Advanced General Physics 323/324

Electricity and Magnetism Unit E2: Dielectrics and Currents

Prerequisite: E1

Overview:

All matter is made up of small charged particles, but, from a macroscopic viewpoint, the electrical properties vary greatly. In some materials, called "conductors", charges flow easily under the influence of an external electric field. In other materials the charges move slightly but do not flow when an external electric field is applied, so no current flows in these "insulators". In this unit we will investigate the effect of electric fields on insulators (also called "dielectrics") and how electric fields produce "currents" in conductors. We also introduce the important idea that a volume of space with an electric field contains an associated energy proportional to the square of the electric field.

SECTION I: Capacitance and Energy

Read:

E. M. Purcell, Electricity and Magnetism, Berkeley Physics Course, Vol. 2, 2nd Ed. , Chapt. 3 - Electric Fields around Conductors, Sec. 3.1-3.2, 3.4-3.5,3.7

CONCEPTS:

First we review the capacitor, a common element in electric circuits. We consider only the definition and the simplest cases of the parallel plate and spherical capacitors. Study the example of the two concentric metal spheres of Sect. 3.4. The potential for a single sphere corresponds to the case Q₂ = 0. Thus for a sphere carrying a uniformly distributed surface charge Q the potential outside of the sphere at a distance R from its center is V = Q/R. This result is summarized again in the discussion centered on Eqn. 11 in Chap. 3. Note that the potential V is proportional to the charge Q. This proportionality is true in general for conductors in electrostatics.

After completing this unit you should understand:

Proportionality of potential and charge on a conductor leads to Q = C V, defining the capacitance C.
Capacitance.
Energy stored in a capacitor: U=QV/2=CV²/2=Q²/(2C).
Energy per volume in an electric field: E²/(2 ε₀).

Problems:

Purcell 3.13 (Assume the capacitance is that of a conducting sphere 1 meter in diameter.)

SECTION II: Currents

Read:

E. M. Purcell, Electricity and Magnetism, Berkeley Physics Course, Vol. 2, 2nd Ed. , Chap. 4 - Electric Currents, Sec. 4.1-4.3, 4.7-4.10

CONCEPTS: In this section you will learn some of the basic facts about electric current: moving electric charge.

After completing this unit you should understand:

The relation between current and the density and velocity of the charged particles .
Charge conservation.
The difference between conductors, semi-conductors, and insulators.
Energy dissipation and electromotive force.

Problems:

Purcell 4.16,4.17,4.18,4.26

SECTION III: Dielectrics

Read:

E. M. Purcell, Electricity and Magnetism, Berkeley Physics Course, Vol. 2, 2nd Ed. , Chapt. 10 - Electric Fields in Matter, Sec. 10.1-10.2, 10.4-10.5,10.7-10.8, 10.11

CONCEPTS: This section introduces the microscopic picture of material in an electric field for the case of dielectrics where the electrons are bound to the atoms, but an electric field can induce a relative displacement of the positive and negative charges (ions and electrons) producing microscopic electric dipoles which result in an additional contribution to the electric field in the material.

After completing this unit you should understand:

Displacements of positive and negative charges in the dielectric.
Dipole density.
Susceptibility of a material and the displacement of charges.
Free and polarization charges.

Questions:

In section 10.5 the induced dipole moment vector p is assumed to be proportional to the perturbing field E. To recapitulate, let a neutral atom be represented by a positive (nuclear) charge +q anchored at the center of a uniform spherical cloud of negative charge -q of radius a. In an external electric field E the negative cloud is displaced a distance dz in the direction opposite to E . Assume the cloud remains spherical after the displacement and that the positive nuclear charge is fixed. The displacement dz is determined by requiring cancellation between the external field and the opposing electrical attraction between the displaced cloud and the nuclear charge. One can show that the magnitude of this opposing electric field is given by E_opp = (dz)³ q/(a³ (dz)²)) , so that E=E_opp requires dz = a³ E /q and thus that the dipole moment p = q dz = a³ E is indeed proportional to the applied electric field.

If all charges are alike, what is the difference between free and polarization charges?

Problems::

Purcell 10.13 (Omit the comparison with magnetic energy!), 10.14, 10.16

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