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

Problem 3.3.2

Interactive: 2-D Laplace / Poisson Field Solver

Problem 3.3.2

Laplace's equation in $ρ$ only

Suppose that there is no source between two infinitely long coaxial circular cylinders of radii $a$ and $b$ . Find the explicit expression of the potential function by assuming that it has known constant values $V_{a}$ and $V_{b}$ on the cylinders as shown in the figure. Determine the electric field between these cylinders. Find the total charge on the outer surface if the length of the cylinders is $L$ .

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 4-8.

Try it yourself first.

Strategy

The same plan in cylindrical coordinates: no charge between the cylinders (Laplace), infinitely long cylinders and $ϕ$ -independent boundary values (so $V = V (ρ)$ ), the radial part of the cylindrical Laplacian (3.3.16), two boundary conditions, then $E = - \nabla V$ and $ρ_{S} = D_{n}$ .

Symmetry and the equation

Laplace's equation in cylindrical coordinates (3.3.16) reads

Δ V = \frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ \frac{\partial V}{\partial ρ}) + \frac{1}{ρ ^{2}} \frac{\partial ^{2} V}{\partial ϕ ^{2}} + \frac{\partial ^{2} V}{\partial z ^{2}} = 0.

The cylinders are infinitely long and the boundary values do not depend on $ϕ$ : sliding along $z$ or rotating about the axis changes nothing. Hence $V = V (ρ)$ , and

\frac{1}{ρ} \frac{d}{d ρ} (ρ \frac{d V}{d ρ}) = 0 \Rightarrow ρ \frac{d V}{d ρ} = C \Rightarrow \frac{d V}{d ρ} = \frac{C}{ρ} \Rightarrow V (ρ) = C ln ρ + D .

Boundary conditions

$V (a) = V_{a}$ and $V (b) = V_{b}$ give

C ln a + D = V_{a}, C ln b + D = V_{b} .

Subtracting, $C (ln b - ln a) = V_{b} - V_{a}$ , and then $D = V_{a} - C ln a$ :

C = \frac{V _{b} - V _{a}}{ln ( b / a )}, D = \frac{V _{a} ln b - V _{b} ln a}{ln ( b / a )} .

Hence

Answer

V (ρ) = \frac{V _{b} - V _{a}}{ln ( b / a )} ln ρ + \frac{V _{a} ln b - V _{b} ln a}{ln ( b / a )} = V_{a} + (V_{b} - V_{a}) \frac{ln ( ρ / a )}{ln ( b / a )} [V]

(The second form makes the boundary values evident: $ln (ρ / a)$ is $0$ at $ρ = a$ and $ln (b / a)$ at $ρ = b$ .)

Field

Only the $ρ$ -component of the gradient survives:

Answer

E = - \nabla V = - \frac{d V}{d ρ} a_{ρ} = \frac{V _{a} - V _{b}}{ln ( b / a )} \frac{1}{ρ} a_{ρ} [V/m]

Charge on the outer cylinder

The charge of the outer conductor sits on its inner face, $ρ = b$ , whose normal pointing out of the conductor into the gap is $- a_{ρ}$ :

ρ_{S} = D \cdot (- a_{ρ})_{ρ = b} = - ε_{0} E_{ρ} (b) = - ε_{0} \frac{V _{a} - V _{b}}{b ln ( b / a )} .

It is uniform over the area $2 π b L$ of a length $L$ , so

Answer

Q_{outer} = 2 π b L ρ_{S} = \frac{2 π ε _{0} L}{ln ( b / a )} (V_{b} - V_{a}) [C]

At $ρ = a$ the outward normal is $+ a_{ρ}$ and $ρ_{S} = ε_{0} E_{ρ} (a)$ , so the inner cylinder carries $2 π a L ε_{0} (V_{a} - V_{b}) / (a ln (b / a)) = - Q_{outer}$ .

Problem 3.3.2 (drawn for V_a>V_b): cross-section with the equipotential cylinders \rho=\text{const} for equal steps of V and the radial field. — **Figure 3.3.23.** Problem 3.3.2 (drawn for $V_{a} > V_{b}$ ): cross-section with the equipotential cylinders $ρ = const$ for equal steps of $V$ and the radial field.

Right: V varies as \ln\rho, E_\rho as 1/\rho. — Right: $V$ varies as $ln ρ$ , $E_{ρ}$ as $1/ ρ$ .

Check

The second form of $V$ gives $V_{a}$ at $ρ = a$ and $V_{b}$ at $ρ = b$ . For $V_{a} > V_{b}$ the field points outwards, from high to low potential. The charges on the two cylinders are equal and opposite, and $2 π ε_{0} L (V_{b} - V_{a}) / ln (b / a)$ has the unit $F V = C$ .

Reading the answer

$V \propto ln ρ$ and $E \propto 1/ ρ$ : between the cylinders the field is that of a line charge on the axis (Problems 3.1.1 and 3.2.1), but now with the voltage as the given quantity. Equal steps of $V$ give equipotential cylinders whose radii form a geometric progression (Figure 3.3.23), and the field is largest at the inner conductor, the place where breakdown would start (Exercise 3.3.10). The ratio $∣ Q ∣/∣ V_{a} - V_{b} ∣ = 2 π ε_{0} L / ln (b / a)$ is the capacitance of a coaxial line of length $L$ (Section 3.4).

The solver below finds the same logarithmic potential numerically: compare the spacing of its equipotentials with the analytic $ln ρ$ law, then shrink the inner conductor and watch the field crowd against it.

InteractiveLoading simulation…

Problem 3.3.2: Laplace's equation in \rho only