In short

The electric potential due to a collection of point charges is the scalar sum of what each charge produces on its own:

V(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{|\vec{r} - \vec{r}_i|}.

For a dipole of moment p = qd, the potential at distance r \gg d is

V \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^2},

so V = +kp/r^2 on the axis (pointing along \vec{p}), V = -kp/r^2 on the opposite axis, and V=0 on the equatorial plane. For continuous distributions, the sum becomes an integral V = \int k\,dq/r: a uniformly charged ring of radius a and charge Q gives V(x) = kQ/\sqrt{x^2+a^2} on its axis; a uniformly charged disc integrates over rings to V(x) = (\sigma/2\varepsilon_0)(\sqrt{x^2+a^2}-|x|); a uniformly charged solid sphere looks like a point charge from outside (V = kQ/r) and has a parabolic profile inside (V = kQ(3R^2-r^2)/2R^3).

Inside an ISRO geostationary satellite, the solar panel strings and the payload electronics sit a few centimetres apart. Before any mission-critical instrument is switched on, an engineer has to ask: what potential does the cabin floor see, given all the charges that have built up on the panels, the harness, the chassis and the thruster nozzles? They cannot answer this by adding forces — forces are vectors, and the arithmetic of keeping track of directions at a thousand different points is murderous. They can answer it instantly by adding potentials, because potential is a scalar. Every charge contributes a number, positive or negative, and the total is the sum of those numbers.

This is the quiet superpower of the electric potential. The field \vec{E} from two charges requires you to decompose each source into x, y, z components, add those, and recombine the magnitude. The potential V from the same two charges requires you to add two real numbers. When you have 10^{20} charges on the surface of a conductor, the difference between scalar and vector addition is the difference between a tractable integral and an unsolvable mess. This article stacks potentials — first from a handful of point charges, then from a dipole, then from the continuous distributions you will meet inside capacitors, on satellite skins, and on the collecting plates of the Koradi and Sipat thermal power plants where electrostatic precipitators clean flue gas.

The superposition rule — potential adds as a scalar

Fix a collection of point charges q_1, q_2, \ldots, q_n at positions \vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n. You want the potential at some field point \vec{r}.

From the previous chapter, the potential due to a single point charge q_i at the field point \vec{r} is

V_i(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_i}{|\vec{r} - \vec{r}_i|},

with the reference taken at infinity (V \to 0 as the distance grows without bound).

Why: this is the work per unit charge to bring a positive test charge from infinity to the point \vec{r}, against the field of q_i alone. It is a single real number — positive if q_i > 0, negative if q_i < 0 — with no direction attached.

Now bring the test charge in with all n source charges present at once. Because the electrostatic force is linear in the sources (Coulomb's law with superposition), the total work done against all of them is the sum of the works you would do against each one in isolation. Dividing by the test charge gives

\boxed{\;V(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{|\vec{r} - \vec{r}_i|}\;} \tag{1}

Why: the force on the test charge was \vec{F} = q_0 \sum_i \vec{E}_i, so the line integral from infinity to \vec{r} factors into q_0 \sum_i \int \vec{E}_i \cdot d\vec{\ell}. Each integral is the work per unit charge against source i alone, which is q_0 V_i(\vec{r}). Dividing by q_0 leaves the scalar sum.

Three features of (1) are worth burning in:

A three-charge warm-up

Three charges sit at the corners of an equilateral triangle of side 1 m: q_1 = +2\ \mu\text{C}, q_2 = +2\ \mu\text{C}, q_3 = -2\ \mu\text{C}. What is V at the centroid?

The centroid is equidistant from all three vertices. For an equilateral triangle of side a, the distance from a vertex to the centroid is r = a/\sqrt{3}. With a = 1 m, r = 1/\sqrt{3}\approx 0.577 m.

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r}+\frac{q_2}{r}+\frac{q_3}{r}\right) = \frac{k}{r}(q_1+q_2+q_3) = \frac{9\times 10^9}{0.577}\times(2\times 10^{-6}) \approx 3.12\times 10^4\ \text{V}.

Why: equal distances let you factor out 1/r and simply add the charges as signed numbers. Two positives and one negative cancel to one positive charge of magnitude 2\ \mu\text{C} worth of potential at the centroid. Contrast with the field at the centroid: the two positive charges' fields point away from them (outward), the negative one's field points toward it (also outward from the centroid towards the negative vertex), and because of the triangle's symmetry, the fields of the two positives cancel while the negative one leaves a net field pointing toward it. The scalar superposition for V was a one-line calculation; the vector superposition for \vec{E} needed a geometric argument.

Three point charges at the corners of an equilateral triangle Two positive charges at the top-left and top-right vertices, one negative charge at the bottom vertex, centroid marked inside. Annotations show equal distances from centroid to each vertex and the scalar sum. +2 +2 −2 centroid r r r charges in μC side = 1 m r = 1/√3 m V = k(q₁+q₂+q₃)/r V ≈ 3.12 × 10⁴ V
Equal distances from the centroid to all three vertices. The scalar sum of charges is $+2\,\mu\text{C}$; multiplied by $k/r$ it gives the potential at the centre directly, with no vector decomposition.

Watching the potentials add

It helps to see the superposition happen as the test point moves. In the figure below, two charges +q and -q sit on a line; the blue curve is the potential of +q alone, the red is the potential of -q alone, and the dark curve is their sum. Drag the position of the test point along the axis and watch the three numbers update and the sum reconstruct itself in real time.

Interactive superposition of potentials from two opposite charges Two point charges at x = -2 and x = +2 with charges +1 and -1 in arbitrary units. Three curves show V from +q, V from -q and V total, with a draggable test point on the x axis reading off each contribution. x (m) V +q −q V₊ V₋ V total drag the red dot along x
Drag the red marker along the $x$-axis. The two dashed curves are the separate potentials of the $+q$ and $-q$ charges; the solid dark curve is their sum. Note the total $V$ passes through zero at the midpoint — not because the field is zero there (it isn't), but because the two scalar contributions cancel exactly.

Potential of a dipole

An electric dipole is two equal and opposite charges +q and -q held a distance d apart. The dipole moment \vec{p} has magnitude p = qd and points from the negative charge to the positive charge. Dipoles appear everywhere: a water molecule has a permanent dipole moment of about 6.2\times 10^{-30}\ \text{C·m}; the charge separation on a thundercloud produces a giant vertical dipole; the molecular dipoles inside a dielectric align in an external field and produce the capacitance enhancement you will meet in the chapters ahead.

Set up coordinates with the midpoint of the dipole at the origin and the axis along \hat{z}. The positive charge sits at +d/2\,\hat{z} and the negative at -d/2\,\hat{z}.

Dipole geometry and the field point Positive charge at z = +d/2 and negative at z = -d/2 along the vertical axis, a field point P at spherical coordinates r, theta, with distances r1 to the positive charge and r2 to the negative charge marked. +z (axis) equator +q −q +d/2 −d/2 P r₊ r₋ r θ p = qd
A dipole along $\hat{z}$ with moment $\vec{p} = qd\,\hat{z}$. The field point $P$ sits at distance $r$ from the midpoint and makes angle $\theta$ with $\vec{p}$. The two source-to-field distances are $r_+$ (to $+q$) and $r_-$ (to $-q$).

General expression at (r,\theta)

From the cosine rule applied to the triangle with sides r, d/2 and the included angle \theta (for r_+) or \pi-\theta (for r_-),

r_+^2 \;=\; r^2 + \tfrac{d^2}{4} - rd\cos\theta, \qquad r_-^2 \;=\; r^2 + \tfrac{d^2}{4} + rd\cos\theta. \tag{2}

Why: the cosine rule c^2 = a^2 + b^2 - 2ab\cos C with a=r, b=d/2, and the angle at the origin between r and the +z side is \theta. For the negative charge, the angle to the origin side toward -z is \pi-\theta, and \cos(\pi-\theta) = -\cos\theta flips the sign of the cross term.

The potential by superposition is

V(r,\theta) \;=\; \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_+}-\frac{q}{r_-}\right). \tag{3}

This is exact. It works at every distance — you have not yet made any approximation. But the formula in (3) is not very insightful as it stands. The revealing move is to look at the far field r \gg d, where almost every application of the dipole concept lives.

Far-field approximation — the p\cos\theta/r^2 form

For r \gg d, the ratio d/r is small. Divide both sides of (2) by r^2:

\frac{r_+^2}{r^2} \;=\; 1 - \frac{d\cos\theta}{r} + \frac{d^2}{4r^2}.

Discard the (d/r)^2 term (it is much smaller than the d/r term when d\ll r):

\frac{r_+}{r} \;\approx\; \sqrt{1 - \frac{d\cos\theta}{r}}\;\approx\; 1 - \frac{d\cos\theta}{2r}.

Why: the binomial approximation (1+x)^{1/2} \approx 1 + x/2 for small x. You are not memorising this — it falls straight out of the Taylor expansion of \sqrt{1+x} around x=0, which is the formal "small distortion of 1" statement.

Invert:

\frac{1}{r_+} \;\approx\; \frac{1}{r}\left(1 + \frac{d\cos\theta}{2r}\right). \tag{4}

Why: (1+x)^{-1}\approx 1-x for small x, with x = -d\cos\theta/(2r). The minus signs conspire to give a +d\cos\theta/(2r) correction.

The same manipulation for the negative charge, with the sign of the cross term flipped, gives

\frac{1}{r_-} \;\approx\; \frac{1}{r}\left(1 - \frac{d\cos\theta}{2r}\right). \tag{5}

Subtract (5) from (4):

\frac{1}{r_+}-\frac{1}{r_-}\;\approx\;\frac{1}{r}\cdot\frac{d\cos\theta}{r} \;=\; \frac{d\cos\theta}{r^2}. \tag{6}

Why: the "1" parts cancel exactly, leaving only the first-order correction. This cancellation is the reason the dipole potential falls off as 1/r^2 rather than 1/r: the two opposite charges almost perfectly erase each other at large distance, and what survives is the small asymmetry they create.

Substitute into (3):

\boxed{\;V(r,\theta) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^2}\;} \tag{7}

with p = qd. This is the far-field dipole potential — the workhorse formula.

Three directions through the dipole

Equation (7) collapses to particularly clean values along three special directions.

On the axis (\theta = 0, along +\vec{p}):

V_\text{axial} \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^2}. \tag{8}

On the opposite axis (\theta = \pi, behind the negative charge):

V \;=\; -\frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^2}.

On the equatorial plane (\theta = \pi/2):

V_\text{equatorial} \;=\; 0. \tag{9}

Why: on the equator, every point is equidistant from both +q and -q (r_+ = r_-), so the two potentials cancel exactly. No approximation is needed for (9) — it is true at every distance, not just in the far field. The field on the equator is not zero, but the potential is — a reminder that V and \vec{E} are not the same quantity.

Comparison with the point-charge potential

A dipole's potential falls off as 1/r^2, one power of r faster than a point charge's 1/r. At ten dipole lengths, a dipole's potential is a hundred times weaker than what a single charge q would produce at the same distance. The reason is in (6): the two opposite charges almost cancel, and only the small "asymmetry" term d\cos\theta/r^2 survives. This pattern continues for higher multipoles — a quadrupole falls as 1/r^3, an octupole as 1/r^4 — and is the content of the multipole expansion you will meet in the going-deeper section.

Exploring the dipole's potential landscape

The interactive figure below plots the exact dipole potential V(r,\theta) on a polar map around a dipole. Drag the test point around the dipole and watch the angle dependence: the potential is large and positive along +\vec{p}, zero on the equator, and large and negative along -\vec{p}.

Interactive dipole potential versus angle Curve V(θ) = cos(θ)/r² for fixed r, with θ from 0° (along +p) to 180° (along -p). A draggable marker reads off the scaled potential at the chosen angle. θ (degrees) V · r²/(kp) +1 0 −1 0° 90° 180° drag along x to set the angle
For fixed $r$ the dipole potential depends on angle as $\cos\theta$: maximum at $\theta = 0$ (along $+\vec{p}$), zero at $\theta = 90°$ (equator), minimum at $\theta = 180°$ (behind the dipole).

Continuous distributions — the integral form

When the charges are packed so densely that you want to treat them as a continuum — a thin ring, a flat disc, a solid sphere — replace the sum in (1) with an integral:

V(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\int\frac{dq}{|\vec{r}-\vec{r}'|}, \tag{10}

where dq is the charge of an infinitesimal piece of the distribution at position \vec{r}'. The element dq is written as a density times a volume, surface, or length:

dq \;=\; \rho\,dV \quad\text{(volume)},\qquad dq \;=\; \sigma\,dA \quad\text{(surface)},\qquad dq \;=\; \lambda\,d\ell \quad\text{(line)}.

This is not a new physics rule — it is (1) with "sum" replaced by "integral." The scalar nature of V means you never worry about which direction dq is pulling from; just integrate the number. The next three subsections apply (10) to the three distributions you will need again and again.

Uniformly charged ring — on the axis

A thin circular ring of radius a carries total charge Q spread uniformly around it. Find V at a point on the ring's axis, a distance x from the centre.

Ring of charge and a field point on its axis A horizontal ring of radius a viewed in cross-section with a small element dq on the ring, a field point P at distance x along the axis, and the straight-line distance sqrt(x squared plus a squared) from dq to P. dq a P (axis) x √(x² + a²) ring radius a, total charge Q
Every infinitesimal element $dq$ of the ring sits at the same distance $\sqrt{x^2 + a^2}$ from the axial field point $P$. That single geometric fact makes the integral trivial.

Step 1. Write the distance from any element dq to the field point.

Every element of the ring is at the same perpendicular distance a from the axis and the same axial distance x from P. By the Pythagorean theorem, the straight-line distance from dq to P is

r' \;=\; \sqrt{x^2 + a^2},

a constant independent of where on the ring the element sits.

Why: the ring's symmetry makes every point on it equivalent for a field point on the axis. This is the reason the ring problem is a one-line integral while the off-axis problem requires elliptic integrals.

Step 2. Substitute into (10).

V(x) \;=\; \frac{1}{4\pi\varepsilon_0}\int\frac{dq}{\sqrt{x^2+a^2}} \;=\; \frac{1}{4\pi\varepsilon_0}\cdot\frac{1}{\sqrt{x^2+a^2}}\int dq.

Why: \sqrt{x^2+a^2} does not depend on which element you are integrating over (only on x and a), so it comes out of the integral. The remaining \int dq is just the total charge on the ring.

Step 3. Evaluate \int dq = Q.

\boxed{\;V_\text{ring}(x) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{\sqrt{x^2+a^2}}\;} \tag{11}

Two limiting checks:

The field on the axis is E_x = -dV/dx = kQx/(x^2+a^2)^{3/2}, which you met in the electric field chapter by direct integration of Coulomb's law. Deriving it from V instead is one derivative instead of a vector integral — the scalar potential has done the hard work.

Uniformly charged disc — build it from rings

A flat disc of radius R carries surface charge density \sigma (C/m²) uniformly over its face. Find V at a point on its axis, a distance x from the centre.

Slice the disc into concentric rings of radius a and thickness da. The charge on one such ring is

dQ \;=\; \sigma\,(2\pi a\,da).

Why: the area of a thin ring of radius a and width da is its circumference 2\pi a times its width da. Multiplying by the surface density \sigma gives the charge on that ring.

By (11), this ring contributes

dV \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{dQ}{\sqrt{x^2+a^2}} \;=\; \frac{\sigma}{2\varepsilon_0}\,\frac{a\,da}{\sqrt{x^2+a^2}}.

Why: 2\pi a\,da in the numerator pairs with the 4\pi\varepsilon_0 in the denominator to produce \sigma/(2\varepsilon_0), which is a constant in the outer integral.

Integrate from a=0 (centre) to a=R (edge):

V(x) \;=\; \frac{\sigma}{2\varepsilon_0}\int_0^R \frac{a\,da}{\sqrt{x^2+a^2}}.

The substitution u = x^2+a^2, du = 2a\,da sends this to

\int \frac{a\,da}{\sqrt{x^2+a^2}} \;=\; \frac{1}{2}\int\frac{du}{\sqrt{u}} \;=\; \sqrt{u} \;=\; \sqrt{x^2+a^2}.

Why: the integrand a\,da/\sqrt{x^2+a^2} is engineered for this substitution — the numerator is exactly half the differential of the expression under the square root.

Evaluate from 0 to R:

V(x) \;=\; \frac{\sigma}{2\varepsilon_0}\Big[\sqrt{x^2+R^2} - \sqrt{x^2}\Big] \;=\; \frac{\sigma}{2\varepsilon_0}\Big[\sqrt{x^2+R^2} - |x|\Big].

Why: \sqrt{x^2} = |x|, not x — the modulus matters when x can be negative (field point on the other side of the disc). For x > 0 (the usual case), drop the bars.

\boxed{\;V_\text{disc}(x) \;=\; \frac{\sigma}{2\varepsilon_0}\Big[\sqrt{x^2+R^2} - |x|\Big]\;} \tag{12}

Two checks:

Uniformly charged solid sphere — use Gauss's law for the field, then integrate

A solid ball of radius R carries total charge Q uniformly through its volume (\rho = 3Q/(4\pi R^3)). Find V(r) at a distance r from the centre, both outside and inside the sphere.

You met the field of this distribution in Applications of Gauss's Law — Spheres and Planes. It is

E(r) \;=\; \begin{cases}\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2} & r > R,\\[6pt]\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qr}{R^3} & r < R.\end{cases}

The potential follows from

V(r) \;=\; -\int_\infty^r E(r')\,dr',

with the reference V(\infty) = 0 and \vec{E} pointing radially outward (so \vec{E}\cdot d\vec{r} = E\,dr).

Outside (r > R):

V(r) \;=\; -\int_\infty^r \frac{1}{4\pi\varepsilon_0}\frac{Q}{r'^2}dr' \;=\; \frac{Q}{4\pi\varepsilon_0}\cdot\frac{1}{r}. \tag{13}

Why: the integral \int dr'/r'^2 = -1/r'. Evaluated between \infty and r, the upper limit kills the -1/\infty = 0 contribution and leaves -1/r; the overall minus sign flips it to +1/r. The sphere at external points looks exactly like a point charge at its centre.

At the surface (r = R):

V(R) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{R}. \tag{14}

Inside (r < R): you cannot integrate from infinity in one step because the field law changes at r = R. Split the path:

V(r) \;=\; -\int_\infty^R E_\text{out}\,dr' \;-\; \int_R^r E_\text{in}\,dr' \;=\; V(R) \;-\; \int_R^r \frac{1}{4\pi\varepsilon_0}\frac{Q r'}{R^3}dr'.

Why: the first piece is just V(R) from (14). The second piece is the line integral of the inside field from the surface inwards; the inside field grows linearly with r'.

Evaluate:

\int_R^r \frac{Q r'}{R^3}dr' \;=\; \frac{Q}{R^3}\cdot\frac{r'^2}{2}\Big|_R^r \;=\; \frac{Q}{2R^3}(r^2 - R^2).

So

V(r) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{R} \;-\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{2R^3}(r^2 - R^2).

Combine the two terms over the common denominator 2R^3:

V(r) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{2R^3}\Big[2R^2 - (r^2 - R^2)\Big] \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q(3R^2 - r^2)}{2R^3}.
\boxed{\;V_\text{inside}(r) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Q(3R^2 - r^2)}{2R^3}\;} \tag{15}

Three features of (13)–(15):

Potential of a uniformly charged solid sphere versus radial distance Curve showing potential V(r) as parabolic for r less than R and 1/r for r greater than R, with markers at r = 0, r = R and a dashed line showing the sphere boundary. r (radial distance) V(r) r = R V(0) = 3kQ/2R V(R) = kQ/R inside: parabolic outside: 1/r
The potential inside a uniformly charged sphere is parabolic (peaks at the centre); outside it decays like $1/r$. The two join smoothly at $r = R$ with matching value and slope.

For a uniformly charged spherical shell (not a solid ball — all charge on the surface, nothing inside), the same calculation gives V = kQ/r outside and V = kQ/R everywhere inside. The interior is an equipotential because \vec{E} = 0 inside a charged shell (Gauss's law). All the workings for this case sit in Applications of Gauss's Law — Spheres and Planes.

Worked examples

Example 1: Potential at the centre of a dipole-like triangle

Two charges +5\,\mu\text{C} and -5\,\mu\text{C} sit at the ends of a 2 cm rod (a physicist's dipole, with d = 2 cm and p = qd = 5\times 10^{-6}\times 0.02 = 10^{-7} C·m). A third charge +3\,\mu\text{C} is placed 5 cm from the rod's midpoint, along the line joining the charges (on the positive side). Find the potential at the point 10 cm from the midpoint on the equatorial plane of the rod.

Configuration for Example 1 A horizontal dipole with +5 microcoulombs on the right and -5 microcoulombs on the left, midpoint at origin, a +3 microcoulombs charge 5 cm to the right, and a field point P 10 cm above the midpoint. −5 +5 d = 2 cm +3 5 cm from midpoint P (field point) 10 cm r₃
The dipole-like pair sits horizontally; a third charge is offset to the right; the field point is 10 cm directly above the midpoint.

Step 1. List all distances.

r_+ = \sqrt{10^2 + 1^2}\ \text{cm} = \sqrt{101}\ \text{cm} \approx 10.050 cm = 0.10050 m.

r_- = \sqrt{10^2 + 1^2}\ \text{cm} = \sqrt{101}\ \text{cm} \approx 10.050 cm = 0.10050 m.

r_3 = \sqrt{10^2 + 5^2}\ \text{cm} = \sqrt{125}\ \text{cm} \approx 11.180 cm = 0.11180 m.

Why: P is on the equatorial plane of the \pm 5 pair, so r_+ = r_-. The distance from P to the third charge uses Pythagoras with a 5 cm horizontal offset and a 10 cm vertical offset.

Step 2. Apply the superposition rule (1).

V \;=\; k\left(\frac{+5\times 10^{-6}}{0.10050} + \frac{-5\times 10^{-6}}{0.10050} + \frac{+3\times 10^{-6}}{0.11180}\right).

Why: scalar superposition — each charge contributes its own kq/r with its own sign. Directions do not enter.

Step 3. The first two terms cancel exactly.

V \;=\; k\cdot\frac{3\times 10^{-6}}{0.11180} \;=\; \frac{9\times 10^{9}\times 3\times 10^{-6}}{0.11180} \;\approx\; 2.415\times 10^{5}\ \text{V}.

Why: the equatorial symmetry of the \pm 5 pair means their potentials at P are equal and opposite — they cancel to zero. Only the third charge contributes. This is equation (9) — the dipole contribution on the equator is zero at every distance, not just far away.

Result: V \approx 2.42\times 10^{5} V = 242 kV at the field point.

What this shows: Even when a dipole is close to your field point (here just 5 cm apart versus 10 cm to the field point, so not in the far-field regime at all), the equatorial plane still sits at zero contribution from the dipole — scalar superposition + symmetry kills it exactly. The only charge that matters is the extra +3\ \mu\text{C}, and its contribution is a single kq/r term.

Example 2: Electron at the centre of a charged ring

A uniformly charged ring of radius a = 5 cm carries a total charge Q = +20 nC. An electron (charge -e, mass m_e = 9.11\times 10^{-31} kg) is released from rest at a point x_0 = 12 cm on the axis. Find (a) the potential at the starting point, (b) the potential at the centre of the ring, and (c) the speed of the electron when it reaches the centre.

Ring of charge with electron released on axis A ring of radius 5 cm viewed edge-on with the electron starting 12 cm from the centre along the axis, potential values at the two ends of the trajectory marked. ring: Q = +20 nC, a = 5 cm centre V = 3600 V e⁻ x₀ = 12 cm V ≈ 1385 V force on e⁻ (accelerates toward ring)
The electron is pulled toward the positive ring. Use scalar potentials at the start and end to find the speed at the centre by energy conservation.

Step 1. Potential at x_0 = 12 cm using (11).

V(x_0) \;=\; \frac{kQ}{\sqrt{x_0^2+a^2}} \;=\; \frac{9\times 10^9 \cdot 20\times 10^{-9}}{\sqrt{0.12^2+0.05^2}} \;=\; \frac{180}{\sqrt{0.0169}} \;=\; \frac{180}{0.13} \;\approx\; 1385\ \text{V}.

Why: \sqrt{x^2+a^2} = \sqrt{0.0144+0.0025} = \sqrt{0.0169} = 0.13 m. Then V = kQ/0.13. The ring looks "roughly" like a point charge Q at 13 cm, but with the correct geometric distance rather than an axial approximation.

Step 2. Potential at the centre, x=0.

V(0) \;=\; \frac{kQ}{a} \;=\; \frac{9\times 10^9\cdot 20\times 10^{-9}}{0.05} \;=\; 3600\ \text{V}.

Why: at x=0 the only relevant distance is the ring's radius. The centre sits closer (on average) to the charge than the axial point did, so the potential is larger there — as expected.

Step 3. Apply conservation of energy. The electron starts at rest; when it reaches the centre, all the potential-energy loss has turned into kinetic energy.

\Delta U + \Delta K \;=\; 0, \qquad \Delta U \;=\; q\Delta V \;=\; (-e)(V_{\text{final}} - V_{\text{initial}}).

With V_\text{final} = 3600 V and V_\text{initial} = 1385 V:

\Delta U \;=\; -e\cdot(3600 - 1385) \;=\; -1.602\times 10^{-19}\times 2215 \;=\; -3.548\times 10^{-16}\ \text{J}.

Why: the electron moves to a region of higher potential V, but because its charge is negative, its potential energy U = qV decreases. Negative \Delta U means positive \Delta K — the electron speeds up, as the arrow in the diagram anticipates.

Step 4. Set \Delta K = -\Delta U and solve for the final speed.

\tfrac{1}{2}m_ev^2 \;=\; 3.548\times 10^{-16}\ \text{J}.
v \;=\; \sqrt{\frac{2\times 3.548\times 10^{-16}}{9.11\times 10^{-31}}} \;\approx\; \sqrt{7.79\times 10^{14}} \;\approx\; 2.79\times 10^{7}\ \text{m/s}.

Why: an electron accelerated through 2215 volts reaches about 2.8\times 10^{7} m/s, which is about 9% of the speed of light. At this speed relativistic corrections are only about 0.5%, so the non-relativistic calculation is still accurate to three digits — but if the potential difference were a few hundred kilovolts (as it is inside the particle accelerator at BARC), you would need relativity.

Result: The electron arrives at the centre with speed v \approx 2.79\times 10^{7} m/s.

What this shows: The scalar potential turns a difficult force-integration problem (the axial component of the field changes at every point of the path) into two plug-ins of equation (11) plus one application of energy conservation. This is the point of working with V instead of \vec{E}: for energy problems involving a test charge moving in a static field, V is all you need.

Common confusions

If all you need is to compute potentials for JEE — dipoles, rings, discs, spheres, and combinations — you have the tools. The rest of this section looks at the multipole expansion, which explains why a dipole's potential falls as 1/r^2 and what comes after it, and at the formal relationship between the continuous sum and the discrete sum.

The multipole expansion — organising the far-field potential

Take an arbitrary finite charge distribution confined within some radius a of the origin. Ask: what does V look like at a far point \vec{r}, where r \gg a?

Write V as the integral form (10). The key step is to expand 1/|\vec{r}-\vec{r}'| in powers of r'/r:

\frac{1}{|\vec{r}-\vec{r}'|} \;=\; \frac{1}{r}\sum_{\ell=0}^{\infty}\left(\frac{r'}{r}\right)^\ell P_\ell(\cos\alpha),

where \alpha is the angle between \vec{r} and \vec{r}' and P_\ell are the Legendre polynomials: P_0 = 1, P_1(\cos\alpha) = \cos\alpha, P_2(\cos\alpha) = (3\cos^2\alpha - 1)/2, and so on.

Why: the Legendre polynomials are the natural functions for axially symmetric problems in spherical coordinates — they appear whenever you expand 1/\text{distance} of two points in terms of their relative angle. You will meet them formally if you take an advanced course in electromagnetism, but the idea is purely geometric: each \ell captures a different "shape" of angular dependence.

Substitute this expansion into the integral for V:

V(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\sum_{\ell=0}^{\infty}\frac{1}{r^{\ell+1}}\int r'^\ell P_\ell(\cos\alpha)\,dq.

The term with \ell = 0 is the monopole contribution:

V_0 \;=\; \frac{1}{4\pi\varepsilon_0 r}\int dq \;=\; \frac{Q_\text{total}}{4\pi\varepsilon_0 r}.

This is the 1/r piece. It is non-zero whenever the distribution has non-zero net charge.

The term with \ell = 1 is the dipole contribution:

V_1 \;=\; \frac{1}{4\pi\varepsilon_0 r^2}\int r'\cos\alpha\,dq \;=\; \frac{\vec{p}\cdot\hat{r}}{4\pi\varepsilon_0 r^2},

where \vec{p} = \int\vec{r}'\,dq is the dipole moment of the distribution. This is the 1/r^2 piece. For our simple two-charge dipole, \vec{p} = qd\hat{z} and \vec{p}\cdot\hat{r} = qd\cos\theta, recovering (7).

The term with \ell = 2 is the quadrupole, falling as 1/r^3. And so on.

Here is the organising principle: for a distribution with net charge, the monopole dominates at large distance and everything else is a small correction. For a net-neutral distribution (total charge zero), the monopole vanishes and the dipole dominates. For a distribution that is both net-neutral and has zero dipole moment (e.g., a symmetric quadrupole — four charges at the corners of a square, alternating sign), the leading term is 1/r^3.

Every real distribution has all these contributions. The multipole expansion is simply saying: from far enough away, the distribution looks like a point charge (monopole); from a bit closer, you can see the slight asymmetry (dipole); from closer still, you can resolve the quadrupole shape; and so on.

From discrete to continuous — a formal comment

The jump from sum to integral in (10) looks purely notational, but it hides a subtle point: the sum \sum_i kq_i/|\vec{r}-\vec{r}_i| is a sum of n terms, each of which is finite (assuming the field point is not located at one of the source positions). The integral \int k\,dq/|\vec{r}-\vec{r}'| is a Riemann sum limit of infinitely many infinitesimal pieces.

When does this limit exist? The integrand has a 1/r singularity at the source. For the integral to converge when the field point sits on (or inside) the distribution, the charge density times r^2\,dr (in three dimensions) — which is the amount of charge in a shell of radius r and thickness dr — must not blow up faster than r^2 as r\to 0.

For a volume distribution \rho(\vec{r}') that is bounded (finite density everywhere), dq = \rho\,dV and dV = r^2\sin\theta\,dr\,d\theta\,d\phi in spherical coordinates — the r^2 in the volume element cancels the 1/r in the Coulomb kernel, leaving an integrand that is \rho r \sin\theta\,dr\,d\theta\,d\phi, perfectly finite at r=0. Volume distributions always give finite potentials.

For a surface distribution \sigma(\vec{r}'), dq = \sigma\,dA, and the 1/r singularity is only cancelled by the r inside the area element in spherical coordinates — so you get a \log divergence if the field point sits on the surface. In practice, the surface charge on a conductor is a mathematical idealisation of a thin volume distribution; real surface charges give finite potentials. But the line charge idealisation (a mathematical line of zero thickness carrying finite \lambda) gives a logarithmically infinite potential at points on the line — which is why you cannot use V(\infty) = 0 as a reference for an infinite line of charge, and instead pick a finite reference distance.

These are not abstractions. When you meet the parallel-plate capacitor in the next chapter, the idealisation of surface charge on an infinite plane gives a potential that grows linearly with distance from the plane, so "zero at infinity" breaks. Instead, you choose V = 0 on one of the plates and measure everything from there.

Why the scalar superposition principle is exactly as deep as the force superposition

The scalar nature of V and the vector superposition of \vec{E} are two sides of the same coin. The underlying statement is:

The electrostatic force is linear in the source charges. Double all source charges and you double every force. Add a new source charge and the total force is the old total plus the new charge's contribution.

From this, \vec{E} = \vec{F}/q_0 and V = -\int \vec{E}\cdot d\vec{\ell} both inherit linearity. You get vector superposition for \vec{E} and scalar superposition for V for the same physical reason. The advantage of V is pragmatic: adding numbers is easier than adding vectors, and once you have V everywhere, a single derivative gives you \vec{E} (the topic of chapter 158 — relation between field and potential).

This is why, in practice, electrostatics problems are almost always attacked via V and not directly via \vec{E}. Even Gauss's law problems, where you compute \vec{E} by symmetry first, are often followed immediately by an integration to get V, because V is what enters energy calculations (next chapter on the energy stored in a capacitor).

Where this leads next