Problem 3.4.1: The dielectric-filled parallel-plate capacitor

Interactive: 2-D Laplace / Poisson Field Solver

Problem 3.4.1

The dielectric-filled parallel-plate capacitor

Let the potentials of the conducting planes $x = 0$ and $x = d$ be $V_{0}$ and $V_{d}$ , respectively. Assuming that the region between these planes is filled with a simple dielectric material of permittivity $ε$ and does not involve any charge, find the explicit expression of the potential function in this region. Find an expression of $C = Q_{d} / (V_{d} - V_{0})$ in terms of $ε$ and $d$ , where $Q_{d}$ stands for the charge in an area $S$ on the plane $x = d$ .

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

Two conducting planes, x=0 at V0 and x=d at Vd, with a dielectric of permittivity epsilon between them

Try it yourself first.

Strategy

The voltages of the plates are given, not their charges, so we solve Laplace's equation with the two plate potentials as boundary conditions (the second route of the capacitance recipe). The charge then follows from the conductor condition $ρ_{S} = D_{n}$ , and $C = Q / V$ .

Laplace's equation

There is no free charge between the plates, so $\nabla \cdot D = 0$ . The medium is simple, so $D = ε E = - ε \nabla V$ with constant $ε$ , and

\nabla \cdot D = - ε \nabla \cdot \nabla V = - ε \nabla^{2} V = 0 ⟹ \nabla^{2} V = \frac{\partial ^{2} V}{\partial x ^{2}} + \frac{\partial ^{2} V}{\partial y ^{2}} + \frac{\partial ^{2} V}{\partial z ^{2}} = 0.

The planes are infinite in the $y$ - and $z$ -directions, so by symmetry $V$ depends on $x$ only and Laplace's equation reduces to an ordinary differential equation,

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

Boundary conditions

The two plate potentials, $V (0) = V_{0}$ and $V (d) = V_{d}$ , give $B = V_{0}$ and $A = (V_{d} - V_{0}) / d$ . Hence

Answer

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

The electric field and the flux density are uniform:

E D = - \nabla V = - \frac{d V}{d x} a_{x} = - \frac{V _{d} - V _{0}}{d} a_{x} [V/m], = ε E = - ε \frac{V _{d} - V _{0}}{d} a_{x} [C/ m^{2}] .

Charge on the plate

x = d

On the lower face of this plate the normal pointing out of the conductor, into the dielectric, is $a_{n} = - a_{x}$ . The conductor condition $D_{n} = ρ_{S}$ gives

ρ_{S} Q_{d} = D \cdot (- a_{x}) = ε \frac{V _{d} - V _{0}}{d}, = ρ_{S} S = \frac{εS}{d} (V_{d} - V_{0}),

and therefore

Answer

C = \frac{Q _{d}}{V _{d} - V _{0}} = \frac{εS}{d} = ε_{r} \frac{ε _{0} S}{d} [F]

(The plate $x = 0$ carries $- Q_{d}$ , as $a_{n} = + a_{x}$ there.)

Bound charge

The polarisation is

P = (ε - ε_{0}) E = - (ε - ε_{0}) \frac{V _{d} - V _{0}}{d} a_{x} .

It is uniform, so $ρ_{b} = - \nabla \cdot P = 0$ . On the upper face of the dielectric, which touches the plate $x = d$ and has the outward normal $+ a_{x}$ ,

ρ_{S b} = P \cdot a_{x} = - (ε - ε_{0}) \frac{V _{d} - V _{0}}{d} = - (1 - \frac{1}{ε _{r}}) ρ_{S},

and the opposite value appears on the lower face. The total charge at $x = d$ is $ρ_{S} + ρ_{S b} = ε_{0} (V_{d} - V_{0}) / d$ : exactly the charge that would produce the same field in vacuum, as it must be, since $ε_{0} E$ sees all charges (Figure 3.4.30).

Problem 3.4.1 (drawn for V_d>V_0). Left: free charge on the plates, bound charge on the faces of the dielectric, and the vectors \vec E, \vec P, \vec D, all pointing in the -x direction. Right: the potential is linear and |\vec E| is the same as in the empty capacitor, while D=\varepsilon|E|=\rho_S is \varepsilon_r times larger than \varepsilon_0|E|. — **Figure 3.4.30.** Problem 3.4.1 (drawn for $V_{d} > V_{0}$ ). Left: free charge on the plates, bound charge on the faces of the dielectric, and the vectors $E$ , $P$ , $D$ , all pointing in the $- x$ direction. Right: the potential is linear and $∣ E ∣$ is the same as in the empty capacitor, while $D = ε ∣ E ∣ = ρ_{S}$ is $ε_{r}$ times larger than $ε_{0} ∣ E ∣$ .

Linear potential V(x), and the constant values of epsilon0 E and D across the gap

Energy check

The field fills the volume $S d$ with the energy density $\frac{1}{2} ε E^{2}$ , so

W = \frac{1}{2} ε E^{2} S d = \frac{1}{2} ε \frac{( V _{d} - V _{0} ) ^{2}}{d ^{2}} S d = \frac{1}{2} \frac{εS}{d} (V_{d} - V_{0})^{2} = \frac{1}{2} C (V_{d} - V_{0})^{2} ✓

Check

Units: $εS / d$ has [F/m]·[m $^{2}$ ]/[m] $=$ [F]. With $ε = ε_{0}$ we recover the empty capacitor $C_{0} = ε_{0} S / d$ .

Reading the answer

$V (x)$ and $E$ are exactly the same as for the empty capacitor (Problem 3.3.1), because the plate voltages are prescribed and Laplace's equation does not contain $ε$ . What the dielectric changes is $D = ε E$ , and with it the free charge that the source must supply: $ε_{r}$ times more, hence $C = ε_{r} C_{0}$ with $C_{0} = ε_{0} S / d$ . The bound charge $- P$ cancels part of the free charge: $ε_{0} E = ρ_{S} - P$ .

Two ways to insert a dielectric

Voltage fixed (capacitor connected to a source): $E = V / d$ is unchanged, while $Q$ , $D$ and $W$ grow by the factor $ε_{r}$ . Charge fixed (isolated capacitor): $D = ρ_{S}$ is unchanged, while $E$ , $V$ and $W = Q^{2} /2 C$ fall by the factor $ε_{r}$ .

Solve the capacitor numerically in the tool below: with the gap completely filled the potential stays linear and $C$ grows by $ε_{r}$ , while a dielectric that fills only part of the gap makes the field jump at its face and acts like two capacitors in series (Worked example 3.4.1).

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