Electricity-And-Magnetism · Unit 4 · Video 3 · Interactive Practice

Electric Field of a Charged Ring and a Charged Disk

IKey Formulas

FormulaNameWhat it says
Ez=14πε0Qz(R2+z2)3/2E_z = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(R^2+z^2)^{3/2}}Ring of radius RR, on its axisZero at the centre, largest at z=±R/2z = \pm R/\sqrt{2} (about 0.385 Q/4πε0R20.385\,Q/4\pi\varepsilon_0 R^2), and ≈Q/4πε0z2\approx Q/4\pi\varepsilon_0 z^2 for ∣z∣≫R|z| \gg R
Ez=σ2ε0[z∣z∣−zz2+R2]E_z = \dfrac{\sigma}{2\varepsilon_0}\left[\dfrac{z}{|z|} - \dfrac{z}{\sqrt{z^2+R^2}}\right]Disk of radius RR, on its axisBuilt from rings with dq=σ 2πr′ dr′dq = \sigma\,2\pi r^{\prime}\,dr^{\prime}; σ=Q/πR2\sigma = Q/\pi R^2, and z2=∣z∣\sqrt{z^2} = |z|
E⃗=±σ2ε0 k^\vec{E} = \pm\dfrac{\sigma}{2\varepsilon_0}\,\hat{k}Infinite plane (R≫∣z∣R \gg |z|)Constant on each side, pointing away from the plane for σ>0\sigma > 0
ΔEn=σε0\Delta E_n = \dfrac{\sigma}{\varepsilon_0}Jump across a charged surfaceThe normal component of E⃗\vec{E} jumps by σ/ε0\sigma/\varepsilon_0; shown here for a flat sheet, and given for any surface, flat or curved (proved later with Gauss’s law)

Key Insight: On the axis, every piece of charge has a partner directly opposite whose sideways field cancels its own, so only the zz components are added. Adding the magnitudes ∣dE⃗∣|d\vec{E}| instead overcounts.

IIVisualization 1 — Why the Ring's Field Rises and Falls

Each opposite pair contributes only 2 dEcos⁡θ2\,dE\cos\theta, giving Ez/E0=(z/R)/[1+(z/R)2]3/2E_z/E_0 = (z/R)/[1+(z/R)^2]^{3/2} with E0=Q/4πε0R2E_0 = Q/4\pi\varepsilon_0 R^2.

IIIVisualization 2 — Building the Disk Out of Rings

Summing rings out to radius r′r^{\prime} gives Ez=σ2ε0[1−zz2+r′2]E_z = \dfrac{\sigma}{2\varepsilon_0}\left[1 - \dfrac{z}{\sqrt{z^2+r^{\prime 2}}}\right] above the disk.

💡 Challenge: find the height at which the whole disk gives half the infinite-plane field; setting z/z2+R2=12z/\sqrt{z^2+R^2} = \tfrac{1}{2} puts it at z=R/3≈0.58Rz = R/\sqrt{3} \approx 0.58R.

IVVisualization 3 — Which Field Jumps at the Centre?

Measured in the unit E0=σ/2ε0E_0 = \sigma/2\varepsilon_0 (not the ring unit of Visualization 1), the disk's field jumps by 2E0=σ/ε02E_0 = \sigma/\varepsilon_0 at z=0z = 0, while a ring with the same total charge passes smoothly through zero.

VQuiz Questions

Problem 1 · Field on the Axis of a Ring

Given: a thin ring of radius R=0.10 mR = 0.10\ \text{m} carrying Q=4.0 nCQ = 4.0\ \text{nC} spread uniformly, with 14πε0=8.99×109 N⋅m2/C2\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 — find the magnitude of the field on the axis at z=0.10 mz = 0.10\ \text{m} above the centre.

✅ Correct! Ez=(8.99×109)(4.0×10−9)(0.10)/(0.020)3/2≈1.27×103 N/CE_z = (8.99 \times 10^{9})(4.0 \times 10^{-9})(0.10)/(0.020)^{3/2} \approx 1.27 \times 10^{3}\ \text{N/C}.
❌ Not quite. That is Q/4πε0z2Q/4\pi\varepsilon_0 z^2, the point-charge field. It applies only for z≫Rz \gg R, and here z=Rz = R: every piece of the ring is R2+z2\sqrt{R^2+z^2} away, not zz.
❌ Not quite. That is Q/4πε0(R2+z2)Q/4\pi\varepsilon_0(R^2+z^2), the sum of the magnitudes ∣dE⃗∣|d\vec{E}|. The sideways parts cancel in pairs, so each piece contributes only dEcos⁡θdE\cos\theta, with cos⁡θ=z/R2+z2=0.707\cos\theta = z/\sqrt{R^2+z^2} = 0.707.
❌ Not quite. The denominator is (R2+z2)3/2(R^2+z^2)^{3/2}, not R2+z2R^2+z^2: one factor of R2+z2\sqrt{R^2+z^2} comes from cos⁡θ\cos\theta, the other two from Coulomb's 1/r21/r^2.
Show solution

Every piece of the ring is r=R2+z2r = \sqrt{R^2+z^2} from the field point, and only its zz component survives, so

Ez=14πε0Qz(R2+z2)3/2.E_z = \frac{1}{4\pi\varepsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}.

With R2+z2=0.010+0.010=0.020 m2R^2 + z^2 = 0.010 + 0.010 = 0.020\ \text{m}^2 and (0.020)3/2=2.83×10−3 m3(0.020)^{3/2} = 2.83 \times 10^{-3}\ \text{m}^3:

Ez=(8.99×109)(4.0×10−9)(0.10)2.83×10−3=3.602.83×10−3≈1.27×103 N/C,E_z = \frac{(8.99 \times 10^{9})(4.0 \times 10^{-9})(0.10)}{2.83 \times 10^{-3}} = \frac{3.60}{2.83 \times 10^{-3}} \approx 1.27 \times 10^{3}\ \text{N/C},

pointing along the axis, away from the ring.

Problem 2 · Below the Disk

Given: a disk of radius RR with uniform σ>0\sigma > 0 in the xyxy plane — find EzE_z on the axis at z=−Rz = -R, one radius below the centre.

✅ Correct! Ez=σ2ε0[−1+12]≈−0.29 σ2ε0E_z = \dfrac{\sigma}{2\varepsilon_0}\left[-1 + \dfrac{1}{\sqrt{2}}\right] \approx -0.29\,\dfrac{\sigma}{2\varepsilon_0}: below a positive disk the field points down, away from it, with the same size as at z=+Rz = +R.
❌ That is the z2=z\sqrt{z^2} = z trap. With z/∣z∣z/|z| replaced by 11, the formula gives 1+1/21 + 1/\sqrt{2} — a field larger than the infinite plane's, which is impossible. Below the disk z/∣z∣=−1z/|z| = -1.
❌ Right size, wrong direction. The field of a positive disk points away from it on both sides, so below the disk Ez<0E_z < 0; the curve is an odd function of zz.
❌ Not quite. The −1-1 is right, but the second term is −z/z2+R2-z/\sqrt{z^2+R^2} with z=−Rz = -R, which is +1/2+1/\sqrt{2}: it partly cancels the −1-1.
Show solution

Use the disk result with z2=∣z∣\sqrt{z^2} = |z| kept honest:

Ez=σ2ε0[z∣z∣−zz2+R2].E_z = \frac{\sigma}{2\varepsilon_0}\left[\frac{z}{|z|} - \frac{z}{\sqrt{z^2+R^2}}\right].

At z=−Rz = -R: z/∣z∣=−1z/|z| = -1 and z/z2+R2=−R/2R2=−1/2z/\sqrt{z^2+R^2} = -R/\sqrt{2R^2} = -1/\sqrt{2}, so

Ez=σ2ε0[−1+12]≈−0.293 σ2ε0.E_z = \frac{\sigma}{2\varepsilon_0}\left[-1 + \frac{1}{\sqrt{2}}\right] \approx -0.293\,\frac{\sigma}{2\varepsilon_0}.

This is minus the value at z=+Rz = +R, as it must be for an odd function: the field points down, away from the disk.

Problem 3 · Where the Ring's Field Peaks

Given: a ring of radius R=6.0 cmR = 6.0\ \text{cm} carrying Q=10 nCQ = 10\ \text{nC}, with 14πε0=8.99×109 N⋅m2/C2\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 — find the height above the centre where the axial field is strongest, and that largest field.

Where is the field strongest?

What is the largest field?

✅ Correct! The peak is at z=R/2≈4.24 cmz = R/\sqrt{2} \approx 4.24\ \text{cm}, where Ez=233Q4πε0R2≈0.385×2.50×104≈9.6×103 N/CE_z = \dfrac{2}{3\sqrt{3}}\dfrac{Q}{4\pi\varepsilon_0 R^2} \approx 0.385 \times 2.50 \times 10^{4} \approx 9.6 \times 10^{3}\ \text{N/C}.
❌ Not quite. The centre is closest to the charge, but there the contributions of opposite pieces cancel exactly: Ez(0)=0E_z(0) = 0.
❌ Not quite. At z=Rz = R the field is Q/(4πε0 22R2)≈0.354 Q/4πε0R2Q/(4\pi\varepsilon_0\,2\sqrt{2}R^2) \approx 0.354\,Q/4\pi\varepsilon_0 R^2, already past the peak. Setting dEz/dz=0dE_z/dz = 0 gives R2−2z2=0R^2 - 2z^2 = 0, so z=R/2z = R/\sqrt{2}.
❌ Not quite. That is z=2Rz = \sqrt{2}R, where the field has fallen to about 0.27 Q/4πε0R20.27\,Q/4\pi\varepsilon_0 R^2. The derivative's numerator R2−2z2R^2 - 2z^2 vanishes at z=R/2z = R/\sqrt{2}, not 2R\sqrt{2}R.
❌ Check the location. Set dEz/dz=0dE_z/dz = 0: the numerator of the derivative is R2−2z2R^2 - 2z^2, so z=R/2z = R/\sqrt{2}.
❌ Not quite. That is the field at z=Rz = R, which is Q/(4πε0 22R2)Q/(4\pi\varepsilon_0\,2\sqrt{2}R^2) — close to the peak but below it. Evaluate at z=R/2z = R/\sqrt{2}.
❌ Not quite. That is the unit E0=Q/4πε0R2E_0 = Q/4\pi\varepsilon_0 R^2 itself. The ring's field never reaches it; the peak is 0.385 E00.385\,E_0.
❌ Not quite. That is Q/4πε0z2Q/4\pi\varepsilon_0 z^2 at z=R/2z = R/\sqrt{2}, the point-charge formula, which only holds for z≫Rz \gg R.
❌ Check the value. Substitute z=R/2z = R/\sqrt{2} into Ez=14πε0Qz(R2+z2)3/2E_z = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(R^2+z^2)^{3/2}}.
Show solution

Location. Differentiate z/(R2+z2)3/2z/(R^2+z^2)^{3/2}:

ddz z(R2+z2)3/2=(R2+z2)−3z2(R2+z2)5/2=R2−2z2(R2+z2)5/2=0  ⇒  z=R2=6.02≈4.24 cm.\frac{d}{dz}\,\frac{z}{(R^2+z^2)^{3/2}} = \frac{(R^2+z^2) - 3z^2}{(R^2+z^2)^{5/2}} = \frac{R^2 - 2z^2}{(R^2+z^2)^{5/2}} = 0 \;\Rightarrow\; z = \frac{R}{\sqrt{2}} = \frac{6.0}{\sqrt{2}} \approx 4.24\ \text{cm}.

Value. There R2+z2=32R2R^2 + z^2 = \tfrac{3}{2}R^2, so

Ez=14πε0Q (R/2)(32R2)3/2=233 Q4πε0R2≈0.385 Q4πε0R2.E_z = \frac{1}{4\pi\varepsilon_0}\frac{Q\,(R/\sqrt{2})}{\left(\tfrac{3}{2}R^2\right)^{3/2}} = \frac{2}{3\sqrt{3}}\,\frac{Q}{4\pi\varepsilon_0 R^2} \approx 0.385\,\frac{Q}{4\pi\varepsilon_0 R^2}.

With Q4πε0R2=(8.99×109)(1.0×10−8)(0.060)2=2.50×104 N/C\dfrac{Q}{4\pi\varepsilon_0 R^2} = \dfrac{(8.99 \times 10^{9})(1.0 \times 10^{-8})}{(0.060)^2} = 2.50 \times 10^{4}\ \text{N/C}, the largest field is 0.385×2.50×104≈9.6×103 N/C0.385 \times 2.50 \times 10^{4} \approx 9.6 \times 10^{3}\ \text{N/C}.

Problem 4 · Reading the Charge from the Jump

Given: just below a charged surface the normal component of the field is 200 N/C200\ \text{N/C}, and just above it is 650 N/C650\ \text{N/C}, both measured along the same normal n^\hat{n} (pointing from below to above), with ε0=8.85×10−12 C2/N⋅m2\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{N}\cdot\text{m}^2 — find the surface charge density σ\sigma there.

✅ Correct! σ=ε0 ΔEn=(8.85×10−12)(650−200)≈4.0×10−9 C/m2\sigma = \varepsilon_0\,\Delta E_n = (8.85 \times 10^{-12})(650 - 200) \approx 4.0 \times 10^{-9}\ \text{C/m}^2.
❌ Not quite. You used ΔEn=σ/2ε0\Delta E_n = \sigma/2\varepsilon_0. The field on each side of a lone sheet is σ/2ε0\sigma/2\varepsilon_0; the jump from one side to the other is twice that, σ/ε0\sigma/\varepsilon_0.
❌ Not quite. You added the two fields. Both components are measured along the same n^\hat{n} and both are positive, so the jump is the difference 650−200650 - 200.
❌ Not quite. That uses only the field above. The rule is about the change across the surface: part of the 650 N/C650\ \text{N/C} comes from other charges, which contribute the same amount just below, so only the difference measures σ\sigma.
Show solution

Across any surface carrying surface charge density σ\sigma, the normal component of the field jumps by σ/ε0\sigma/\varepsilon_0 (given: the disk shows it for a flat sheet, and Gauss’s law will prove it in general):

ΔEn=En,above−En,below=650−200=450 N/C.\Delta E_n = E_{n,\text{above}} - E_{n,\text{below}} = 650 - 200 = 450\ \text{N/C}. σ=ε0 ΔEn=(8.85×10−12)(450)≈3.98×10−9 C/m2≈4.0 nC/m2.\sigma = \varepsilon_0\,\Delta E_n = (8.85 \times 10^{-12})(450) \approx 3.98 \times 10^{-9}\ \text{C/m}^2 \approx 4.0\ \text{nC/m}^2.

Think of it as superposition: the sheet adds +σ/2ε0=225 N/C+\sigma/2\varepsilon_0 = 225\ \text{N/C} above and −225 N/C-225\ \text{N/C} below to an outside field of 425 N/C425\ \text{N/C}, giving 650650 and 200200.

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