EE3321 Electromagnetic Field Theory

Potential Energy, Energy DensityCapacitance, Polarization

Boundary Conditions

Potential EnergyThe electric field and potential energy are directly

related: As a test charge +q moves in the direction that the

field opposed it, its potential energy increases.

The electrostatic potential energy is the energy of an electrically charged particle (at rest) in an electric field. The energy difference between two potentials is given

by

U = q(V2 – Vref) Joules (VAs)

Example

Test Charge +q

Charge Q

Potential V = kQ R

Potential Energy U = +qV

RV(∞→R) =Vref = 0

ObservationsThe potential at infinity is zeroA positive test charge +q gains potential as it

gets closer to the charge +QA negative test charge –q loses potential as it

gets closer to the charge +Q.

Cathode Ray Tube The CRT is a vacuum tube

containing an electron gun (a source of electrons) and a fluorescent screen used to create images in the form of light emitted from the fluorescent screen.

The image may represent electrical waveforms (oscilloscope), pictures (television, computer monitor), radar targets and others.

CRT Simplified Set UpOnce the electrons leave the cathode, they

accelerate toward the grid.Electrons entering the deflecting plate region

and change directions depending on the voltage between the plates.

ExerciseAn electron moves at a constant velocity v =

vo ax. Assume that the electron enters in a field E = - 1 a z (V/m) at x =0.

Compute the potential energy U the electron loses as it moves from A (at z = 0 cm) to B (at z = 0.5 cm). Recall e = 1.602 x 10 – 19 As.

Energy Stored in an E FieldUsing Gauss’ Law in differential form and the

Divergence theorem and it can be shown that the energy density or energy per unit volume (J/m3) of the electric field is:

u = ½ є |E|2 (Joules/m3)

The total energy stored in the electrostatic field is

U = ∭ u dV (Joules)where dV is the volume differential.

ExampleLet E = 9 V/mm in between the plates.

Suppose that the area A = 1 cm2 and the dielectric thickness is d = 1 mm. Find the energy stored by the capacitor for a relative permittivity of 2.8. Neglect (field) fringing effects.Notice that the field is constantCalculate the energy densityCalculate the volume between the

capacitor plates

ExerciseCalculate the energy stored in the field

produced by a metal sphere of radius a holding a charge Q.Determine the electric field EFind the energy density uSet up the integral for UIntegrate over space a<R< ∞

CapacitanceAs shown above a capacitor consists of two

conductors separated by a non-conductive region. The non-conductive substance is called the

dielectric medium. The conductors contain equal and opposite

charges on their facing surfaces, and the dielectric contains an electric field.

A capacitor is assumed to be self-contained and isolated, with no net electric charge and no influence from an external electric field.

CapacitanceAn ideal capacitor is wholly

characterized by its capacitance C (in Farads), defined as the ratio of charge ±Q on each conductor to the voltage V between them

C = Q/V

More generally, the capacitance is defined in terms of incremental changes

C = dq/dv

Parallel Plate CapacitorFrom Gauss’ Law the charge and the electric

field between the plates is related by

Likewise, the line integral relating the potential and the electric field simplifies to

Thus the capacitance is given by

ExerciseConsider a parallel plate capacitor. Derive an

expression for the stored energy U in terms of the capacitance C and the potential V.

PolarizationSuppose that a capacitor is charged up by

connecting it to a voltage source V which is then removed.

A fixed charge Q is placed on its upper plate and –Q on the lower plate.

Suppose the capacitor is air filled. In this case,

The capacitance is

PolarizationNext assume that the

capacitor is filled with dielectric material as illustrated here.

Since the charge does not change, the electric flux D is the same as before.

However, the electric field E changes to

PolarizationThe decrease of E is said to be due to the

polarization P of the dielectric molecules which opposes E:

But the capacitance increases to

ExerciseThe relative permittivity of air is 1.0 and that

of quartz is about 4.5. Calculate the difference in capacitance for two capacitors with identical geometry using these two dielectric materials.

Electric Boundary ConditionsOn a perfect conductor

The component of E parallel to the conducting surface is zero

The component of D normal to the conducting surface is numerically equal to the charge density

On a perfect dielectric materialThe component of E parallel to the interface is

continuousThe component of D normal to the interface is

continuous

Perfect Dielectric Medium

WATER DROPPLET

Medium 1 (air)

Medium 2

Tangential components of E are continuousE1t = E2t

Perfect Dielectric Medium

WATER DROPPLET

Medium 1 (air)

Medium 2

Normal components of E are discontinuousε1E1n = ε2E2n

no free charges

Perfect Conductor Medium

WATER DROPPLET

Medium 1 (air)

Medium 2

Tangential components of E are zeroE1t = E2t = 0

Short circuit

X

Perfect Conductor Medium

WATER DROPPLET

Medium 1 (air)

Medium 2

Normal components of E are discontinuousE1n≠ 0

E2n= 0

Examples of Field Lines

ExerciseConsider a dielectric interface at z =

constant. Let єr1 = 2, єr2 = 5, and E1 = 2ax + 3ay + 5az

Find E2

єr2 = 5

єr1 = 2

HomeworkRead textbook sections 4-8, 4-9, 4-10, 4-11Solve problems 4.43, 4.45, 4.50, 4.51, 4.52,

4.54