Fields of Charged Rods, Rings and Disks

Cut a charged object into pieces small enough to count as point charges, add their fields as vectors, and let the pieces shrink: that sum is the integral. Symmetry does most of the work. Each piece has a mirror partner whose sideways push cancels its own, so on a symmetric shape only one component needs integrating. Choose a shape, move P, and set how many pieces the page adds up.

Add the pieces as vectors	$\vec E=k\int\frac{dq}{r^{2}}\,\hat r,\qquad k=\frac{1}{4\pi\varepsilon_0}$
Or add them as numbers, then differentiate	$V=k\int\frac{dq}{r},\qquad E_z=-\frac{\partial V}{\partial z}$

Worked example — Fields of Charged Rods, Rings and Disks

Charge per metre	1.592×10⁻⁸ C/m
Field at P, by integration	3,178 N/C
The same, as 12 point charges	3,178 N/C
Potential at P	635.5 V
Field from the potential, −dV/dz	3,178 N/C
All of Q at the centre	8,988 N/C

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Fields of Charged Rods, Rings and Disks

The field of a continuous charge distribution is the vector sum of the fields of its pieces, in the limit where each piece dq is small enough to be a point charge: E = k∫dq r̂/r², with k = 1/(4πε₀) = 8.988 × 10⁹ N·m²/C². The potential is the same sum without the directions, V = k∫dq/r.

What each symbol means

dq: the charge on one piece: λ dx on a line, σ dA on a surface
λ, σ: charge per unit length and per unit area; Q/L and Q/(πR²) when the charge is spread evenly
r, θ: the distance from the piece to P, and the angle between that line and the axis

Where the formula comes from

Cut the charge into pieces. Each piece carries dq and acts at P like a point charge: a field k dq/r² along the line from the piece to P.
$dq=\lambda\,dx,\qquad dE=\frac{k\,dq}{r^{2}}$
Pair each piece with its mirror image. The partner across the axis gives a field of the same size with the opposite sideways part, so the sideways parts cancel and only the component along the axis, dE cos θ = dE·z/r, is left to add.
$dE_z=dE\cos\theta=\frac{k\,dq}{r^{2}}\cdot\frac{z}{r}$
Integrate that component. On a ring every piece is the same distance √(z² + R²) from P, so the integrand is the same for every dq and the integral is Q times it.
$E=\int\frac{kz\,dq}{\left(z^{2}+R^{2}\right)^{3/2}}=\frac{kQz}{\left(z^{2}+R^{2}\right)^{3/2}}$
Build the disk from rings. A ring of radius r and width dr carries dq = σ·2πr dr. Add the rings' fields, each from step 3, from r = 0 to R.
$E=\int_{0}^{R}\frac{kz\,\sigma\,2\pi r\,dr}{\left(z^{2}+r^{2}\right)^{3/2}}=\frac{\sigma}{2\varepsilon_0}\Bigl(1-\frac{z}{\sqrt{z^{2}+R^{2}}}\Bigr)$
Or integrate the potential and differentiate. V is a sum of numbers with no directions, so nothing has to be resolved into components; the field along the axis is minus the slope of V.
$V=k\int\frac{dq}{r}=\frac{kQ}{\sqrt{z^{2}+R^{2}}},\qquad E_z=-\frac{dV}{dz}=\frac{kQz}{\left(z^{2}+R^{2}\right)^{3/2}}$

Assumes: The charge is fixed and spread evenly, unless a problem gives λ or σ as a function of position, in which case dq carries that function into the integral. Vacuum, or air, whose permittivity is larger by 0.06%.

Problems solved in full

A rod, on its own line 3 steps
A rod 10 cm long carries 10 nC spread evenly. Find the field on the rod's line, 5 cm beyond one end.

Open this problem in the tool
1. Put the rod on the x-axis from x = 0 to x = L and P at x = L + d. A piece dx at x is L + d − x from P, and every piece's field points the same way, along the rod, so nothing cancels and nothing needs resolving:
  
  $dE=\frac{k\lambda\,dx}{(L+d-x)^{2}},\qquad\lambda=\frac{Q}{L}=1{.}0\times10^{-7}\ \mathrm{C/m}$
2. Integrate over the rod:
  
  $E=k\lambda\int_{0}^{L}\frac{dx}{(L+d-x)^{2}}=k\lambda\Bigl(\frac{1}{d}-\frac{1}{d+L}\Bigr)=\frac{kQ}{d\,(d+L)}$
3. With d = 0.05 m and L = 0.10 m:
  
  $E=\frac{(8{.}988\times10^{9})(1{.}0\times10^{-8})}{(0{.}05)(0{.}15)}=1{.}198\times10^{4}\ \mathrm{N/C}$
Answer

1.198 × 10⁴ N/C, along the rod's line, away from it. When d is much larger than L the formula becomes kQ/d², the field of a point charge.

$E=\frac{kQ}{d\,(d+L)}\approx1{.}198\times10^{4}\ \mathrm{N/C}$
A rod, beside its middle, and the infinite line 4 steps
A rod 20 cm long carries 10 nC. Find the field 10 cm from its midpoint on the perpendicular, and compare it with an infinitely long line carrying the same charge per metre.

Open this problem in the tool
Show solution
1. Put the rod on the x-axis from −L/2 to L/2 and P at (0, y). The pieces at x and −x are the same distance from P; their x-components cancel and their y-components add, so integrate the y-component:
  
  $dE_y=\frac{k\lambda\,dx}{x^{2}+y^{2}}\cdot\frac{y}{\sqrt{x^{2}+y^{2}}}$
2. Integrate, using ∫dx/(x² + y²)^(3/2) = x/(y²√(x² + y²)):
  
  $E=k\lambda y\int_{-L/2}^{L/2}\frac{dx}{\left(x^{2}+y^{2}\right)^{3/2}}=\frac{k\lambda L}{y\sqrt{y^{2}+L^{2}/4}}=\frac{kQ}{y\sqrt{y^{2}+L^{2}/4}}$
3. With λ = 5.0 × 10⁻⁸ C/m, y = 0.10 m and L = 0.20 m:
  
  $E=\frac{(8{.}988\times10^{9})(1{.}0\times10^{-8})}{(0{.}10)\sqrt{0{.}10^{2}+0{.}10^{2}}}\approx6355\ \mathrm{N/C}$
4. Let L grow without limit: the square root tends to L/2, and
  
  $E\to\frac{2k\lambda}{y}=\frac{2\,(8{.}988\times10^{9})(5{.}0\times10^{-8})}{0{.}10}\approx8988\ \mathrm{N/C}$
Answer

6,355 N/C, perpendicular to the rod. The infinite line gives 8,988 N/C, so the finite rod delivers 70.7% of it, which is sin 45°: each half of the rod reaches 45° from the perpendicular.

$E\approx6355\ \mathrm{N/C}=0{.}707\cdot\frac{2k\lambda}{y}$
A ring on its axis, and where its field peaks 4 steps
A ring of radius 10 cm carries 10 nC. Find the field on its axis 10 cm from the centre, and the distance at which the axial field is greatest.

Open this problem in the tool
Show solution
1. Every piece is the same distance √(z² + R²) from P. The components across the axis cancel round the ring, and each piece adds dE·z/r along it:
  
  $E=\int\frac{k\,dq}{r^{2}}\cdot\frac{z}{r}=\frac{kz}{\left(z^{2}+R^{2}\right)^{3/2}}\int dq=\frac{kQz}{\left(z^{2}+R^{2}\right)^{3/2}}$
2. At z = R = 0.10 m:
  
  $E=\frac{(8{.}988\times10^{9})(1{.}0\times10^{-8})(0{.}10)}{(0{.}02)^{3/2}}\approx3178\ \mathrm{N/C}$
3. For the maximum, set dE/dz = 0:
  
  $\frac{dE}{dz}=kQ\,\frac{\left(z^{2}+R^{2}\right)-3z^{2}}{\left(z^{2}+R^{2}\right)^{5/2}}=0\ \Rightarrow\ z=\frac{R}{\sqrt{2}}\approx7{.}07\ \mathrm{cm}$
4. There:
  
  $E_{\max}=\frac{kQ\,(R/\sqrt{2})}{\left(\tfrac{3}{2}R^{2}\right)^{3/2}}=\frac{2kQ}{3\sqrt{3}\,R^{2}}\approx3459\ \mathrm{N/C}$
Answer

3,178 N/C at 10 cm; the maximum is 3,459 N/C, at z = 7.07 cm. At the centre the field is zero.

$E(0{.}10\ \mathrm{m})\approx3178\ \mathrm{N/C},\qquad E\Bigl(\frac{R}{\sqrt{2}}\Bigr)=E_{\max}\approx3459\ \mathrm{N/C}$
A disk on its axis, built from rings 3 steps
A disk of radius 10 cm carries 10 nC spread evenly over its face. Find the field on its axis 1 cm from the centre, and compare it with an infinite sheet with the same charge per square metre.

Open this problem in the tool
Show solution
1. σ = Q/(πR²) = 3.183 × 10⁻⁷ C/m². Slice the disk into rings of radius r and width dr; each carries dq = σ·2πr dr and gives the ring's field along the axis:
  
  $dE=\frac{kz\,\sigma\,2\pi r\,dr}{\left(z^{2}+r^{2}\right)^{3/2}}$
2. Integrate from r = 0 to R, substituting u = z² + r², du = 2r dr:
  
  $E=\pi k\sigma z\int_{z^{2}}^{z^{2}+R^{2}}u^{-3/2}\,du=2\pi k\sigma\Bigl(1-\frac{z}{\sqrt{z^{2}+R^{2}}}\Bigr)=\frac{\sigma}{2\varepsilon_0}\Bigl(1-\frac{z}{\sqrt{z^{2}+R^{2}}}\Bigr)$
3. At z = 0.01 m:
  
  $E=\frac{3{.}183\times10^{-7}}{2\,(8{.}854\times10^{-12})}\Bigl(1-\frac{0{.}01}{\sqrt{0{.}01^{2}+0{.}10^{2}}}\Bigr)=(1{.}798\times10^{4})(0{.}9005)\approx1{.}619\times10^{4}\ \mathrm{N/C}$
Answer

1.619 × 10⁴ N/C, which is 90.0% of the infinite sheet's σ/2ε₀ = 1.798 × 10⁴ N/C.

$E\approx1{.}619\times10^{4}\ \mathrm{N/C}\approx0{.}900\cdot\frac{\sigma}{2\varepsilon_0}$
A semicircle, at its centre 3 steps
A thin rod bent into a semicircle of radius 10 cm carries 10 nC. Find the field at the centre of the circle.

Open this problem in the tool
Show solution
1. λ = Q/(πR) = 3.183 × 10⁻⁸ C/m. Measure the angle φ from the diameter. The piece at φ carries λR dφ at distance R, and its partner at π − φ cancels the component along the diameter, which leaves the component across it:
  
  $dE_\perp=\frac{k\lambda R\,d\varphi}{R^{2}}\sin\varphi$
2. Integrate over the half-circle:
  
  $E=\frac{k\lambda}{R}\int_{0}^{\pi}\sin\varphi\,d\varphi=\frac{2k\lambda}{R}=\frac{2kQ}{\pi R^{2}}$
3. In numbers:
  
  $E=\frac{2\,(8{.}988\times10^{9})(1{.}0\times10^{-8})}{\pi\,(0{.}10)^{2}}\approx5722\ \mathrm{N/C}$
Answer

5,722 N/C, along the line of symmetry, away from the middle of the arc. An infinite line with the same λ at distance R gives 2kλ/R, the same number.

$E=\frac{2k\lambda}{R}\approx5722\ \mathrm{N/C}$
A quarter circle, by components 3 steps
A quarter circle of radius 10 cm carries 10 nC. Find the field at its centre by integrating both components.

Open this problem in the tool
Show solution
1. Put the arc from φ = 0 to φ = π/2 with its centre at the origin; λ = Q/(πR/2) = 6.366 × 10⁻⁸ C/m. A piece at φ pushes P directly away from itself, along −(cos φ, sin φ), and here no partner cancels anything:
  
  $dE_x=-\frac{k\lambda}{R}\cos\varphi\,d\varphi,\qquad dE_y=-\frac{k\lambda}{R}\sin\varphi\,d\varphi$
2. Integrate from 0 to π/2:
  
  $E_x=E_y=-\frac{k\lambda}{R}=-\frac{(8{.}988\times10^{9})(6{.}366\times10^{-8})}{0{.}10}\approx-5722\ \mathrm{N/C}$
3. Combine the two components:
  
  $|E|=\sqrt{2}\,\frac{k\lambda}{R}\approx8092\ \mathrm{N/C}$
Answer

8,092 N/C, pointing away from the arc along the line at 45° that bisects it. The symmetric route gives the same: E = (2kλ/R) sin(θ₀/2) with θ₀ = 90°.

$|E|=\sqrt{2}\,\frac{k\lambda}{R}=\frac{2k\lambda}{R}\sin45^{\circ}\approx8092\ \mathrm{N/C}$
The potential of a ring, and the field from it 2 steps
Find the potential on the axis of the ring of problem 3 at z = 10 cm, and find the field from the potential.

Open this problem in the tool
Show solution
1. Potentials add as numbers, and every piece is the same distance away:
  
  $V=\int\frac{k\,dq}{\sqrt{z^{2}+R^{2}}}=\frac{kQ}{\sqrt{z^{2}+R^{2}}}=\frac{89{.}88}{\sqrt{0{.}02}}\approx635{.}5\ \mathrm{V}$
2. Differentiate along the axis:
  
  $E_z=-\frac{dV}{dz}=\frac{kQz}{\left(z^{2}+R^{2}\right)^{3/2}}\approx3178\ \mathrm{N/C}$
Answer

V = 635.5 V, and −dV/dz gives back problem 3's 3,178 N/C without resolving anything into components.

$V\approx635{.}5\ \mathrm{V},\qquad E_z=-\frac{dV}{dz}\approx3178\ \mathrm{N/C}$
A rod whose charge is not spread evenly 4 steps
A rod lies on the x-axis from x = 0 to x = 0.50 m and carries λ = αx, with α = 4.0 × 10⁻⁸ C/m². Find its total charge and the field at x = −0.10 m.
Show solution
1. λ changes along the rod, so the total charge is an integral too:
  
  $Q=\int_{0}^{L}\alpha x\,dx=\frac{\alpha L^{2}}{2}=\frac{(4{.}0\times10^{-8})(0{.}50)^{2}}{2}=5{.}0\times10^{-9}\ \mathrm{C}$
2. The piece at x is x + d from P, with d = 0.10 m, and every piece pushes P the same way, in −x:
  
  $E=k\alpha\int_{0}^{L}\frac{x\,dx}{(x+d)^{2}}$
3. Write x as (x + d) − d and integrate each part:
  
  $\int_{0}^{L}\frac{x\,dx}{(x+d)^{2}}=\ln\frac{L+d}{d}-\frac{L}{L+d}=\ln6-\frac{5}{6}\approx0{.}9584$
4. In numbers:
  
  $E=(8{.}988\times10^{9})(4{.}0\times10^{-8})(0{.}9584)\approx344{.}6\ \mathrm{N/C}$
Answer

Q = 5.0 nC, and E = 344.6 N/C in the −x direction. The same 5.0 nC spread evenly would give kQ/(d(d + L)) = 749 N/C; this rod keeps most of its charge at the far end.

$Q=5{.}0\ \mathrm{nC},\qquad E\approx344{.}6\ \mathrm{N/C}$

Supplementary problems, answers only

A rod 30 cm long carries −6.0 nC spread evenly. Find the field on its line, 10 cm beyond one end.

Show answer

1,348 N/C, pointing towards the rod.

$E=\frac{k|Q|}{d\,(d+L)}\approx1348\ \mathrm{N/C}$
A very long straight wire carries 2.0 μC per metre. Find the field 4.0 cm from it.

Show answer

8.988 × 10⁵ N/C, pointing away from the wire.

$E=\frac{2k\lambda}{r}\approx8{.}988\times10^{5}\ \mathrm{N/C}$
Show that a ring's axial field approaches kQ/z² when z is much larger than R, and find how far out it is within 1% of that value.

Show answer

The ratio is (1 + R²/z²)^(−3/2), which tends to 1; it reaches 0.99 at z = 12.2R.

$\frac{E}{kQ/z^{2}}=\Bigl(1+\frac{R^{2}}{z^{2}}\Bigr)^{-3/2}\ge0{.}99\ \Leftrightarrow\ z\ge12{.}2R$
A disk of radius 5.0 cm carries σ = 2.0 μC/m². Find the potential at its centre and 5.0 cm along its axis.

Show answer

5,647 V at the centre and 2,339 V at z = 5.0 cm.

$V(0)=\frac{\sigma R}{2\varepsilon_0}\approx5647\ \mathrm{V},\qquad V(R)=\frac{\sigma R}{2\varepsilon_0}\bigl(\sqrt{2}-1\bigr)\approx2339\ \mathrm{V}$
A ring of radius R carries +Q spread evenly over one half and −Q over the other. Find the field at its centre.

Show answer

4kQ/(πR²), from the positive half towards the negative half: each half gives 2kQ/(πR²), and the two point the same way.

$E=2\cdot\frac{2kQ}{\pi R^{2}}=\frac{4kQ}{\pi R^{2}}$
A rod of length L carries Q spread evenly. Find the potential on its perpendicular bisector, a distance y from it.

Show answer

V = (2kQ/L) ln[(L/2 + √(y² + L²/4))/y].

$V=\frac{2kQ}{L}\ln\frac{L/2+\sqrt{y^{2}+L^{2}/4}}{y}$

A ring pushes hardest at z = R/√2, and not at all at its centre

Press Where a ring's field peaks. At the centre every piece of the ring has a partner directly across, so the field there is exactly zero; far out along the axis the ring looks like a point charge and the field falls as 1/z². In between it has to rise and then fall. Setting dE/dz = 0 for kQz/(z² + R²)^(3/2) gives z² + R² = 3z², so the peak is at z = R/√2 ≈ 0.707R, where E = 2kQ/(3√3 R²). For 10 nC on a 10 cm ring that is 3,459 N/C, only 38.5% of what the same charge would give 10 cm away as a point. Close to the ring, most of each piece's push is sideways, and sideways pushes cancel.

Near a large charged disk the field barely changes with distance

Press A disk up close. 1 cm above a disk of radius 10 cm the field is 90.0% of σ/2ε₀, the field of an infinite sheet, and at 1 mm it is 99.0%. The formula hides the reason. Write 1 − z/√(z² + R²) as 1 − cos α, where α is the angle from the axis to the rim, and the field becomes E = kσΩ, with Ω = 2π(1 − cos α) the solid angle the disk fills as seen from P. For a flat, evenly charged surface the field across it depends only on how much of the sky the charge covers. An infinite sheet always covers half of it, Ω = 2π, so its field is the same 1 mm away as 1 km away: every piece is farther and weaker, and exactly enough more of the sheet comes into view at a useful angle to make up for it.

A semicircle and an infinite line give the same field

Press A semicircle at its centre. 10 nC spread over a half-circle of radius 10 cm gives 5,722 N/C at the centre, and the table prints 5,722 N/C again for an infinite straight line with the same charge per metre, 10 cm away. The two agree piece by piece. Seen from P, a stretch dx of the line at angle θ off the perpendicular is R/cos θ away and fills an angle dθ = cos²θ dx/R, so its field, kλ dx cos²θ/R², comes to kλ dθ/R and depends on the angle alone. The stretch of arc filling the same dθ is R dθ long, sits at distance R, and gives the same kλ dθ/R in the same direction. So a straight rod and the arc of radius R that fills the same angles give equal fields; set the arc's θ₀ below 180° and the table's last row is that rod.

Learning path

Where fields come from

References (3)

The method and four of the page's closed forms, each derived there: the line segment above its midpoint and the infinite line as its limit (Examples 5.5, 5.6), the ring on its axis (5.7), the disk on its axis and the infinite plane σ/(2ε₀) (5.8, Eq. 5.15): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 5.5, “Calculating Electric Fields of Charge Distributions”.
The potentials on the axis of a ring and a disk and on the bisector of a finite line (Examples 7.13–7.15), and the ring's field recovered as E_z = −∂V/∂z (Example 7.18): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), sections 7.3, “Calculations of Electric Potential”, and 7.4, “Determining Field from Potential”.
The standard graduate-entry text in which the field of a continuous distribution is written as the integral of k dq r̂/r²: D. J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press (2017). DOI 10.1017/9781108333511

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.