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Topic 4 of 9

Dielectrics & Polarization

In a dielectric material, an applied electric field displaces positive and negative charges within each atom, creating a net polarization. This bound charge modifies the total field. The displacement field $\vec D=\varepsilon_0\vec E+\vec P$ satisfies a simpler form of Gauss's law involving only free charges.

Key Concepts

Polarization

\vec P

is the electric dipole moment per unit volume. Bound surface charge density:

\sigma_b=\vec P\cdot\hat n

. Bound volume charge density:

\rho_b=-\nabla\cdot\vec P

Displacement Field

\vec D=\varepsilon_0\vec E+\vec P

. Gauss's law for

\vec D

\oint\vec D\cdot d\vec a=Q_{\rm free,enc}

. In linear dielectrics:

\vec P=\varepsilon_0\chi_e\vec E

and

\vec D=\varepsilon\vec E=\varepsilon_0\varepsilon_r\vec E

Dielectric Constant

Relative permittivity

\varepsilon_r=1+\chi_e\ge1

. Inside a capacitor filled with dielectric:

C=\varepsilon_r C_0

. The field is reduced to

E=E_0/\varepsilon_r

compared to vacuum.

Boundary Conditions

At a dielectric interface:

D_{1\perp}-D_{2\perp}=\sigma_f

(free surface charge) and

E_{1\parallel}=E_{2\parallel}

(tangential

E

is continuous). These replace the vacuum conditions.

Key Equations

Displacement field

\vec{D} = \varepsilon_0\vec{E} + \vec{P} = \varepsilon_0\varepsilon_r\vec{E}

In linear isotropic dielectrics; εr = relative permittivity.

Gauss's law for D

\oint \vec{D}\cdot d\vec{a} = Q_{\rm free}

Involves only free charges; simplifies analysis.

Capacitance with dielectric

C = \varepsilon_r C_0 = \varepsilon_r \frac{\varepsilon_0 A}{d}

Dielectric multiplies capacitance by εr.

Worked Example

Parallel-Plate Capacitor with Dielectric

Problem

A parallel-plate capacitor ( $A=0.020$ m², $d=1.0$ mm) is filled with a dielectric ( $\varepsilon_r=4.5$ ). Find $C$ and the voltage for $Q=2.0\times10^{-7}$ C.

Solution

Vacuum capacitance: $C_0=\varepsilon_0A/d=8.85\times10^{-12}\times0.020/10^{-3}=1.77\times10^{-10}$ F.

C = \varepsilon_r C_0 = 4.5\times1.77\times10^{-10} = 7.965\times10^{-10}\text{ F}\approx797\text{ pF}

V = Q/C = 2.0\times10^{-7}/7.965\times10^{-10} = 251\text{ V}

Answer

C\approx797

pF,

V\approx251

Practice

Exercises

7 problems

1 of 7

Drag the plate-separation slider — watch $C_0 = \varepsilon_0 A/d$ fall as the gap opens. For $A = 0.010$ m² and $d = 0.5$ mm, find C₀ (in pF).

Parallel-plate capacitor

Plate separation d 0.50 mm → 177.0 pF

Increase plate separation — watch capacitance fall as 1/d. At d = 0.5 mm with A = 0.010 m², find C₀.

Capacitance

C₀ = pF

Formula: C₀ = ε₀A/d | ε₀ = 8.85×10⁻¹² F/m, A = 0.010 m², d = 0.5 mm

2 of 7

Slide the dielectric ($\varepsilon_r$) into the gap — bound dipoles align and capacitance grows. Starting from $C_0 = 177$ pF, find C at $\varepsilon_r = 3.0$ (in pF).

Dielectric insertion

Relative permittivity εr 3.0 → 531.0 pF

Slide εr — the dielectric polarizes, bound charges reduce the internal field, and capacitance grows. At εr = 3.0, C = ?

Capacitance

C = pF

Formula: C = εr × C₀ | εr = 3.0, C₀ = 177 pF

3 of 7

A dielectric sphere (radius $R=0.10$ m, $\varepsilon_r=2.5$ ) in a uniform external field $E_0=1000$ N/C. The field inside a dielectric sphere is $E_{\rm in}=3E_0/(\varepsilon_r+2)$ (in N/C).

Unlock Exercise 3