Exercise 3.1.8: Disk with a radially increasing charge

Exercise 3.1.8

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 3.1.8

Disk with a radially increasing charge

cf. Problem 3.1.3

A circular disk of radius $b$ in the $x y$ -plane, centred at the origin, carries the surface charge density $ρ_{S} = ρ_{0} ρ^{'} / b$ C/m², $ρ_{0}$ being a constant.

a) Find the total charge on the 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 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 now the density grows with $ρ^{'}$ . Because $ρ_{S}$ does not depend on $ϕ^{'}$ , the symmetry argument of Problem 3.1.3 is unchanged: only $E_{z}$ survives and the $ϕ^{'}$ -integral gives $2 π$ . The radial integral contains $(ρ^{'})^{2}$ in the numerator, which is not one of the standard integrals (3.1.30), so one algebraic step is needed to split it into standard forms.

Part a): charge element

On the disk the area element is $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ (Figure 3.1.15c, and do not forget the factor $ρ^{'}$ ). With $ρ_{S} = ρ_{0} ρ^{'} / b$ the charge of one element is

d q^{'} = ρ_{S} d S^{'} = ρ_{0} \frac{ρ ^{'}}{b} ρ^{'} d ρ^{'} d ϕ^{'} = \frac{ρ _{0}}{b} (ρ^{'})^{2} d ρ^{'} d ϕ^{'} .

Part a): integrate

The integrand is a product of a function of $ρ^{'}$ and a constant in $ϕ^{'}$ , so the two integrals separate:

Q = \int_{ϕ^{'} = 0}^{2 π} \int_{ρ^{'} = 0}^{b} \frac{ρ _{0}}{b} (ρ^{'})^{2} d ρ^{'} d ϕ^{'} = \frac{ρ _{0}}{b} 2 π \int_{0}^{2 π} d ϕ^{'} b^{3} /3 \int_{0}^{b} (ρ^{'})^{2} d ρ^{'} = \frac{ρ _{0}}{b} \cdot 2 π \cdot \frac{b ^{3}}{3},

Answer

Q = \frac{2 π ρ _{0} b ^{2}}{3} [C]

Check of part a): $ρ_{0} b^{2}$ is $(C/ m^{2}) m^{2} = C$ . If the density were the constant $ρ_{0}$ everywhere, the charge would be $π b^{2} ρ_{0}$ . Here the density equals $ρ_{0}$ only at the rim and is smaller inside, and the ratio $\frac{2}{3}$ of the two results is the area-weighted average of $ρ^{'} / b$ .

Part b): symmetry

Rotating the disk about the $z$ -axis leaves it unchanged, because $ρ_{S}$ depends on $ρ^{'}$ only. The point P lies on this axis, so the field cannot single out a horizontal direction.

The element at $(ρ^{'}, ϕ^{'})$ and the one at $(ρ^{'}, ϕ^{'} + π)$ have the same density and the same distance from P. Their horizontal contributions are equal and opposite, while their $z$ -contributions are identical and add. Hence $E_{x} = E_{y} = 0$ and $E = E_{z} a_{z}$ .
All elements on the ring of radius $ρ^{'}$ are at the same distance $R = (ρ^{'})^{2} + h^{2}$ from P and carry the same density. The $ϕ^{'}$ -integral therefore only multiplies by $2 π$ , and the density $ρ_{0} ρ^{'} / b$ enters the one remaining integral, over $ρ^{'}$ .

Part b): points and displacement vector

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

In polar coordinates $x^{'} = ρ^{'} cos ϕ^{'}$ and $y^{'} = ρ^{'} sin ϕ^{'}$ , so that

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

The charge element is $d q^{'} = ρ_{0} (ρ^{'} / b) ρ^{'} d ρ^{'} d ϕ^{'}$ , as in part a). The electric field at P is

E (P) = \frac{1}{4 π ε _{0}} \int_{ρ^{'} = 0}^{b} \int_{ϕ^{'} = 0}^{2 π} ρ_{0} \frac{ρ ^{'}}{b} 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, collecting the factors of $ρ^{'}$ and separating the components,

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

Part b): integrate

First the angular integrals. Since $\int_{0}^{2 π} cos ϕ^{'} d ϕ^{'} = \int_{0}^{2 π} sin ϕ^{'} d ϕ^{'} = 0$ and $\int_{0}^{2 π} d ϕ^{'} = 2 π$ , the $x$ - and $y$ -terms vanish, as predicted in the symmetry step, and

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

The numerator of $I$ is $(ρ^{'})^{2}$ , whereas the first standard integral (3.1.30) needs $ρ^{'}$ there. We therefore write $(ρ^{'})^{2} = [(ρ^{'})^{2} + h^{2}] - h^{2}$ , which splits the integrand into two simpler terms:

\frac{( ρ ^{'} ) ^{2}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} = \frac{1}{( ρ ^{'} ) ^{2} + h ^{2}} - \frac{h ^{2}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} .

The second term is $h^{2}$ times the second standard integral (3.1.30):

\int \frac{h ^{2} d ρ ^{'}}{[ ( ρ ^{'} ) ^{2} + h ^{2} ] ^{3/2}} = h^{2} \frac{ρ ^{'}}{h ^{2} ( ρ ^{'} ) ^{2} + h ^{2}} = \frac{ρ ^{'}}{( ρ ^{'} ) ^{2} + h ^{2}} .

The first term is a new standard form. Its antiderivative is $ln (ρ^{'} + (ρ^{'})^{2} + h^{2})$ , which we check by differentiation:

\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}} .

Putting the two pieces together and inserting the limits,

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

Hence

Answer

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

Interpretation: two limits

Put $u = b / h$ . Then $ln [(b + b^{2} + h^{2}) / h] = ln (u + 1 + u^{2}) = sinh^{- 1} u$ and $b / b^{2} + h^{2} = u / 1 + u^{2}$ , so the bracket is $sinh^{- 1} u - u / 1 + u^{2}$ .

$h ≫ b$ (far away, $u ≪ 1$ ): with $sinh^{- 1} u ≃ u - u^{3} /6$ and $u / 1 + u^{2} ≃ u - u^{3} /2$ the bracket tends to $u^{3} /3 = b^{3} / (3 h^{3})$ , and

$E \to \frac{ρ _{0} h}{2 ε _{0} b} \frac{b ^{3}}{3 h ^{3}} a_{z} = \frac{ρ _{0} b ^{2}}{6 ε _{0} h ^{2}} a_{z} = \frac{Q}{4 π ε _{0} h ^{2}} a_{z}, Q = \frac{2 π ρ _{0} b ^{2}}{3} :$

the disk looks like a point charge carrying its total charge from part a).
$h \to 0$ (very close to the centre): the bracket tends to $ln (2 b / h) - 1$ , which grows only logarithmically, so $h ln (2 b / h) \to 0$ and $E \to 0$ . This is reasonable: just above the centre the disk looks locally like an infinite sheet with field $ρ_{S} / (2 ε_{0})$ , and $ρ_{S} = 0$ at the centre.

Check

Units: $ρ_{0} / ε_{0}$ is $(C/ m^{2}) / (F/m) = V/m$ , and $h / b$ and the bracket are pure numbers, so the result is in V/m. Direction: along $+ z$ above a positive disk (and, by the mirror symmetry about the plane of the disk, along $- z$ below it), that is, away from the disk, with no horizontal component. Far away: the point-charge field of $Q = 2 π ρ_{0} b^{2} /3$ .