Introduction: the ring is the building block for disks, coils, and beam optics

A uniformly charged ring is the first continuous charge distribution most physics courses integrate by hand, and it stays useful long after the exam. Slice a charged disk into concentric rings and integrate; stack rings along an axis and you have a charged cylinder; swap charge for current and the identical geometry gives the on-axis field of a circular coil. Getting the ring right is therefore worth more than one result.

This page computes the electric field on the symmetry axis of a thin ring carrying a uniform total charge $Q$ , at an axial distance $z$ from the centre, for a ring of radius $R$ . It also inverts that relationship: leave any one of the four boxes blank and the calculator solves for it.

The reason a closed form exists at all is symmetry. Pair each charge element with the element diametrically opposite it. Their radial contributions are equal and opposite and cancel exactly; their axial contributions are equal and add. Every element sits the same slant distance $\sqrt{z^{2} + R^{2}}$ from the field point, and each contributes an axial share proportional to $\cos θ = z / \sqrt{z^{2} + R^{2}}$ . The integral collapses to a multiplication, which is why the answer is algebraic rather than a special function. That collapse is also exactly what fails off the axis, and it is the single largest limitation of everything on this page.

How to use the charged ring calculator and read the field profile

The four boxes are linked by one equation, so exactly three of them get filled in. Enter the total ring charge $Q$ in coulombs, the ring radius $R$ in metres, the axial distance $z$ in metres, and the field magnitude $E$ in newtons per coulomb for whichever three you know, leave the fourth completely empty, and press Compute Missing Quantity.

Every box takes zero or positive numbers, because the tool reports a magnitude. A negative charge does not make $E$ negative; it reverses the direction the field points. Enter the size of the charge, read the magnitude, and attach the sign yourself once you have fixed a positive direction for the axis. Scientific notation works everywhere: type 2e-6 rather than 0.000002.

After each successful solve, the curve below the form plots $E (z)$ for your $Q$ and $R$ , marks the axial maximum, and drops a pin on your operating point. The address bar also updates with a link that reproduces the whole state, which is handy for problem sets and lab notes.

Mixing centimetres with metres is the single most common error here. Because $E$ carries a net inverse-square dependence at large $z$ , entering 10 instead of 0.10 changes the answer by four orders of magnitude, and the result still looks perfectly plausible.

Formula for the axial field, its derivative, and the axial maximum

Integrating Coulomb's law around the loop with the cosine projection described above gives the standard result:

E = \frac{k Q z}{{(z^{2} + R^{2})}^{\frac{3}{2}}}

$E$ = axial electric field magnitude, in N/C
$Q$ = total ring charge, in C
$R$ = ring radius, in m
$z$ = axial distance from the ring centre, in m
$k = 1 / 4 π ε_{0}$ = Coulomb constant, 8.9875517862×10⁹ N·m²/C²

Differentiating with respect to $z$ gives the slope of the curve, which is what makes the maximum easy to locate:

\frac{d E}{d z} = k Q \frac{R^{2} - 2 z^{2}}{{(z^{2} + R^{2})}^{\frac{5}{2}}}

The numerator vanishes when $R^{2} = 2 z^{2}$ , so the field peaks at

z_{\max} = \frac{R}{\sqrt{2}} \approx 0.7071 R

and substituting back gives a peak value that depends only on $Q$ and $R$ :

E_{\max} = \frac{2 k Q}{3 \sqrt{3} R^{2}} \approx 0.3849 \frac{k Q}{R^{2}}

Inverting for the charge or the radius is straightforward algebra. Given $E$ , $R$ and a nonzero $z$ :

Q = \frac{E {(z^{2} + R^{2})}^{\frac{3}{2}}}{k z}

and, solving the same relation for the radius,

R = \sqrt{{(\frac{k Q z}{E})}^{\frac{2}{3}} - z^{2}}

If the quantity under the square root is negative, no real ring radius produces that field at that distance, and the calculator says so rather than returning a complex number.

Why solving for z has two answers, and what most calculators get wrong

The axial distance is the awkward one, because $z$ appears both in the numerator and inside the three-halves power. The function $E (z)$ rises from zero at the centre, peaks at $R / \sqrt{2}$ , and decays back toward zero. It is therefore not one-to-one, and a requested field below the peak has two valid solutions: one on the rising inner branch and one on the falling outer branch.

A naive Newton iteration converges to whichever root happens to lie in its basin of attraction and reports it as if it were the answer. That is a silent error, and it is why this calculator brackets each branch separately and bisects, returning both roots when they exist. Three cases are possible for a given $Q$ , $R$ , and requested $E$ :

$E > E_{\max}$ : no solution exists at any distance. The calculator reports the peak field this ring can reach and stops.
$E = E_{\max}$ : exactly one solution, the tangency point at $R / \sqrt{2}$ .
$E < E_{\max}$ : two solutions, reported as a near root and a far root.

Physically the two roots are not interchangeable. On the inner branch the field is dominated by the geometry of the loop and scales roughly as $k Q z / R^{3}$ ; on the outer branch the ring is already behaving like a point charge and the field scales as $k Q / z^{2}$ . If your problem says "far from the ring", you want the far root; if it describes a probe inside the loop's own scale, you want the near one.

Worked example: a 2 μC ring of 10 cm radius

Take the kind of hoop you might clamp to a stand for a demonstration: 2 microcoulombs spread evenly around a 20-centimetre diameter ring, with the probe on the axis 5 cm from the centre.

Ring charge $Q$ = 2 μC = 2e-6 C
Ring radius $R$ = 0.10 m
Axial distance $z$ = 0.05 m

The denominator first: $z^{2} + R^{2}$ = 0.0025 + 0.0100 = 0.0125 m², and raising that to the three-halves power gives 1.3975×10⁻³ m³. The numerator is $k Q z$ = 8.98755×10⁹ × 2×10⁻⁶ × 0.05 = 898.76. Dividing:

E = \frac{898.76}{1.3975 \times 10^{- 3}} = 6.43 \times 10^{5} N/C

This ring peaks at $z_{\max} = 0.0707$ m with $E_{\max} = 0.3849 \times 8.98755 \times 10^{9} \times 2 \times 10^{- 6} / 0.01$ = 6.92×10⁵ N/C, so the 5 cm reading sits just below the crest on the rising branch.

Now run the inverse. Ask for $E$ = 6.43×10⁵ N/C with the same $Q$ and $R$ and leave $z$ blank. The calculator returns two distances: 0.0500 m on the inner branch and 0.0976 m on the outer one. Both are correct answers to the question as posed, and choosing between them is a physics decision, not an arithmetic one.

Push the probe to $z$ = 0.30 m and the field falls to about 1.7×10⁵ N/C. Compare that with the point-charge estimate $k Q / z^{2}$ = 2.0×10⁵ N/C: still 17% high at three radii out, which is a useful reminder of how slowly the ring converges to the point-charge limit.

Behaviour along the axis and sanity checks

Three limiting behaviours cover almost every sanity check you will need.

Close to the centre ( $z ≪ R$ ): expanding the denominator gives $E \approx k Q z / R^{3}$ , a linear rise. The restoring-force form of that expression is why a charge threaded on the axis oscillates harmonically about the centre for small displacements.
Far from the ring ( $z ≫ R$ ): the field approaches $E \approx \frac{k Q}{z^{2}}$ . The leading correction is a factor of $(1 - \frac{3}{2} R^{2} / z^{2})$ , so the error is about 1.5% at ten radii and 17% at three.
At the centre itself: $E (0) = 0$ exactly, for any charge. The potential there is not zero, however: it is $V = k Q / R$ , which is a useful independent check on your numbers.

Dimensional analysis catches most remaining slips. The expression is $k Q$ (N·m²/C) times a length divided by a length cubed, which leaves N/C. If a centimetre value got entered as metres, the field is off by factors of 100 or 10,000, and nothing in the output looks obviously wrong.

Representative axial field values for a 2 μC, 10 cm ring

Holding $Q$ at 2 μC and $R$ at 10 cm, the table walks the probe outward. The field climbs from zero at the centre, crests near 7.07 cm, then tapers. Every row is reproducible by entering $Q$ , $R$ , and $z$ and solving for $E$ .

z (cm)	z / R	E (N/C)	Point-charge estimate kQ/z² (N/C)
0	0	0	—
2	0.20	3.39×10⁵	4.49×10⁷
5	0.50	6.43×10⁵	7.19×10⁶
7.07	0.707	6.92×10⁵ (peak)	3.60×10⁶
10	1.00	6.36×10⁵	1.80×10⁶
30	3.00	1.71×10⁵	2.00×10⁵
100	10.0	1.77×10⁴	1.80×10⁴

The last column is the point-charge approximation. Watch it converge: it is useless inside a radius, 16% high at three radii, and within 1.5% by ten radii.

Units and conversions for typical classroom values

The calculator works in strict SI. These conversions cover most textbook problems:

1 μC (microcoulomb) = 1e-6 C
1 nC (nanocoulomb) = 1e-9 C
1 cm = 0.01 m and 1 mm = 0.001 m
1 kV/m = 1000 N/C (the two units are identical)
Linear charge density to total charge: $Q = λ \cdot 2 π R$

Limitations and assumptions of the on-axis ring model

The closed form is exact, but only inside a narrow set of assumptions. Check each one before trusting a number.

On-axis only. This is the binding limitation. One millimetre off the symmetry axis and the radial cancellation fails; the off-axis field requires elliptic integrals or a numerical evaluation, and no rearrangement of this formula will produce it.
Uniform linear charge density. Real conductors do not hold charge uniformly on a ring unless something forces them to. On an isolated conducting torus, charge redistributes toward regions of higher curvature, and the uniform assumption is an idealisation appropriate to an insulating ring with painted-on charge.
Zero cross-section. The ring is treated as a mathematical loop. A ring of finite wire thickness $a$ deviates from this result once $z$ approaches $a$ .
Vacuum or air. The Coulomb constant used here is the free-space value. Immersing the ring in a dielectric of relative permittivity $ε_{r}$ divides every field by that factor.
Electrostatics. Charges are stationary and nothing radiates. Time-varying charge on the ring produces induced fields this model does not contain.
Magnitude only. The output is nonnegative; direction is yours to assign from the sign of the charge and your choice of positive axis.
Inverse solutions are not always unique or existent. Solving for $z$ yields two roots below the peak field and none above it; solving for $R$ fails when the requested field exceeds what any radius can produce at that distance.
No superposition of multiple rings. Each solve treats a single isolated ring with no external field and no nearby conductors to induce image charges.

Sources. The axial-field expression is the standard textbook result for a uniformly charged ring, derived above from Coulomb's law and the cosine projection; the peak location and peak magnitude follow from setting the derivative to zero, and both are reproduced in full so they can be checked by hand.

Coulomb constant: derived from the CODATA 2022 recommended vacuum electric permittivity $ε_{0} = 8.8541878188 (14) \times 10^{- 12}$ F/m, giving $k = 8.9875517862 \times 10^{9}$ N·m²/C². See NIST CODATA: vacuum electric permittivity.
Ring geometry and the on-axis derivation: HyperPhysics, Georgia State University, "Electric field of a ring of charge".

Questions that come up about ring fields

Why does solving for the axial distance return two answers?

Because the axial field is not a one-to-one function of distance. It rises from zero at the ring centre, peaks at z = R divided by the square root of 2, and then decays. Any field magnitude below that peak is therefore reached twice, once on the rising inner branch and once on the falling outer branch, and both distances are physically valid. The calculator brackets and bisects each branch separately so neither root is silently discarded.

What happens if I ask for a field larger than the ring can produce?

No distance satisfies the request, and the calculator says so rather than returning a spurious number. The largest field a ring of charge Q and radius R can produce anywhere on its axis is 2kQ divided by 3 times the square root of 3 times R squared, which is about 0.3849 times kQ over R squared. If your target exceeds that, you need more charge or a smaller ring.

Does the calculator handle negative charge?

The input boxes accept nonnegative values because the output is a magnitude. In electrostatics a negative charge reverses the direction of the field rather than the size of it. Compute the magnitude here, then apply a negative sign according to the positive direction you have chosen for the axis.

Why can I not solve for the charge when z is zero and E is nonzero?

At the exact centre of a uniformly charged ring, symmetry forces the axial field to be exactly zero for any finite charge. No value of Q can produce a nonzero axial field there, so the inverse problem has no solution and the calculator reports that instead of dividing by zero.

Is this the same as the field on the axis of a charged disk?

No. A disk carries charge over an area rather than around a circumference, so its axial field has a different closed form that approaches a constant near the surface instead of falling to zero at the centre. A disk can be assembled by integrating this ring result over radii from zero to the disk radius, but the final expression is not the same function.

What if I know the linear charge density instead of the total charge?

Convert first. For a uniform ring the total charge is the linear charge density multiplied by the circumference, so Q equals lambda times 2 pi R. Enter that total charge in coulombs. Doing the conversion yourself also keeps the units honest, because a density in coulombs per metre and a radius in centimetres is a common source of error.

Solve for a quantity to plot the axial field profile E(z) for your ring.

Arcade Mini-Game: Electric Field of a Charged Ring Calibration Run

Use this quick arcade run to practice separating useful scenario inputs from common planning mistakes before you rely on the calculator output.

Score: 0 Timer: 30s Best: 0

Start the game, then use your pointer or arrow keys to catch useful inputs and avoid bad assumptions.

Electric Field of a Charged Ring

Introduction: the ring is the building block for disks, coils, and beam optics

How to use the charged ring calculator and read the field profile

Formula for the axial field, its derivative, and the axial maximum

Why solving for z has two answers, and what most calculators get wrong

Worked example: a 2 μC ring of 10 cm radius

Behaviour along the axis and sanity checks

Representative axial field values for a 2 μC, 10 cm ring

Units and conversions for typical classroom values

Limitations and assumptions of the on-axis ring model

Questions that come up about ring fields

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