On axis of a uniformly charged disk of radius R with surface density σ at height z, E_z(z) equals which expression?

On axis of a uniformly charged disk of radius R with surface density σ at height z, E_z(z) equals which expression?

• A. E_z(z) = (σ/(2 ε0)) [1 − z/√(z^2 + R^2)]
• B. E_z(z) = (σ/(2 ε0)) [z/√(z^2 + R^2) − 1]
• C. E_z(z) = (σ/(4π ε0)) [z/(z^2+R^2)^(3/2)]
• D. E_z(z) = (σ/(2 ε0)) [1 + z/√(z^2 + R^2)]

Answer: (A)

The on-axis field from a uniformly charged disk comes only in the z-direction due to symmetry, so you sum contributions from concentric rings. Consider a ring of radius r and width dr with charge dq = σ (2π r dr). The axial field from this ring at a point a distance z above the disk is dEz = (1/(4πε0)) dq · z /(r^2 + z^2)^{3/2}. Integrating r from 0 to R gives Ez(z) = ∫0^R (σ z)/(2ε0) · r dr /(r^2 + z^2)^{3/2}. Let u = r^2 + z^2, du = 2r dr. This yields Ez(z) = (σ z)/(2ε0) · ∫_{z^2}^{z^2+R^2} (1/2) u^{-3/2} du = (σ z)/(2ε0) [ -u^{-1/2} ]_{z^2}^{z^2+R^2} = (σ/(2ε0)) [1 − z/√(z^2 + R^2)]. So the axial field at height z is Ez(z) = (σ/(2ε0)) [1 − z/√(z^2 + R^2)]. This also shows Ez → 0 as z → ∞ and Ez → σ/(2ε0) as z → 0.

The on-axis field from a uniformly charged disk comes only in the z-direction due to symmetry, so you sum contributions from concentric rings.

Consider a ring of radius r and width dr with charge dq = σ (2π r dr). The axial field from this ring at a point a distance z above the disk is dEz = (1/(4πε0)) dq · z /(r^2 + z^2)^{3/2}. Integrating r from 0 to R gives

Ez(z) = ∫0^R (σ z)/(2ε0) · r dr /(r^2 + z^2)^{3/2}.

Let u = r^2 + z^2, du = 2r dr. This yields

Ez(z) = (σ z)/(2ε0) · ∫_{z^2}^{z^2+R^2} (1/2) u^{-3/2} du

= (σ z)/(2ε0) [ -u^{-1/2} ]_{z^2}^{z^2+R^2}

= (σ/(2ε0)) [1 − z/√(z^2 + R^2)].

So the axial field at height z is Ez(z) = (σ/(2ε0)) [1 − z/√(z^2 + R^2)]. This also shows Ez → 0 as z → ∞ and Ez → σ/(2ε0) as z → 0.