as energy per unit test charge: $V = U/q$ .

Step 3. Divide and cancel:

V = \frac{1}{q}\cdot\frac{kQq}{r} = \frac{kQ}{r}

The test charge is gone, as it must be — $V$ describes the location, and the location is built by $Q$ alone. The sign of $V$ is now the sign of the source: bring a positive test charge toward positive $Q$ and you climb ( $V > 0$ ); toward negative $Q$ and you descend into the well ( $V < 0$ ). Why zero at infinity came along for free: the pair formula was constructed with $U = 0$ at infinite separation, so the potential map inherits $V(\infty) = 0$ without a fresh decision. Any problem that redefines the zero shifts the whole map by a constant and changes no physics. *Verified: units $\mathrm{N\,m^2/C^2 \times C / m} = \mathrm{N\,m/C} = \mathrm{J/C} = \mathrm{V}$ ✓; limits $V \to 0$ as $r \to \infty$ ✓ and $V \to \pm\infty$ at the charge ✓ (the point-charge idealisation talking); numbers $kQ/r = (9.0\times10^9)(1.0\times10^{-6})/1.0 = 9.0\times10^3$ V, halving to 4500 V at 2.0 m ✓.*

Many charges: the census rule

For an arrangement, superpose. Each charge builds its own hill or well independently, and the composite landscape is the sum of altitudes at every point:

V_P = k\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + \cdots\right) \qquad (2)

where: $r_i$ — the distance from charge $q_i$ to the point $P$ ; each term carries its charge's sign. This is the same superposition principle that governed fields — the sources do not interfere with each other's contributions — but expressed in a currency with no direction to track. The practical consequences stack up fast. Symmetric arrangements collapse: equal charges equidistant from a centre contribute identical terms, so the census is one term times a count. Cancellations are legal: a positive and a negative contribution can sum to zero at a point where both fields are strong — the trap the overview promised, now visible as ordinary arithmetic. Energy is one multiplication away: once $V_P$ is known, the work to bring any charge $q$ from infinity to $P$ is $W = qV_P$ , by the previous module's ledger — the single most common cash-in on board papers.

Go deeper · Derivation B · 1 minRunning the census at the centre of a charged square

Derivation BRunning the census at the centre of a charged square

We want to see equation (2) collapse a four-charge problem to one line, and to certify the answer with an independent check. The setup is the classic: four equal charges $q = +2.0\ \mu$ C at the corners of a square of side $a = 30$ cm; find $V$ at the centre. The strategy is symmetry-first: measure once, count four. Watch what we never compute — a single angle.

Step 1. Geometry once: the centre sits at half the diagonal from each corner. Diagonal $= a\sqrt{2}$ , so $r = \dfrac{a\sqrt{2}}{2} = \dfrac{0.30 \times 1.414}{2} = 0.212$ m — the same for all four charges.

Step 2. Census: four identical contributions, all positive. $V = 4 \times \dfrac{kq}{r} = 4 \times \dfrac{(9.0\times10^9)(2.0\times10^{-6})}{0.212}$ .

Step 3. Arithmetic: $\dfrac{(9.0\times10^9)(2.0\times10^{-6})}{0.212} = 8.49\times10^4$ V per charge, so $V = 3.4\times10^5$ V at the centre. Why the field at the same point is zero: pair up opposite corners — their field vectors at the centre cancel exactly, by the symmetry logic of the E1 modules. So here is a point with $E = 0$ and $V = 3.4\times10^5$ V: a flat hilltop, high but level. The two maps disagree because they answer different questions — altitude versus slope.

Step 4. Flip two charges on one diagonal to $-2.0\ \mu$ C. The census becomes $(+ + - -)$ at equal distances: $V = 0$ exactly — while the field is now nonzero (the cancelling pairs no longer cancel). Same geometry, opposite disagreement. Verified: $r = 0.30/\sqrt{2} = 0.2121$ m; $4 \times 9.0\times10^9 \times 2.0\times10^{-6}/0.2121 = 3.39\times10^5$ V, recomputed numerically; the $E = 0$ claim at the symmetric centre matches the null-point logic of ARCHETYPE: Null Points & the Third-Charge Equilibrium.

Reading V–r graphs

Boards increasingly test the map itself. For a single positive charge, the graph of $V$ against $r$ is the top curve of the hero figure: steep near the charge, halving whenever $r$ doubles, approaching zero from above. A negative charge gives its mirror below the axis. Two skills earn the marks. First, identify the sign and relative size of charges from graphed curves — a curve approaching zero from below is a negative source; a curve twice as high at every $r$ is twice the charge. Second, read crossings: when the potentials of two charges are plotted along the line joining them, the composite $V$ crosses zero wherever the census balances — between unequal opposite charges, that crossing sits closer to the smaller charge, and there is a second zero beyond it on the smaller charge's side. The worked example below computes exactly such a crossing, and the coming module Equipotentials & the Field–Potential Link will add the deeper graph skill: reading the slope as the field.

Limiting cases

case	result	meaning
$r \to \infty$	$V \to 0$	the agreed zero; every map fades to sea level
very far from any cluster	$V \approx kQ_{\text{total}}/r$	from a distance, an arrangement impersonates one point charge of its total
$Q \to -Q$	$V \to -V$	the map flips upside-down; hills become wells
double all charges	$V$ doubles everywhere	the map is linear in its sources (unlike energy, which was quadratic)
equal + and − at equal distances	$V = 0$	cancellation point — field generally not zero there
symmetric ring/polygon centre	$V = k(\sum q)/r$ , one term	direction-blindness collapses the sum

Common trapV equals zero here, so nothing would happen to a charge placed here

The most reliably planted trap on this topic. $V = 0$ at a point means one thing only: the signed census balances — the work to bring a charge there from infinity is zero net. It says nothing about the field, which is the slope of the map, not its value. Midway between $+q$ and $-q$ : $V = 0$ while $E$ is strong and the charge would accelerate immediately. Centre of the four-charge square: $E = 0$ while $V = 3.4\times10^5$ V — a high, flat plateau where a charge sits unpushed but arrived expensively. The two zeros live at different places because the two maps encode different information; only where both vanish is a point truly inert. When a question mentions either zero, your first reflex is to ask about the other map explicitly.

Worked examples

Board level — the charged square (CBSE/ISC pattern)

Four charges of $+2.0\ \mu\mathrm{C}$ each are fixed at the corners of a square of side $30\ \mathrm{cm}$ . (a) Find the electric potential at the centre. (b) How much work is needed to bring a charge of $+1.0\ \mu\mathrm{C}$ from infinity to the centre? (c) What is the electric field at the centre? Take $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ .

Step 1. (a) Distance from each corner to the centre: half the diagonal, $r = \dfrac{0.30\sqrt{2}}{2} = 0.212$ m. Same for all four — write this line; it is the symmetry mark.

Step 2. Census, all signs positive: $V = \dfrac{4kq}{r} = \dfrac{4 \times (9.0\times10^9)(2.0\times10^{-6})}{0.212} = \boxed{3.4\times10^5\ \mathrm{V}}$ .

Step 3. (b) Cash-in: $W = qV = (1.0\times10^{-6})(3.4\times10^5) = \boxed{0.34\ \mathrm{J}}$ — the work equals charge times destination potential, because the trip starts at the $V = 0$ reference.

Step 4. (c) By symmetry the four field vectors at the centre cancel in opposite-corner pairs: $\boxed{E = 0}$ . State the pairing — "fields of diagonally opposite charges are equal and opposite at the centre" — that sentence is the mark.

Step 5. The examiner's favourite follow-up, pre-empted: the answers to (b) and (c) are not in tension. Zero field means no push at the destination; 0.34 J is the toll for the journey from infinity through everywhere the field was not zero.

Competitive level — the deceptive zero (JEE-Main/NEET pattern)

Charge $Q_1 = +4.0\ \mu\mathrm{C}$ is at the origin and $Q_2 = -2.0\ \mu\mathrm{C}$ is at $x = 6.0\ \mathrm{cm}$ . Point P lies at $x = 12.0\ \mathrm{cm}$ . (a) Find the potential at P. (b) Find the work needed to bring $q = +1.0\ \mu\mathrm{C}$ from infinity to P. (c) Is the electric field at P zero? (d) Find the other point on the x-axis between the charges where $V = 0$ .

Step 1. (a) Census with signs and distances: $r_1 = 0.12$ m, $r_2 = 0.06$ m. $V_P = k\left(\dfrac{4.0\times10^{-6}}{0.12} - \dfrac{2.0\times10^{-6}}{0.06}\right) = k\left(3.33\times10^{-5} - 3.33\times10^{-5}\right) = \boxed{0}$ . The near small charge exactly outvotes the far large one.

Step 2. (b) $W = qV_P = \boxed{0}$ . Zero net work — the field pushed and pulled along the way, and the books closed even.

Step 3. (c) Fields at P: $E_1 = \dfrac{k(4.0\times10^{-6})}{(0.12)^2} = 2.5\times10^6$ N/C pointing $+x$ (away from $Q_1$ ); $E_2 = \dfrac{k(2.0\times10^{-6})}{(0.06)^2} = 5.0\times10^6$ N/C pointing $-x$ (toward the negative $Q_2$ ). Net: $\boxed{2.5\times10^6\ \mathrm{N/C}}$ in the $-x$ direction — emphatically not zero. $V = 0$ , $W = 0$ , and yet the charge released at P accelerates at once.

Step 4. (d) Between the charges, both distances shrink; set $\dfrac{4.0}{x} = \dfrac{2.0}{0.06 - x}$ (μC and metres understood): $4.0(0.06 - x) = 2.0x$ gives $x = \boxed{0.04\ \mathrm{m}}$ from the origin. Two zeros of $V$ on one axis — one between (at 4.0 cm), one beyond the smaller charge (P itself) — exactly the double-crossing the V–r graph section promised. Timing note: 2–2.5 minutes for all four parts. Parts (a), (b), (d) are pure arithmetic if you write distances first; part (c) is where the trap lives — never infer $E$ from $V$ at a single point. If a JEE option says "V = 0 therefore E = 0", it is the distractor.

Exam traps

Squaring the distance. $V = kQ/r$ , not $kQ/r^2$ . Under pressure, Coulomb's-law muscle memory inserts the square — and one planted option always contains it. The limiting-cases row "double $r$ → half $V$ " is your two-second check.

Dropping a sign in the census. Every term carries its source's sign. Write the signed list before evaluating anything: $+\,kq_1/r_1 - kq_2/r_2 \dots$ . A census with all bars positive is only legal when every charge is positive.

Resolving components for V. If your potential calculation contains $\cos\theta$ , stop — you have drifted into field superposition. Geometry may be needed to find distances (diagonals, Pythagoras), but never to weight contributions.

$V = 0 \Rightarrow E = 0$ (and the converse). Both directions false; both tested every year. Zero of the map ≠ flat map. Compute the other quantity explicitly when asked — it costs thirty seconds and dodges the chapter's favourite ambush.

Forgetting the cash-in direction. $W_{\text{ext}} = qV_{\text{destination}}$ for a trip from infinity (rest to rest). For trips between two finite points, $W_{\text{ext}} = q(V_{\text{final}} - V_{\text{initial}})$ — subtract in the right order, and keep $q$ 's sign: carrying a negative charge to a positive- $V$ point does negative external work (the field pulls it in — wait, check the signs: $q < 0$ , $V > 0$ , so $qV < 0$ ✓, the field does the work).

Units discipline. μC → C ( $\times10^{-6}$ ) and cm → m ( $\times10^{-2}$ ) before the census. $k = 9.0\times10^9$ accepted on all boards here; answers land in volts and joules cleanly when the inputs are SI.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · Paris

Laplace turned the potential from a bookkeeping convenience into a branch of mathematics. Studying how gravity acts inside and around spheroidal planets, he championed a single scalar function — one number per point of space — whose slopes reproduce the force everywhere, and discovered the equation that function obeys in empty space (the "Laplace equation", waiting for you at university). The payoff was industrial: instead of summing force vectors from every particle of a planet, one solves for the scalar landscape once and differentiates. That strategic retreat from vectors to scalars is precisely the move you practised in this article, with $kq/r$ terms instead of planetary integrals. Napoleon, reviewing Laplace's great Mécanique céleste, reportedly asked why God appeared nowhere in it; Laplace's reply — "I had no need of that hypothesis" — is the most famous mission statement determinism ever got.

History

1773

— Lagrange, computing planetary pulls, quietly introduces the function whose slopes give the force — the potential in embryo.

1782

— Laplace and Legendre develop the mathematics of the potential around spheroids; the landscape view of attraction matures.

1785

— Coulomb pins down the inverse-square law: the force whose landscape is the 1/r hill.

1813

— Poisson extends Laplace's equation to points inside the charge, tying the map directly to its sources.

1828

— Green names the "potential function" for electricity and founds the theory this chapter's UG deep-dives inherit.

Going further

Texts. NCERT Physics Part 1 (Class 12), Ch. 2 §2.4–2.5 — the CBSE/NEET canonical statement of $kQ/r$ and superposition · H.C. Verma, Concepts of Physics 2, Ch. 29 · Halliday, Resnick & Walker Ch. 24 (AP-2/A-Level voice) · CIE 9702 topic 18.5 for the A2 phrasing of potential at a point.

Provenance. All results standard (NCERT Ch. 2, HRW Ch. 24). A verifier should recompute: $kQ/r$ ladder $9000/4500/3000$ V at 1/2/3 m from $+1.0\ \mu$ C; square centre $r = 0.30/\sqrt{2} = 0.212$ m, $V = 4\times(9.0\times10^9)(2.0\times10^{-6})/0.212 = 3.4\times10^5$ V, $W = 0.34$ J, $E = 0$ by pairing; deceptive zero $V_P = k(4\times10^{-6}/0.12 - 2\times10^{-6}/0.06) = 0$ , $E_P = 2.5\times10^6$ N/C toward the negative charge, inner zero at $x = 4.0$ cm.

Prerequisites. Electric Potential & Potential Difference · Coulomb's Law · Electrostatic Potential Energy. Leads to. Potential of a Dipole (the neutral cluster's map) · Equipotentials & the Field–Potential Link (reading slopes) · ARCHETYPE: Assembling & Moving Charges (the gym) · Potential of a Localized Charge Distribution (UG deep dive).

Why does the potential use 1/r when Coulomb's law uses 1/r²? I keep mixing them.

Because potential is accumulated force — work gathered over the whole approach from infinity — and accumulation always trades one power of distance. The force at each instant goes as $1/r'^2$ ; adding up force-times-step over the trip produces a total that shrinks only as $1/r$ (the same way the area under the spring's force line $kx$ builds the quadratic $\tfrac12kx^2$ ). Quick anchors: field, force → square; energy, potential → single power. And use the mixing danger productively — if halving the distance quadruples your quantity, you are holding a force; if it merely doubles it, a potential.

How can the potential be non-directional when the charges pull in definite directions?

Because potential answers an energy question, and energy has no direction. "How much work per coulomb to bring a charge here from infinity?" has a single-number answer regardless of the tug-of-war of directions en route — the route does not even matter, since the Coulomb force is conservative. Direction has not been lost, though: it is stored in how the potential changes from point to point. The steepest-drop direction of the map is exactly where a positive charge gets pushed, which is the next module's whole subject. Value of the map: scalar. Slope of the map: the vector you gave up.

Two charges of opposite sign — where exactly does V = 0, and why are there two places?

On the line through unequal opposite charges the census $k q_1/r_1 - k q_2/r_2$ balances twice. Between the charges, both distances shrink together and there is always a balance point, sitting nearer the smaller charge (its smallness must be offset by proximity). Outside the pair, on the smaller charge's far side, distance to the big charge grows slowly enough that a second balance occurs — the worked example's point P. For like charges the census never balances anywhere (all terms share one sign): no $V = 0$ point at any finite distance. Compare null points of the field, which follow different rules — that contrast is a favourite JEE construction.

Does V = 0 at a point mean it takes no energy to keep a charge there?

Careful — three different questions hide in that sentence. Bringing a charge from infinity to a $V = 0$ point costs zero net work: true by definition. Keeping it there costs nothing in electrostatics anyway — holding a charge stationary does no work regardless of the local field (force without displacement). But whether it stays is the field's question, not the potential's: at the deceptive zero of the worked example, a released charge accelerates immediately, because $E \neq 0$ . So: free to deliver, free to hold, but not necessarily willing to stay. Only a point with $E = 0$ lets go of a charge gently — and only a stable one keeps it.

When I bring the +1 μC charge to the square's centre, doesn't it change the potential there?

It changes the potential everywhere else — but the census at any point never includes the charge sitting at that very point (a charge feels the others' landscape, not its own). The 0.34 J you paid is work against the four corner charges' map, and that map is what $V$ at the centre meant. What the newcomer does change: the potential energy of the system now has four new pair terms (newcomer with each corner), which is exactly the 0.34 J, booked by the pair-sum of Electrostatic Potential Energy. The two modules' bookkeepings agree perfectly — $qV$ is just the fast way to total the newcomer's pairs.

Is there a fastest safe route through symmetric-arrangement questions?

Yes, and it is worth rehearsing until it is reflex. First: find the common distance — centres of squares, rings, polygons put every charge at one $r$ (half-diagonal for the square: $a/\sqrt{2}$ ). Second: sum the charges algebraically first, then multiply once: $V = k(\sum q_i)/r$ . Equal and opposite charges kill each other in the sum before any division happens — a hexagon with three $+q$ and three $-q$ gives $V = 0$ at the centre in one glance. Third: state the field by symmetry pairing separately if asked; never derive it from your $V$ value. This pipeline turns five-charge problems into fifteen-second wins.

Very far from a messy cluster of charges, what does the potential look like?

Like a single point charge carrying the cluster's total charge: $V \approx kQ_{\text{total}}/r$ . From far away, the individual distances $r_i$ all look like the one distance $r$ , so the census factorises into $k(\sum q_i)/r$ . This "impersonation" is the first rung of a beautiful ladder: if the total happens to be zero — a neutral molecule — the leading term dies and the map is ruled by the next correction, which falls as $1/r^2$ and depends on how the + and − are separated inside the cluster. That correction is the dipole potential, and it is precisely the subject of Potential of a Dipole, two modules ahead.