Exercise 3.1.10: Quarter annulus with a nonuniform charge

Exercise 3.1.10

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.

Exercise 3.1.10

Quarter annulus with a nonuniform charge

cf. Home exercise 3.1.2

The annular arc of Home exercise 3.1.2,

a \leq ρ \leq b, 0 \leq ϕ \leq π /2, z = 0,

now carries the surface charge density

ρ_{S} (ρ^{'}, ϕ^{'}) = ρ_{0} sin 2 ϕ^{'} C/ m^{2},

$ρ_{0}$ being a constant.

a) Find the total charge on the annular arc.

b) Find the electric field at $P (0, 0, h)$ .

Try it yourself first.

Strategy

Part a) is a surface integral of $ρ_{S}$ over the quarter annulus with $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ . Part b) is the surface recipe of Problem 3.1.3 with $d q^{'} = ρ_{S} d S^{'}$ and the limits of Home exercise 3.1.2. A mirror symmetry about the plane $y = x$ shows that $E_{x} = E_{y}$ , so only two components have to be computed. With $sin 2 ϕ^{'} = 2 sin ϕ^{'} cos ϕ^{'}$ each angular integral is a power of $sin ϕ^{'}$ or $cos ϕ^{'}$ times its derivative, and the radial integrals are the standard integrals (3.1.30) (the second one after a split of the numerator).

Part a): charge element

On the quarter annulus $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ (Figure 3.1.15c), so

d q^{'} = ρ_{S} d S^{'} = ρ_{0} sin 2 ϕ^{'} ρ^{'} d ρ^{'} d ϕ^{'} .

Part a): integrate

The radial and angular integrals separate:

Q = \int_{ρ^{'} = a}^{b} \int_{ϕ^{'} = 0}^{π /2} ρ_{0} sin 2 ϕ^{'} ρ^{'} d ϕ^{'} d ρ^{'} = ρ_{0} (b^{2} - a^{2}) /2 \int_{a}^{b} ρ^{'} d ρ^{'} [- \frac{1}{2} c o s 2 ϕ^{'}]_{0}^{π /2} = \frac{1}{2} (1 + 1) = 1 \int_{0}^{π /2} sin 2 ϕ^{'} d ϕ^{'} = ρ_{0} \frac{b ^{2} - a ^{2}}{2} \cdot 1,

Answer

Q = ρ_{0} \frac{b ^{2} - a ^{2}}{2} [C]

Check of part a): $ρ_{0} b^{2}$ is $(C/ m^{2}) m^{2} = C$ , and $Q > 0$ because $sin 2 ϕ^{'} \geq 0$ for $0 \leq ϕ^{'} \leq π /2$ .

Part b): symmetry

The quarter annulus is not symmetric about the $z$ -axis, so the field at P has horizontal parts. One symmetry survives.

The mirror in the plane $y = x$ exchanges $x^{'}$ and $y^{'}$ , that is, $ϕ^{'} \to π /2 - ϕ^{'}$ . It maps the quarter annulus onto itself, and it leaves the density unchanged because $sin (2 (π /2 - ϕ^{'})) = sin (π - 2 ϕ^{'}) = sin 2 ϕ^{'}$ . The point P lies on the $z$ -axis and therefore in this mirror plane. The element at $(x^{'}, y^{'})$ and its partner at $(y^{'}, x^{'})$ carry the same charge, and the $x$ -component of the first contribution equals the $y$ -component of the second one. Summing over all pairs gives $E_{x} = E_{y}$ .
All the charge lies in $x^{'} \geq 0$ and $y^{'} \geq 0$ , so nothing cancels the horizontal parts: they point away from the charge, that is, towards $- x$ and $- y$ , and $E_{x} = E_{y} < 0$ .
The $z$ -components of all (positive) elements point along $+ z$ and add up.

Hence $E = E_{x} (a_{x} + a_{y}) + E_{z} a_{z}$ .

Part b): points and displacement vector

The source point is $Q (x^{'}, y^{'}, 0)$ on the annular arc 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} .

With $x^{'} = ρ^{'} cos ϕ^{'}$ and $y^{'} = ρ^{'} sin ϕ^{'}$ (now $a \leq ρ^{'} \leq b$ and $0 \leq ϕ^{'} \leq π /2$ ),

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

Part b): charge element and integral

With $d q^{'} = ρ_{0} sin 2 ϕ^{'} ρ^{'} d ρ^{'} d ϕ^{'}$ the electric field at P is

E (P) = \frac{1}{4 π ε _{0}} \int_{ρ^{'} = a}^{b} \int_{ϕ^{'} = 0}^{π /2} ρ_{0} sin 2 ϕ^{'} 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}},

that is, separating the components,

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

Part b): integrate

First the angular integrals, using $sin 2 ϕ^{'} = 2 sin ϕ^{'} cos ϕ^{'}$ :

\int_{0}^{π /2} sin 2 ϕ^{'} cos ϕ^{'} d ϕ^{'} = 2 \int_{0}^{π /2} cos^{2} ϕ^{'} sin ϕ^{'} d ϕ^{'} = [- \frac{2}{3} cos^{3} ϕ^{'}]_{0}^{π /2} = \frac{2}{3}, \int_{0}^{π /2} sin 2 ϕ^{'} sin ϕ^{'} d ϕ^{'} = 2 \int_{0}^{π /2} sin^{2} ϕ^{'} cos ϕ^{'} d ϕ^{'} = [\frac{2}{3} sin^{3} ϕ^{'}]_{0}^{π /2} = \frac{2}{3}, \int_{0}^{π /2} sin 2 ϕ^{'} d ϕ^{'} = [- \frac{1}{2} cos 2 ϕ^{'}]_{0}^{π /2} = 1.

The first two are equal, which confirms $E_{x} = E_{y}$ . Then

E (P) = \frac{ρ _{0}}{4 π ε _{0}} {- \frac{2}{3} (a_{x} + a_{y}) I_{2} + h a_{z} I_{1}},

where

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

The integral $I_{1}$ is the first standard integral (3.1.30):

I_{1} = - \frac{1}{( ρ ^{'} ) ^{2} + h ^{2}}_{ρ^{'} = a}^{b} = \frac{1}{a ^{2} + h ^{2}} - \frac{1}{b ^{2} + h ^{2}} .

For $I_{2}$ we write $(ρ^{'})^{2} = [(ρ^{'})^{2} + h^{2}] - h^{2}$ , which splits the integrand into $1/ (ρ^{'})^{2} + h^{2}$ and $- h^{2} / [(ρ^{'})^{2} + h^{2}]^{3/2}$ . The second piece is $h^{2}$ times the second standard integral (3.1.30), which gives $- ρ^{'} / (ρ^{'})^{2} + h^{2}$ . The first piece has the antiderivative $ln (ρ^{'} + (ρ^{'})^{2} + h^{2})$ , because

\frac{d}{d ρ ^{'}} ln (ρ^{'} + (ρ^{'})^{2} + h^{2}) = \frac{1 + ρ ^{'} / ( ρ ^{'} ) ^{2} + h ^{2}}{ρ ^{'} + ( ρ ^{'} ) ^{2} + h ^{2}} = \frac{1}{( ρ ^{'} ) ^{2} + h ^{2}} .

Therefore, between the limits $a$ and $b$ ,

I_{2} = [ln (ρ^{'} + (ρ^{'})^{2} + h^{2}) - \frac{ρ ^{'}}{( ρ ^{'} ) ^{2} + h ^{2}}]_{ρ^{'} = a}^{b} = ln \frac{b + b ^{2} + h ^{2}}{a + a ^{2} + h ^{2}} - \frac{b}{b ^{2} + h ^{2}} + \frac{a}{a ^{2} + h ^{2}} .

The prefactor of the horizontal part is $\frac{ρ _{0}}{4 π ε _{0}} \cdot \frac{2}{3} = \frac{ρ _{0}}{6 π ε _{0}}$ , and that of the $z$ -part is $\frac{ρ _{0} h}{4 π ε _{0}}$ . Finally,

Answer

E (P) = - \frac{ρ _{0}}{6 π ε _{0}} (a_{x} + a_{y}) [ln \frac{b + b ^{2} + h ^{2}}{a + a ^{2} + h ^{2}} - \frac{b}{b ^{2} + h ^{2}} + \frac{a}{a ^{2} + h ^{2}}] + \frac{ρ _{0} h}{4 π ε _{0}} [\frac{1}{a ^{2} + h ^{2}} - \frac{1}{b ^{2} + h ^{2}}] a_{z} [V/m]

Interpretation: limits and far field

$a \to 0$ (a full quarter disk): since $a + a^{2} + h^{2} \to h$ and $a / a^{2} + h^{2} \to 0$ , the bracket of the horizontal part becomes $ln [(b + b^{2} + h^{2}) / h] - b / b^{2} + h^{2}$ , the function met in Exercise 3.1.8.
$h ≫ b$ (far away): $I_{1} ≃ (b^{2} - a^{2}) / (2 h^{3})$ and $I_{2} ≃ (b^{3} - a^{3}) / (3 h^{3})$ , so that

$E_{z} E_{x} = E_{y} \to \frac{ρ _{0}}{4 π ε _{0}} \frac{b ^{2} - a ^{2}}{2 h ^{2}} = \frac{Q}{4 π ε _{0} h ^{2}}, \to - \frac{ρ _{0}}{6 π ε _{0}} \frac{b ^{3} - a ^{3}}{3 h ^{3}} = - \frac{ρ _{0} ( b ^{3} - a ^{3} )}{18 π ε _{0} h ^{3}} .$

The $z$ -component is the point-charge field of the total charge of part a). The horizontal components decay faster, as $1/ h^{3}$ . They are controlled by the first moment of the charge: for $h ≫ b$ the $x$ -part of $R$ is $- x^{'}$ and $R ≃ h$ , so $E_{x} ≃ - \int x^{'} d q^{'} / (4 π ε_{0} h^{3})$ . With

$\int x^{'} d q^{'} = ρ_{0} \int_{a}^{b} (ρ^{'})^{2} d ρ^{'} \times \int_{0}^{π /2} sin 2 ϕ^{'} cos ϕ^{'} d ϕ^{'} = \frac{2}{3} ρ_{0} (b^{3} - a^{3}) /3$

this gives $E_{x} ≃ - ρ_{0} (b^{3} - a^{3}) / (18 π ε_{0} h^{3})$ , in agreement.

Check

Units: $ρ_{0} / ε_{0}$ is $(C/ m^{2}) / (F/m) = V/m$ . The bracket of the horizontal part is a pure number (the argument of the logarithm and $b / b^{2} + h^{2}$ are ratios of lengths), and in the $z$ -part $h [1/ a^{2} + h^{2} - 1/ b^{2} + h^{2}]$ is a pure number as well, so every term is in V/m. Direction: $E_{x} = E_{y}$ by the mirror symmetry about $y = x$ (compare Home exercise 3.1.2, where $cos ϕ^{'}$ favoured the $x$ -axis and gave $∣ E_{x} ∣ > ∣ E_{y} ∣$ ), both are negative because the positive charge lies in the first quadrant, and $E_{z} > 0$ . Far away: the point-charge field of $Q = ρ_{0} (b^{2} - a^{2}) /2$ .