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

Problem 3.3.1

Laplace's equation in one variable

Consider a parallel-plate capacitor with the plates located at $x = 0$ and $x = d$ with constant potentials $V_{0}$ volts and $V_{d}$ volts, respectively. Assuming no charge between the plates, determine the electrostatic potential and the electric field inside the capacitor. Find the total charge on the plate above if the surface area of the plates is $S m^{2}$ each.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 4-1.

Try it yourself first.

Strategy

There is no charge between the plates, so $V$ obeys Laplace's equation. Plates much larger than their spacing make $V$ depend on $x$ only, which turns Laplace's equation into an ordinary differential equation, and the two plate voltages fix its two constants. The charge then follows from $ρ_{S} = D_{n}$ (recipe of Section 3.3.5).

Symmetry and the equation

There is no charge between the plates, so the potential satisfies Laplace's equation,

Δ V = \frac{\partial ^{2} V}{\partial x ^{2}} + \frac{\partial ^{2} V}{\partial y ^{2}} + \frac{\partial ^{2} V}{\partial z ^{2}} = 0.

The plates are taken to be infinite in the $y$ - and $z$ -directions (their size is much larger than their spacing $d$ , and we ignore the fringing at the edges, Figure 3.3.13a). Shifting the observation point parallel to the plates then changes nothing, so $V$ depends on $x$ only, and Laplace's equation reduces to an ordinary differential equation:

\frac{d ^{2} V}{d x ^{2}} = 0 \Rightarrow \frac{d V}{d x} = A \Rightarrow V (x) = A x + B .

Boundary conditions

The two plates are conductors at given potentials:

V (0) = V_{0} and V (d) = V_{d} .

The first gives $B = V_{0}$ , the second $A d + V_{0} = V_{d}$ , i.e. $A = (V_{d} - V_{0}) / d$ . Hence

Answer

V (x) = V_{0} + \frac{V _{d} - V _{0}}{d} x [V]

Field

The gradient has only an $x$ -component:

Answer

E = - \nabla V = - \frac{d V}{d x} a_{x} = - \frac{V _{d} - V _{0}}{d} a_{x} [V/m]

The field is uniform. For $V_{d} > V_{0}$ it points in the $- x$ direction, from the plate at the higher potential to the plate at the lower one, as it must.

Charge on the upper plate

The charge resides on the lower surface of the upper plate, facing the gap. There the normal pointing out of the conductor into the gap is $- a_{x}$ , so

ρ_{S} = D \cdot (- a_{x}) = ε_{0} (- \frac{V _{d} - V _{0}}{d}) (a_{x} \cdot (- a_{x})) = ε_{0} \frac{V _{d} - V _{0}}{d} .

Since $ρ_{S}$ is uniform over the area $S$ ,

Answer

Q_{above} = ρ_{S} S = \frac{ε _{0} S}{d} (V_{d} - V_{0}) [C]

On the upper surface of the lower plate the outward normal is $+ a_{x}$ and $ρ_{S} = - ε_{0} (V_{d} - V_{0}) / d$ : the lower plate carries $- Q_{above}$ .

Problem 3.3.1, drawn for V_d>V_0 with \Delta V=V_d-V_0: equally spaced equipotential planes x=\text{const}, a uniform field from the high to the low potential, and surface charges \pm Q on the inner faces of the plates. — **Figure 3.3.22.** Problem 3.3.1, drawn for $V_{d} > V_{0}$ with $Δ V = V_{d} - V_{0}$ : equally spaced equipotential planes $x = const$ , a uniform field from the high to the low potential, and surface charges $\pm Q$ on the inner faces of the plates.

Right: V is linear in x, |E_x| constant. — Right: $V$ is linear in $x$ , $∣ E_{x} ∣$ constant.

Check

$V (0) = V_{0}$ and $V (d) = V_{d}$ are reproduced, $E$ has the unit V/m and points from the plate at the higher potential to the other, the two plates carry $\pm Q_{above}$ , and the unit of $ε_{0} S (V_{d} - V_{0}) / d$ is $(F/m) m^{2} V / m = F V = C$ .

Reading the answer

$V$ is linear and $E$ uniform: the equipotentials are equally spaced planes parallel to the plates (Figure 3.3.22). The magnitude of the field is simply "voltage divided by spacing", the familiar rule for parallel plates. The charge is proportional to the voltage, and the ratio $Q_{above} / (V_{d} - V_{0}) = ε_{0} S / d$ is the capacitance of the parallel-plate capacitor (Section 3.4). The same uniform field accelerates the electrons of the gun in Application 3.3.1.

Problem 3.3.1: Laplace's equation in one variable