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

Exercise 3.1.9

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 3.1.9

Half disk with a nonuniform charge

cf. Home exercise 3.1.2

A half disk described by

0 \leq ρ \leq b, 0 \leq ϕ \leq π, z = 0

carries the nonuniform surface charge density

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

$ρ_{0}$ being a constant.

a) Find the total charge on the half disk.

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 half disk with $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ . Part b) is the surface recipe of Problem 3.1.3 with $d q^{'} = ρ_{S} d S^{'}$ , but the charge now fills only half of the disk and its density depends on $ϕ^{'}$ . A mirror symmetry removes one horizontal component, the angular integrals decide which of the others survive, and the radial integrals are the two standard integrals (3.1.30) (the second one after a split of the numerator).

Part a): charge element

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

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

Part a): integrate

The radial and angular integrals separate:

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

Answer

Q = ρ_{0} b^{2} [C]

Check of part a): $ρ_{0} b^{2}$ is $(C/ m^{2}) m^{2} = C$ , and $Q > 0$ because $sin ϕ^{'} \geq 0$ on the whole half disk.

Part b): symmetry

The half disk and its density are not symmetric about the $z$ -axis, so the field at P will have a horizontal part. Still, one symmetry survives.

The mirror $x \to - x$ (that is, $ϕ^{'} \to π - ϕ^{'}$ ) maps the half disk onto itself, and it leaves the density unchanged because $sin (π - ϕ^{'}) = sin ϕ^{'}$ . The element at $(x^{'}, y^{'})$ and its mirror partner at $(- x^{'}, y^{'})$ carry the same charge. Their $x$ -components are equal and opposite, while their $y$ - and $z$ -components are identical and add. Hence $E_{x} = 0$ .
The charge lies entirely in $y^{'} \geq 0$ , so nothing cancels the $y$ -components of the contributions: they all point away from the charge, that is, along $- y$ , and $E_{y} \neq = 0$ .
The $z$ -components of all (positive) elements point along $+ z$ and add up, so $E_{z} \neq = 0$ .

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

Part b): points and displacement vector

The source point is $Q (x^{'}, y^{'}, 0)$ on the half disk 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 $0 \leq ϕ^{'} \leq π$ ),

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

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

Part b): integrate

First the three angular integrals:

\int_{0}^{π} sin ϕ^{'} cos ϕ^{'} d ϕ^{'} \int_{0}^{π} sin^{2} ϕ^{'} d ϕ^{'} \int_{0}^{π} sin ϕ^{'} d ϕ^{'} = [\frac{1}{2} sin^{2} ϕ^{'}]_{0}^{π} = 0, = \int_{0}^{π} \frac{1 - cos 2 ϕ ^{'}}{2} d ϕ^{'} = [\frac{ϕ ^{'}}{2} - \frac{sin 2 ϕ ^{'}}{4}]_{0}^{π} = \frac{π}{2}, = [- cos ϕ^{'}]_{0}^{π} = 2.

The first one confirms $E_{x} = 0$ . The remaining two give

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

where

I_{1} I_{2} = \int_{0}^{b} \frac{ρ ^{'} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}}, = \int_{0}^{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^{'} = 0}^{b} = \frac{1}{h} - \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

I_{2} = [ln (ρ^{'} + (ρ^{'})^{2} + h^{2}) - \frac{ρ ^{'}}{( ρ ^{'} ) ^{2} + h ^{2}}]_{ρ^{'} = 0}^{b} = ln \frac{b + b ^{2} + h ^{2}}{h} - \frac{b}{b ^{2} + h ^{2}} .

Now the numerical prefactors. For the $y$ -component, $\frac{ρ _{0}}{4 π ε _{0}} \cdot \frac{π}{2} = \frac{ρ _{0}}{8 ε _{0}}$ . For the $z$ -component,

\frac{ρ _{0}}{4 π ε _{0}} 2 h I_{1} = \frac{ρ _{0}}{2 π ε _{0}} h I_{1} = \frac{ρ _{0}}{2 π ε _{0}} [1 - \frac{h}{b ^{2} + h ^{2}}] .

Finally,

Answer

E (P) = - \frac{ρ _{0}}{8 ε _{0}} a_{y} \times [ln \frac{b + b ^{2} + h ^{2}}{h} - \frac{b}{b ^{2} + h ^{2}}] + \frac{ρ _{0}}{2 π ε _{0}} [1 - \frac{h}{b ^{2} + h ^{2}}] a_{z} [V/m]

Interpretation: far field

Put $u = b / h$ . The bracket in $E_{y}$ is $sinh^{- 1} u - u / 1 + u^{2} ≃ u^{3} /3$ for $h ≫ b$ (the expansion used in Exercise 3.1.8), and the bracket in $E_{z}$ is $1 - (1 + u^{2})^{- 1/2} ≃ u^{2} /2$ . So, far away,

E_{z} E_{y} \to \frac{ρ _{0}}{2 π ε _{0}} \frac{b ^{2}}{2 h ^{2}} = \frac{ρ _{0} b ^{2}}{4 π ε _{0} h ^{2}} = \frac{Q}{4 π ε _{0} h ^{2}}, \to - \frac{ρ _{0}}{8 ε _{0}} \frac{b ^{3}}{3 h ^{3}} = - \frac{ρ _{0} b ^{3}}{24 ε _{0} h ^{3}} .

The $z$ -component is the point-charge field of the total charge $Q = ρ_{0} b^{2}$ from part a). The $y$ -component decays faster, as $1/ h^{3}$ , and it exists only because the charge is not centred on the $z$ -axis. This can be confirmed from the displacement vector of part b). For $h ≫ b$ we have $R ≃ h$ and the $y$ -part of the vector $R$ is $- y^{'}$ , so $E_{y} ≃ - \int y^{'} d q^{'} / (4 π ε_{0} h^{3})$ . With

\int y^{'} d q^{'} = ρ_{0} \int_{0}^{b} (ρ^{'})^{2} d ρ^{'} \int_{0}^{π} sin^{2} ϕ^{'} d ϕ^{'} = π ρ_{0} b^{3} /6

this gives $E_{y} ≃ - ρ_{0} b^{3} / (24 ε_{0} h^{3})$ , in agreement.

Check

Units: $ρ_{0} / ε_{0}$ is $(C/ m^{2}) / (F/m) = V/m$ , and both brackets are pure numbers. Direction: there is no $x$ -component because the density is symmetric about the $y$ -axis. The $y$ -component is negative because the positive charge lies in $y^{'} \geq 0$ and pushes the test charge at P towards $- y$ , and $E_{z} > 0$ because P is above a positive charge. Far away: the point-charge field of $Q = ρ_{0} b^{2}$ .

Exercise 3.1.9: Half disk with a nonuniform charge