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

Exercise 3.1.7

About the further exercises of Section 3.1

Ten further exercises in the spirit of Problems 3.1.1 to 3.1.3 and Home exercises 3.1.1 and 3.1.2. Every one is solved with the four steps of Section 3.1.8, and the two standard integrals (3.1.30) appear again and again.

Surface charges

Exercise 3.1.7

Annular disk (a washer)

cf. Problem 3.1.3

An annular disk $a \leq ρ \leq b$ , $z = 0$ , carries a uniform surface charge density $ρ_{S}$ . Find the electric field at $P (0, 0, h)$ . What happens when $b \to \infty$ ?

Adapted from: B. M. Notaroš, Electromagnetics, Problem 1.25.

Try it yourself first.

Strategy

This is Problem 3.1.3 with the lower limit of $ρ^{'}$ changed from $0$ to $a$ . We use the recipe with a surface source, $d q^{'} = ρ_{S} d S^{'}$ and $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ . The $ϕ^{'}$ -integration removes the horizontal components and the $ρ^{'}$ -integration is the first standard integral (3.1.30). The same result follows from superposition with a negative piece, a full disk of radius $b$ minus a small disk of radius $a$ , which we use as a check.

Symmetry

The washer is symmetric about the $z$ -axis and P lies on it (compare Figure 3.1.27).

Rotating the washer about the $z$ -axis leaves it and P unchanged, so the field on the axis cannot single out a horizontal direction and $E$ is along $z$ .
Equivalently, the element at $ϕ^{'}$ and the one at $ϕ^{'} + π$ (same $ρ^{'}$ ) give equal and opposite horizontal parts, while their $z$ -parts are equal and add.
All elements on the ring of radius $ρ^{'}$ are at the same distance $(ρ^{'})^{2} + h^{2}$ from P, and for $ρ_{S} > 0$ every element pushes P upwards, so $E_{z} > 0$ for $h > 0$ .

Points and displacement vector

The source point is $Q (x^{'}, y^{'}, 0)$ somewhere on the washer and the field point is $P (0, 0, h)$ :

R R = (0 - x^{'}) a_{x} + (0 - y^{'}) a_{y} + (h - 0) a_{z} = - x^{'} a_{x} - y^{'} a_{y} + h a_{z}, = (x^{'})^{2} + (y^{'})^{2} + h^{2} .

In polar coordinates $x^{'} = ρ^{'} cos ϕ^{'}$ and $y^{'} = ρ^{'} sin ϕ^{'}$ , with $a \leq ρ^{'} \leq b$ and $0 \leq ϕ^{'} \leq 2 π$ , so that

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

Charge element and integral

The area element is $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ (Figure 3.1.15c, and do not forget the factor $ρ^{'}$ ), so $d q^{'} = ρ_{S} ρ^{'} d ρ^{'} d ϕ^{'}$ and

E (P) = \frac{1}{4 π ε _{0}} \int_{ρ^{'} = a}^{b} \int_{ϕ^{'} = 0}^{2 π} ρ_{S} ρ^{'} d ϕ^{'} d ρ^{'} \times \frac{- ρ ^{'} cos ϕ ^{'} a _{x} - ρ ^{'} sin ϕ ^{'} a _{y} + h a _{z}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} .

Separating the components,

E (P) = \frac{ρ _{S}}{4 π ε _{0}} {- a_{x} \int_{a}^{b} \frac{( ρ ^{'} ) ^{2} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} \int_{0}^{2 π} cos ϕ^{'} d ϕ^{'} - a_{y} \int_{a}^{b} \frac{( ρ ^{'} ) ^{2} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} \int_{0}^{2 π} sin ϕ^{'} d ϕ^{'} + h a_{z} \int_{a}^{b} \frac{ρ ^{'} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} \int_{0}^{2 π} d ϕ^{'}},

where the integrand factorises because the $ρ^{'}$ and $ϕ^{'}$ dependences are separate and the limits are constants.

Integrate

First the angular integrals, one line at a time:

\int_{0}^{2 π} cos ϕ^{'} d ϕ^{'} \int_{0}^{2 π} sin ϕ^{'} d ϕ^{'} \int_{0}^{2 π} d ϕ^{'} = [sin ϕ^{'}]_{0}^{2 π} = 0, = [- cos ϕ^{'}]_{0}^{2 π} = - 1 + 1 = 0, = 2 π .

The $x$ - and $y$ -components vanish, as predicted in Step 1, and

E (P) = \frac{ρ _{S}}{4 π ε _{0}} 2 π h a_{z} \int_{a}^{b} \frac{ρ ^{'} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} = \frac{ρ _{S} h}{2 ε _{0}} a_{z} \int_{a}^{b} \frac{ρ ^{'} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} .

This is the sum of the fields of the thin rings of radius $ρ^{'}$ between $a$ and $b$ (Problem 3.1.2 with $b \to ρ^{'}$ and $ρ_{L} \to ρ_{S} d ρ^{'}$ ). With the first standard integral (3.1.30) ( $s = ρ^{'}$ ),

\int_{a}^{b} \frac{ρ ^{'} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} = - \frac{1}{( ρ ^{'} ) ^{2} + h ^{2}}_{ρ^{'} = a}^{b} = - \frac{1}{b ^{2} + h ^{2}} + \frac{1}{a ^{2} + h ^{2}},

and the result is

Answer

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

The limit

b \to \infty

For $b \to \infty$ the washer becomes an infinite sheet with a circular hole of radius $a$ . Then $1/ b^{2} + h^{2} \to 0$ and

Answer

E (P) = \frac{ρ _{S}}{2 ε _{0}} \frac{h}{a ^{2} + h ^{2}} a_{z} [V/m]

Interpretation

The limiting result (Figure 3.1.32) shows what a hole does to the sheet field.

$h \to 0$ (P at the centre of the hole): $E_{z} \to 0$ , linearly, $E_{z} ≃ ρ_{S} h / (2 ε_{0} a)$ . At the centre of the hole the elements all around P cancel in pairs, and the nearest charge is a distance $a$ away.
$h ≫ a$ : $h / a^{2} + h^{2} = (1 + a^{2} / h^{2})^{- 1/2} ≃ 1 - a^{2} / (2 h^{2})$ , so $E_{z} \to ρ_{S} / (2 ε_{0})$ , the infinite-sheet value. The hole no longer matters.

Superposition gives the same result in another way:

(sheet with hole) = (full sheet) - (disk of radius a) ⟹ E_{z} = \frac{ρ _{S}}{2 ε _{0}} - \frac{ρ _{S}}{2 ε _{0}} [1 - \frac{h}{a ^{2} + h ^{2}}] = \frac{ρ _{S}}{2 ε _{0}} \frac{h}{a ^{2} + h ^{2}},

where the bracket is the result of Problem 3.1.3 with $b = a$ . For a washer of finite size, the same expansion with $h ≫ b$ gives $h / a^{2} + h^{2} - h / b^{2} + h^{2} ≃ (b^{2} - a^{2}) / (2 h^{2})$ and so

E_{z} Q \to \frac{ρ _{S} ( b ^{2} - a ^{2} )}{4 ε _{0} h ^{2}} = \frac{Q}{4 π ε _{0} h ^{2}}, = π (b^{2} - a^{2}) ρ_{S},

the field of a point charge carrying the total charge of the washer.

Exercise 3.1.7 with b\to\infty: on the axis, the field of a sheet with a hole (solid) is the field of the full sheet (dashed) minus that of the missing disk (dotted). At the centre of the hole the two cancel exactly, and far from the hole (h\gg a) the hole no longer matters. — **Figure 3.1.32.** Exercise 3.1.7 with $b \to \infty$ : on the axis, the field of a sheet with a hole (solid) is the field of the full sheet (dashed) minus that of the missing disk (dotted). At the centre of the hole the two cancel exactly, and far from the hole ( $h ≫ a$ ) the hole no longer matters.

Check

Units: $ρ_{S} / ε_{0}$ is $(C/ m^{2}) / (F/m) = V/m$ and $h / a^{2} + h^{2}$ is a pure number. Direction: along $+ a_{z}$ above a positively charged washer, away from the charge. Limits: for $a \to 0$ the first result becomes $ρ_{S} h / (2 ε_{0}) [1/ h - 1/ b^{2} + h^{2}] = ρ_{S} / (2 ε_{0}) [1 - h / b^{2} + h^{2}]$ , the result of Problem 3.1.3, and far away it is the point-charge field of the total charge $Q$ .

Exercise 3.1.7: Annular disk (a washer)

Surface charges