Physics II Class 04/02


Gauss' Law

Using symbols, what is the flux of a point charge q at the center of an imaginary sphere of radius r?

The flux is 4 pi k q.

Using symbols, what is the surface area of the surrounding sphere?

The sphere’s area is 4 pi r^2.

Symmetry arguments prove that the field is uniform at the surface of the sphere and at every point perpendicular to the surface so what is the electric field at the surface of the sphere?

The field is E = flux / area = 4 pi k q / 4 pi r^2 = k q / r^2.

This is consistent with Coulomb’s law, since the force on a test charge Q at this point will be

·         Force = Q E = Q * k q / r^2 = k q Q / r^2.

If we have a long line of uniformly distributed charge with charge density lambda (measured in charge / unit length) then how much charge is enclosed in a cylinder of length L and radius r which surrounds the wire?  What is the flux of this charge?

The cylinder encloses charge q = lambda * L (charge / length * length).  The flux of the charge is therefore

·         Flux = 4 pi k q = 4 pi k (lambda L)

If the line charge coincides with the axis of the cylinder then do the conditions for Gauss’ Law hold?

By symmetry arguments the electric field is of constant magnitude and radial, so it is everywhere uniform in magnitude perpendicular to the curved surface of the cylinder.

So what is the magnitude of the field at the curved surface?

The area of the curved surface is circumference * length = 2 pi r * L

·         E = flux / area = 4 pi k lambda L / (2 pi r L) = 2 k lambda / r.

Note that this is an inverse first power, as opposed to the inverse square of a point charge.

Note also that there is no flux through the ends of the cylinder, since the ends are parallel to the field.

 

What is the charge enclosed within an area A of an infinity plane distribution of charge having uniform charge density sigma (measured in charge / unit area)?  What is the flux of this charge?

The charge enclosed is sigma * A.  The flux is 4 pi k ( sigma * A).

If we construct a rectangular box with sides perpendicular to the plane, with the top of the box above and the bottom of the box below the plane, and with cross-sectional area A then what is the electric field at the top and at the bottom?

The flux exits thru top and bottom, total area 2 A, so the field is

·         E = 4 pi k sigma A / (2 A) = 2 pi k sigma.

   

What is the electrical field in a uniform wire of length L across which a potential difference Vab is maintained?  How does this field affect the drift velocity of charge carriers?

The field is E = Vab / L.

Greater field implies greater acceleration of charge carriers between collisions.  

Note that current is defined as the time rate of charge drift past a point:

If two uniform wires have identical cross-section, are made of the same material but one is twice as long as the other.  If both are subjected to the same voltage then how will the electric fields in the wires compare?

The field is E = V / `ds.  If V is the same for both, then E1 = V / L is twice as great as E2 = V / (2 L).

The average drift velocity of charge carriers is proportional to the electric field.  So which wire carries more current?

The field in the first wire is double that in the second, so the average drift velocity is higher in the first.

Since both wires, being of identical material and equal cross-section, have the same number of charge carriers per unit length, the greater drift velocity will bring more charge carriers past a given point than for the second wire and the current will be greater.  

Why does it make sense to define resistance to flow of current as V / I, where V is potential difference or voltage and I is current?

High resistance means less current flow for a given voltage.  Less current flow means that the denominator of V / I will be less, resulting in a larger quotient.  The lower the current I, the higher will be V / I, for a given voltage.

How does the resistance of the second wire (the one twice as long) compare to that of the first?  How therefore does length affect electrical resistance, defined as the ratio V / I ?  

The shorter wire had the greater electric field, therefore the greater current.  So the second wire had the greater length, lesser current and therefore greater resistance.

 

If both wires are of the same length but the second has double the cross-sectional area of the first, then how will the currents compare?

Since lengths are the same, the electric fields will be the same and therefore the drift average velocities will be the same.

Since the second wire has double the cross-sectional area it has twice as many charge carriers available per unit length.  It follows that, drift velocities being identical, the second wire will carry twice the charge of the first.  

How does the resistance of the second wire compare to that of the first?  How therefore does cross-sectional area affect resistance?

The second wire carries more current for the given voltage so it has the lesser resistance.

All other things being equal, more c.s. area implies less resistance.  

For a given material what proportionality relates resistance to cross-sectional area and length?

Resistance appears to be proportional to length and inversely proportional to c.s. area.

This is written as

R = constant * L / A, where ‘constant’ is a proportionality constant.

We generally call the proportionality constant ‘resistivity’ and denote it by the Greek letter rho.  Our equation is therefore

·         R = rho * L / A.

rho is a property of the material, and in general changes with temperature.

If a wire runs for a ways, narrows seriously for a ways, then widens back to its original c.s. area for a ways, what will happen to the electron flow if a constant voltage is applied over the whole thing.

The charge carriers are pretty much like an incompressible fluid.  The flow is much faster in the narrower wire.  

What does this tell you about the electric field and about the voltages across the three segments of wire?

Drift velocity is proportional to electric field, so the field in the smaller wire is greater.

Potential difference, or voltage, is the product of electric field and length.  So if the segments are of fairly equal lengths, the voltage across the narrow wire will be much greater than the voltage across the wide wire.  

In a circuit consisting of a source, a wire lead and a flashlight bulb (in which the current passes through a very thin filament), where do you think most of the voltage drop occurs?

The wire lead is much, much thicker than the filament.  The speed of the charge carriers is therefore much, much greater in the filament.  Most of the voltage drop therefore occurs in the filament.

 

Wires that Narrow (-> series circuits)

Wires that Branch (-> parallel circuits)

Potential Energy Change around a Loop (Kirchoff's #? law)

Parallel-Plate Capacitors

Discharge of a Parallel-Plate Capacitor

Cylindrical Capacitors

 

 

 

 

 

 

 

Assignment 14:

·         `gChapter 12 Problems 42, 46, 48, 52, 53 //

·         `uUniversity Physics:  Ch. 21 Problems 23, 24, 26, 27, 30

Assignment 15:

·         `gCh. 23 Problems 4, 11, 14, 17, 20//

·         `uUniversity Physics:  Ch. 34, Problems 25, 28, 35, 40, 42//

Assignment 16:

·         `gCh. 23 Problems 23, 34, 37, 43, 52//

·         `uUniversity Physics:  Ch. 35, Problems 45, 52, 59, 62, 68, 75//

·         `gCh. 23 Problems 58, 61; Ch. 24 Problems 4, 7, 10  Submit Ch. 24 #10//

·         `uUniversity Physics:  Ch. 37, Problems 37, 40, 46, 49, 54  Submit Ch. 37 #44//

Assignment 18:

·         `gCh. 24 Problems 15, 23, 28, 31, 34    //

·         `uUniversity Physics:  Ch. 38, Problems 40, 45, 46, 49, 51//

Assignment 19:

·         `gCh. 24 Problems 40, 43, 46, 51, 57, 60    Submit Ch. 24 #43//

Assignment 25:

·         `gText Chapter 16, Problems 8, 11, 14, 18

·         `uUniversity Physics:  Chapter 22 Problems 52, 54, 57, 60, 63, 66, 67, 74, 79, 83, 84

Assignment 26: 

·         `gText Chapter 16, Problems 26, 29, 32, 35

·         `uUniversity Physics:  Chapter 23 Problems 20, 21, 25, 27, 33, 34, 40, 41

Assignment 27: 

·         `gText Chapter 17, Problems 9, 12, 16, 19, 22

·         `uUniversity Physics:  Chapter 24 Problems 49, 50, 55, 57, 58, 61, 65, 72, 75

Assignment 28:    

·         `gText Chapter 17, Problems 34, 37, 39, 45, 48

·         `uUniversity Physics:  Chapter 25 Problems 36, 37, 42, 45, 46, 49, 52

Assignment 29: 

·         `gText Chapter 18, Problems 6, 7, 9, 12, 29, 36    

·         `uUniversity Physics:  Chapter 26 Problems 44, 47, 50, 55, 60, 61 

Assignment 30: 

·         `gText Chapter 20, Problems 11, 13, 17, 21, 32    Submit #32//

·         `uUniversity Physics:  Chapter 28 Problems 41, 42, 46, 52, 57, 65, 66; Chapter 29 Problems 43, 46, 51, 56

Assignment 31:    

·         `gText Chapter 21, Problems 10, 12, 15, 22    

·         `uUniversity Physics:  Chapter 30 Problems 31, 36, 41, 46