Field at axial distance x from centre of a disc of radius R and surface charge density σ. Reduces to infinite plane result (σ/2ε₀) as R→∞.
Setup
A disc of radius R R R σ \sigma σ r ′ r' r ′ d r ′ dr' d r ′
Charge on a ring of radius r ′ r' r ′
d q = σ ⋅ 2 π r ′ d r ′ dq = \sigma \cdot 2\pi r'\, dr' d q = σ ⋅ 2 π r ′ d r ′ Using the ring result
From the axial field of a ring (formula ef22-field-ring-axis), the contribution from ring r ′ r' r ′ x x x
d E = d q ⋅ x 4 π ε 0 ( x 2 + r ′ 2 ) 3 / 2 = σ x r ′ d r ′ 2 ε 0 ( x 2 + r ′ 2 ) 3 / 2 dE = \frac{dq \cdot x}{4\pi\varepsilon_0(x^2 + r'^2)^{3/2}} = \frac{\sigma x\, r'\, dr'}{2\varepsilon_0(x^2 + r'^2)^{3/2}} d E = 4 π ε 0 ( x 2 + r ′2 ) 3/2 d q ⋅ x = 2 ε 0 ( x 2 + r ′2 ) 3/2 σ x r ′ d r ′ Integration
E = ∫ 0 R σ x r ′ 2 ε 0 ( x 2 + r ′ 2 ) 3 / 2 d r ′ E = \int_0^R \frac{\sigma x\, r'}{2\varepsilon_0(x^2 + r'^2)^{3/2}}\, dr' E = ∫ 0 R 2 ε 0 ( x 2 + r ′2 ) 3/2 σ x r ′ d r ′ Substitute u = x 2 + r ′ 2 u = x^2 + r'^2 u = x 2 + r ′2 d u = 2 r ′ d r ′ du = 2r'\, dr' d u = 2 r ′ d r ′
E = σ x 4 ε 0 ∫ x 2 x 2 + R 2 u − 3 / 2 d u = σ x 4 ε 0 [ − 2 u − 1 / 2 ] x 2 x 2 + R 2 E = \frac{\sigma x}{4\varepsilon_0}\int_{x^2}^{x^2+R^2} u^{-3/2}\, du = \frac{\sigma x}{4\varepsilon_0}\left[-2u^{-1/2}\right]_{x^2}^{x^2+R^2} E = 4 ε 0 σ x ∫ x 2 x 2 + R 2 u − 3/2 d u = 4 ε 0 σ x [ − 2 u − 1/2 ] x 2 x 2 + R 2 E = σ x 2 ε 0 [ 1 x − 1 x 2 + R 2 ] E = \frac{\sigma x}{2\varepsilon_0}\left[\frac{1}{x} - \frac{1}{\sqrt{x^2+R^2}}\right] E = 2 ε 0 σ x [ x 1 − x 2 + R 2 1 ] E = σ 2 ε 0 ( 1 − x x 2 + R 2 ) \boxed{E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{x^2+R^2}}\right)} E = 2 ε 0 σ ( 1 − x 2 + R 2 x ) Limiting cases
Infinite plane (R → ∞ R \to \infty R → ∞ x / x 2 + R 2 → 0 x/\sqrt{x^2+R^2} \to 0 x / x 2 + R 2 → 0 E → σ / 2 ε 0 E \to \sigma/2\varepsilon_0 E → σ /2 ε 0
Far field (x ≫ R x \gg R x ≫ R ( 1 + R 2 / x 2 ) − 1 / 2 ≈ 1 − R 2 / 2 x 2 (1 + R^2/x^2)^{-1/2} \approx 1 - R^2/2x^2 ( 1 + R 2 / x 2 ) − 1/2 ≈ 1 − R 2 /2 x 2
E ≈ σ 2 ε 0 ⋅ R 2 2 x 2 = σ π R 2 4 π ε 0 x 2 = Q 4 π ε 0 x 2 E \approx \frac{\sigma}{2\varepsilon_0} \cdot \frac{R^2}{2x^2} = \frac{\sigma \pi R^2}{4\pi\varepsilon_0 x^2} = \frac{Q}{4\pi\varepsilon_0 x^2} E ≈ 2 ε 0 σ ⋅ 2 x 2 R 2 = 4 π ε 0 x 2 σ π R 2 = 4 π ε 0 x 2 Q The disc looks like a point charge at large distances.
At the surface (x → 0 x \to 0 x → 0 E → σ / 2 ε 0 E \to \sigma/2\varepsilon_0 E → σ /2 ε 0
Remember
This derivation illustrates the general strategy for continuous distributions: reduce to a known result (here, the ring) then integrate over the parameter describing the distribution (here, $r'$).