Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 3.3.2

Interactive: Ring, Disk and Annulus on Axis

Home exercise 3.3.2

Circular disk source

A circular disk of radius $b$ in the $x y$ -plane, centred at the origin, carries a uniform surface charge density $ρ_{S}$ . Find the electrostatic potential at a point $P (0, 0, z)$ on its axis ( $z > 0$ ), and from it the electric field. Evaluate both at $z = h$ and compare with Problem 3.1.3.

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 3-9.

Try it yourself first.

Strategy

Superposition over the disk, cut into rings (cylindrical coordinates, $d S^{'} = ρ^{'} d ϕ^{'} d ρ^{'}$ ). Every point of a ring of radius $ρ^{'}$ is at the same distance $ρ^{'2} + z^{2}$ from P, so the integral is elementary. On the axis the field is along $z$ by symmetry: $E = - (d V / d z) a_{z}$ .

Potential

A source point of the disk is $Q (x^{'}, y^{'}, 0)$ with $x^{'} = ρ^{'} cos ϕ^{'}$ , $y^{'} = ρ^{'} sin ϕ^{'}$ , so

R = - ρ^{'} cos ϕ^{'} a_{x} - ρ^{'} sin ϕ^{'} a_{y} + z a_{z} \Rightarrow R = (ρ^{'})^{2} + z^{2} .

With $d S^{'} = ρ^{'} d ϕ^{'} d ρ^{'}$ the electrostatic potential is

V (z) = \frac{1}{4 π ε _{0}} \int_{ρ^{'} = 0}^{b} \int_{ϕ^{'} = 0}^{2 π} \frac{ρ _{S} ρ ^{'} d ϕ ^{'} d ρ ^{'}}{( ρ ^{'} ) ^{2} + z ^{2}} = \frac{ρ _{S}}{2 ε _{0}} \int_{ρ^{'} = 0}^{b} \frac{ρ ^{'} d ρ ^{'}}{( ρ ^{'} ) ^{2} + z ^{2}},

the $ϕ^{'}$ -integration simply giving $2 π$ because the integrand does not depend on $ϕ^{'}$ . With the change of variables $t = (ρ^{'})^{2} + z^{2}$ , $d t = 2 ρ^{'} d ρ^{'}$ , the integrand becomes $d t / (2 t)$ , whose integral is $t$ :

V (z) = \frac{ρ _{S}}{2 ε _{0}} (ρ^{'})^{2} + z^{2}_{ρ^{'} = 0}^{b} = \frac{ρ _{S}}{2 ε _{0}} [b^{2} + z^{2} - z] [V] (z > 0) .

Field

On the axis the field is along $z$ by symmetry, and

E (z) = - \nabla V = - \frac{d V}{d z} a_{z} = \frac{ρ _{S}}{2 ε _{0}} [1 - \frac{z}{b ^{2} + z ^{2}}] a_{z} [V/m] .

At $z = h$ :

Answer

V (h) E (h) = \frac{ρ _{S}}{2 ε _{0}} [b^{2} + h^{2} - h], = \frac{ρ _{S}}{2 ε _{0}} [1 - \frac{h}{b ^{2} + h ^{2}}] a_{z}

Check

Just above the disk ( $z \to 0^{+}$ ) the field tends to $ρ_{S} / (2 ε_{0})$ , the field of a charged sheet, as it must very close to the surface, and far away it becomes the point-charge field (remark below). $ρ_{S} b / ε_{0}$ has the unit $(C/ m^{2}) m / (F/m) = V$ .

Remark

The field agrees with Problem 3.1.3, where it was obtained by integrating $d E$ . At the centre of the disk $V (0) = ρ_{S} b / (2 ε_{0})$ . Far away ( $z ≫ b$ ), $b^{2} + z^{2} - z ≃ b^{2} / (2 z)$ and $V \to ρ_{S} π b^{2} / (4 π ε_{0} z)$ , the potential of a point charge $Q = π b^{2} ρ_{S}$ . For $b \to \infty$ the field tends to $ρ_{S} / (2 ε_{0})$ , that of an infinite sheet, while the potential again diverges.

Move the field point from the disk surface to far away: near the disk $E_{z}$ approaches the sheet value $ρ_{S} /2 ε_{0}$ , far away both $V$ and $E_{z}$ follow the point-charge curves.

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Home exercise 3.3.2: Circular disk source