Linear Dielectrics (Part 2)

In Part 1, we introduced polarization, permittivity, and the displacement field. Here we use those ideas to see why a dielectric increases a capacitor’s capacitance, then solve two standard capacitor problems.

Why a dielectric increases capacitance

Consider a large parallel-plate capacitor with plate area \(A\), separation \(d\), and free charges \(+Q\) and \(-Q\). Fringing is neglected. Let

\[ \sigma_f=\frac{Q}{A}. \]

With the free charge held fixed, planar symmetry and Gauss’s law for \(\mathbf D\) give the same displacement field between the plates whether the gap is vacuum or is filled by a homogeneous dielectric:

\[ \lvert\mathbf D\rvert=\sigma_f. \]

This statement depends on the fixed-free-charge condition and the parallel-plate symmetry. More generally, \(\nabla\cdot\mathbf D=\rho_f\); \(\mathbf D\) is an auxiliary field whose flux is governed by free charge, not an electric field produced only by free charge.

In vacuum,

\[ \mathbf D=\varepsilon_0\mathbf E_0, \qquad E_0=\frac{\sigma_f}{\varepsilon_0}. \]

If a linear, isotropic dielectric of permittivity \(\varepsilon\) completely fills the gap,

\[ \mathbf D=\varepsilon\mathbf E, \qquad E=\frac{\sigma_f}{\varepsilon}. \]

For an ordinary dielectric with \(\varepsilon\gt\varepsilon_0\), the field is reduced by

\[ \frac{E}{E_0}=\frac{\varepsilon_0}{\varepsilon}. \]

Because the field is uniform, the voltage magnitudes are \(V_0=E_0d\) and \(V=Ed\). Thus, at the same fixed free charge \(Q\),

\[ \frac{V}{V_0}=\frac{\varepsilon_0}{\varepsilon}. \]

Using \(C=Q/V\),

\[ \boxed{ \frac{C}{C_0}=\frac{\varepsilon}{\varepsilon_0} } \qquad\Longrightarrow\qquad C=\frac{\varepsilon A}{d}. \]

The dimensionless ratio

\[ \boxed{ \varepsilon_r=\frac{\varepsilon}{\varepsilon_0}=1+\chi_e } \]

is the relative permittivity, often called the dielectric constant. It compares the material’s permittivity with the vacuum permittivity, in that order. The constitutive relation can therefore be written as

\[ \boxed{ \mathbf D=\varepsilon\mathbf E =\varepsilon_0\varepsilon_r\mathbf E }. \]

Real capacitors often use thin dielectric films between rolled or stacked conducting foils to obtain a large capacitance in a compact volume. A battery, by contrast, is primarily an electrochemical energy-storage device and should not be modeled as a folded capacitor with a gel dielectric.

If the capacitor remains connected to an ideal voltage source while the dielectric is inserted, \(V\) rather than \(Q\) is fixed. The capacitance still increases by \(\varepsilon_r\), but additional free charge flows from the source so that \(Q=CV\).

Worked example: two dielectric layers in series

Two homogeneous, linear dielectric slabs completely fill a parallel-plate capacitor. Each slab has thickness \(a\) and plate area \(A\), so the plate separation is \(2a\). The upper slab has

\[ \varepsilon_1=2\varepsilon_0, \]

and the lower slab has

\[ \varepsilon_2=1.5\varepsilon_0=\frac{3}{2}\varepsilon_0. \]

The top and bottom electrodes carry free surface-charge densities \(+\sigma\) and \(-\sigma\), respectively. Choose \(+\hat{\mathbf z}\) upward, so the field points in the \(-\hat{\mathbf z}\) direction. Neglect fringing and assume there is no free charge at the dielectric interface.

We will find \(\mathbf D\), \(\mathbf E\), \(\mathbf P\), the voltage, the capacitance, and all bound charges.

Displacement field

The normal component of \(\mathbf D\) jumps only where free surface charge is present. Because the interface between the slabs carries no free charge,

\[ \boxed{ \mathbf D_1=\mathbf D_2=-\sigma\hat{\mathbf z} }. \]

This equality follows from the free-charge boundary condition and the planar symmetry. It does not follow merely from saying that the two materials have the same free volume-charge density.

Electric field

Using \(\mathbf D_i=\varepsilon_i\mathbf E_i\),

\[ \boxed{ \mathbf E_1=-\frac{\sigma}{2\varepsilon_0}\hat{\mathbf z}, \qquad \mathbf E_2=-\frac{2\sigma}{3\varepsilon_0}\hat{\mathbf z} }. \]

The normal electric field is larger in slab 2 because its permittivity is smaller.

Polarization

For a linear dielectric,

\[ \mathbf P_i=\varepsilon_0\chi_{e,i}\mathbf E_i, \qquad \varepsilon_{r,i}=1+\chi_{e,i}. \]

Here

\[ \chi_{e,1}=1, \qquad \chi_{e,2}=\frac{1}{2}, \]

so

\[ \boxed{ \mathbf P_1=-\frac{\sigma}{2}\hat{\mathbf z}, \qquad \mathbf P_2=-\frac{\sigma}{3}\hat{\mathbf z} }. \]

Voltage and capacitance

The voltage drop magnitude across each slab is \(V_i=E_i a\):

\[ V_1=\frac{\sigma a}{2\varepsilon_0}, \qquad V_2=\frac{2\sigma a}{3\varepsilon_0}. \]

Therefore

\[ \boxed{ V=V_1+V_2=\frac{7\sigma a}{6\varepsilon_0} }. \]

Since \(Q=\sigma A\), the capacitance is

\[ \boxed{ C=\frac{Q}{V}=\frac{6\varepsilon_0 A}{7a} }. \]

This agrees with treating the two layers as capacitors in series:

\[ \frac{1}{C} =\frac{a}{\varepsilon_1A} +\frac{a}{\varepsilon_2A}. \]

Bound charge

Because each slab is homogeneous and \(\mathbf P_i\) is uniform,

\[ \boxed{ \rho_b=-\nabla\cdot\mathbf P=0 } \]

within both slabs. Bound surface charge obeys

\[ \sigma_b=\mathbf P\cdot\hat{\mathbf n}, \]

where \(\hat{\mathbf n}\) points outward from the dielectric under consideration.

At the top face of slab 1, \(\hat{\mathbf n}=+\hat{\mathbf z}\), so

\[ \sigma_{b,\mathrm{top}}=-\frac{\sigma}{2}. \]

At the interface, the lower face of slab 1 contributes \(+\sigma/2\), while the upper face of slab 2 contributes \(-\sigma/3\). The net bound sheet charge at the interface is therefore

\[ \sigma_{b,\mathrm{interface}} =\frac{\sigma}{2}-\frac{\sigma}{3} =\frac{\sigma}{6}. \]

At the bottom face of slab 2,

\[ \sigma_{b,\mathrm{bottom}}=+\frac{\sigma}{3}. \]

The three net sheets sum to zero per unit area:

\[ -\frac{\sigma}{2}+\frac{\sigma}{6}+\frac{\sigma}{3}=0, \]

as expected for initially neutral dielectric material.

Which partial filling gives the larger capacitance?

Now compare two ways to place a dielectric of permittivity \(\varepsilon\) in a parallel-plate capacitor of plate area \(A\) and separation \(d\). Fringing is neglected.

Configuration A: layers along the field direction

The dielectric spans the full plate area and occupies thickness \(d/2\). Vacuum occupies the remaining total thickness \(d/2\); it may be split into two gaps without changing the result. The dielectric and vacuum regions act as capacitors in series:

\[ \frac{1}{C_A} =\frac{d/2}{\varepsilon A} +\frac{d/2}{\varepsilon_0A}. \]

Hence

\[ \boxed{ C_A=\frac{2\varepsilon\varepsilon_0A} {d(\varepsilon+\varepsilon_0)} }. \]

Configuration B: regions side by side

The dielectric fills the full separation \(d\) over half the plate area, while vacuum fills the other half. The two regions share the same voltage and act as capacitors in parallel:

\[ \boxed{ C_B =\frac{\varepsilon(A/2)}{d} +\frac{\varepsilon_0(A/2)}{d} =\frac{(\varepsilon+\varepsilon_0)A}{2d} }. \]

For reference, the empty capacitor has \(C_0=\varepsilon_0A/d\). Therefore

\[ \frac{C_A}{C_0}=\frac{2\varepsilon}{\varepsilon+\varepsilon_0}, \qquad \frac{C_B}{C_0}=\frac{\varepsilon+\varepsilon_0}{2\varepsilon_0}. \]

Their difference is

\[ \boxed{ C_A-C_B =-\frac{A(\varepsilon-\varepsilon_0)^2} {2d(\varepsilon+\varepsilon_0)} \lt 0 } \]

for \(\varepsilon\gt\varepsilon_0\). Thus

\[ \boxed{C_B\gt C_A}. \]

The side-by-side arrangement wins because the dielectric-covered region provides an additional parallel path with high capacitance. The layered arrangement forces the electric flux through a lower-capacitance vacuum section in series.