Welcome… …to Physics 104.. Prologue Some things you will recall from “last”* semester… *or whenever you took your previous physics class Newton’s Laws

Welcome…

…to Physics 104.

Prologue

Some things you will recall from “last”* semester…

*or whenever you took your previous physics class

Newton’s Laws

F ma

energy and its conservation

21KE mv

2 gravU mgy

f i other i fE E W

E K U

2ospring xxk

2

1U

momentum and its conservation (linear and angular)

p mv

z zL I

f iP P��

z,f z,iL L

These “things” aren’t going to go away!

like charges repel

unlike charges attract

charges can move but charge is conserved

“Law” of conservation of charge: the net amount of electric charge produced in any process is zero.

There are two kinds of charge. + -

Electric Charge

Static Electricity

Properties of charges

Coulomb’s Law

1 22

12

q qF k12 r

Coulomb’s “Law” gives the force (in Newtons) between charges q1 and q2, where r12 is the distance in meters between the

charges, and k=9x109 N·m2/C2.

Coulomb’s “Law” quantifies the magnitude of the electrostatic force.

1 22

12

q qF k ,12 r

attractive for unlike repulsive for like

Force is a vector quantity. The equation on the previous slide gives the magnitude of the force. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/40,

where 0=8.85x10-12 C2/N·m2.

Remember, a vector has a magnitude and a direction.

I could write Coulomb’s “Law” like this…

The equation is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r12 is the distance between the centers of the spheres.

If more than one charge is involved, the net force is the vector sum of all forces (superposition). For objects with complex shapes, you must add up all the forces acting on each separate charge (turns into calculus!).

+ -

r12

++

+-

--

We could have agreed that in the formula for F, the symbols q1

and q2 stand for the magnitudes of the charges. In that case,

the absolute value signs would be unnecessary.

1 22

12

q qF k ,12 r

However, in later equations the sign of the charge will be important, so we really need to keep the magnitude part.

On your diagrams, show both the magnitudes and signs of q1

and q2.

Your starting equation is this version of the equation:

which gives you the magnitude F12 and tells you that you need to figure out the direction separately.

Solving Problems Involving Coulomb’s “Law” and Vectors

You may wish to review vectors (on your own).

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

Example: Calculate the net electrostatic force on charge Q3 due to the charges Q1 and Q2.

Step 0: Think!

This is a Coulomb’s “Law” problem (all we have to work with, so far).

We only want the forces on Q3. Don’t worry about other forces.

Forces are additive, so we can calculate F32 and F31 and add the two.

If we do our vector addition using components, we must resolve our forces into their x- and y-components.

Draw and label forces (only those on Q3).

Draw components of forces which are not along axes.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

F31

F32Draw a representative sketch—done.

Draw and label relevant quantities—done.

Draw axes, showing origin and directions—done.

Step 1: Diagram

1 22

12

q qF k12 r

<complaining> “Do I have to put in the absolute value signs?”

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

F31

F32

Yes. Unless you like losing points.

Step 2: Starting Equation

3 2232

Q QF k ,32 r

repulsive

3 2232

Q QF k32, y r

F 032, x

(from diagram)

Can you put numbers in at this point? OK for this problem. You would get F32,y = 330 N and F32,x = 0 N.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

r31 =60 cm

r 32=

30

cm

=30º

F31

F32

Step 3: Replace Generic Quantities by Specifics

3 1231

Q QF k ,31 r

attractive

3 1231

Q QF k cos31, x r

Can you put numbers in at this point? OK for this problem. You would get F31,x = +120 N and F31,y = -70 N.

(- sign comes from diagram)3 1231

Q QF k sin31, y r

(+ sign comes from diagram)

x

y

Q2=+50C

Q3=+65C

Q1=-86C

=30º

F31

F32

Step 3 (continued)

r 32=

30

cm r

31 =60 cm

52 cm

F3x = F31,x + F32,x = 120 N + 0 N = 120 N

F3y = F31,y + F32,y = -70 N + 330 N = 260 N

You know how to calculate the magnitude F3 and the angle between F3 and the x-axis. (If not, holler!)

F3The net force is the vector sum of all the forces on Q3.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

F31

F32

Step 3: Complete the Math

I did a sample Coulomb’s law calculation using three point charges.

How do you apply Coulomb’s law to objects that contain distributions of charges?

We’ll use another tool to do that…

Today’s agendum:

The electric field.You must be able to calculate the force on a charged particle in an electric field.

Electric field of due to a point charge.You must be able to calculate electric field of a point charge.

Motion of a charged particle in a uniform electric field.You must be able to solve for the trajectory of a charged particle in a uniform electric field.

The electric field due to a collection of point charges.You must be able to calculate electric field of a collection of point charges.

The electric field due to a continuous line of charge.You must be able to calculate electric field of a continuous line of charge.

Coulomb's “Law” quantifies the interaction between charged particles.

+ -

Coulomb’s “Law”:The Big Picture

Q1 Q2

r121 2

20 12

q q1F = ,12 4πε r

Coulomb’s “Law” was discovered through decades of experiment.

By itself, it is just “useful." Is it part of something bigger?

You experienced gravitational fields in Physics 103.

Gravitational Fields

1 22

12

m mF =G , attractiveG r

GFg(r) =

m

is the local gravitational field. On earth, it is 9.8 N/kg, directed towards the center of the earth. What we calledg = 9.8 m/sec2 is the magnitude of the gravitational field.

g(r)

Units of g are actually N/kg!

The Electric Field

Coulomb's “Law” (demonstrated in 1785) shows that charged particles exert forces on each other over great distances.How does a charged particle "know" another one is “there?”

Electric, magnetic, and gravitational forces result of direct and instantaneous interaction between particles.

Action At A Distance Viewpoint

Relativity theory shows why this viewpoint is wrong. Faraday developed the correct explanation.

Faraday, beginning in 1830's, was the leader in developing the idea of the electric field. Here's the idea:

A charged particle emanates a "field" into all space.

Another charged particle senses the field, and “knows” that the first one is there.

+

+-

like charges

repel


F12

F21

F31

F13

The idea of an electric field is good for a number of reasons:

It makes us feel good, like we’ve actually explained something.

We can develop a theory based on this idea. From this theory may spring unimagined inventions.

+

+-

like charges

repel


F12

F21

F31

F13

OK, that was a flippant remark. There are serious reasons why the idea is “good.”

If the theory explains past observations and leads to new predictions, the idea was “good.”

We define the electric field by the force it exerts on a test charge q0:

0

0

FE =

q

This is your second starting equation. By convention the direction of the electric field is the direction of the force exerted on a POSITIVE test charge. The absence of absolute value signs around q0 means you must include the sign of q0 in your work.

If the test charge is "too big" it perturbs the electric field, so the “correct” definition is

0

0

q 00

FE = lim

q

The subscript “0” reminds you the force is on the “test charge.” I won’t require the subscripts when you use this equation for boardwork or on exams.

Any time you know the electric field, you can use this equation to calculate the force on a charged particle in that electric field.

You won’t be required to use this version of the equation.

The units of electric field are newtons/coulomb.

0

0

F NE = =

q C

Subsequently, you will learn that the units of electric field can also be expressed as volts/meter:

N VE = =

C m

The electric field exists independent of whether there is a charged particle around to “feel” it.

+Remember: the electric field direction is the direction a + charge would feel a force.

A + charge would be repelled by another + charge.

Therefore the direction of the electric field is away from positive (and towards negative).

http://regentsprep.org/Regents/physics/phys03/afieldint/default.htm

http://regentsprep.org/Regents/physics/phys03/afieldint/default.htm

Today’s agendum:






The Electric FieldDue to a Point Charge

Coulomb's law says

... which tells us the electric field due to a point charge q is

1 22

12

q qF =k ,12 r

This is your third starting equation.

q 2

qE =k , away from +

r

…or just…

2

qE=k

r

The equation for the electric field of a point charge then becomes:

2

qÊ=k r

r

We define as a unit vector from the source point to the field point:

r̂

+source point

field point

r̂

You may start with either equation for the electric field (this one or the one on the previous slide).

Example: calculate the electric field at the electron’s distance away from the proton in a hydrogen atom (5.3x10-11 m).To be worked at the blackboard.

For comparison, air begins to break down and conduct electricity at about 30 kV/cm, or 3x106 V/m.

Today’s agendum:






Motion of a Charged Particlein a Uniform Electric Field

A charged particle in an electric field experiences a force, and if it is free to move, an acceleration.

- - - - - - - - - - - - -

+ + + + + + + + + + + + +

-F

If the only force is due to the electric field, then

F ma qE.

If E is constant, then a is constant, and you can use the equations of kinematics* (remember way back to the beginning of Physics 103?).

E

Example: a proton and an electron enter a region of uniform electric field. Describe their motion.

Direction of forces?

Magnitudes of accelerations?

Shape of trajectories?

Example: an electron moving with velocity v0 in the positive x direction enters a region of uniform electric field that makes a right angle with the electron’s initial velocity. Express the position and velocity of the electron as a function of time.

- - - - - - - - - - - - -

+ + + + + + + + + + + + +

E

x

y

-

v0

To be worked at the blackboard.

Today’s agendum:






Example: calculate the electric field at position P due to the two protons shown.

To be worked at the blackboard.

+ +

DD

P

Example: field of an electric dipole.

No time to work today. Will work in Lecture 3. Study example 21.9 in your text.

Today’s agendum:






Electric Field Due To A Line of Charge

Think of a line of charge as a collection of very tiny point charges all lined up. The “official” term for “as tiny as you can imagine” is “infinitesimal.”

We get the electric field for the line of charge by adding the electric fields for all the infinitesimal point charges.

These words are meant to remind you of things you have learned in calculus.

+ + + + + + + + + + + + + + + +

If charge is distributed along a straight line segment parallel to the x-axis, the amount of charge dq on a segment of length dx is dx.

is the linear density of charge (amount of charge per unit length). may be a function of position.

Think length. times the length of line segment is the total charge on the line segment.

x

dx

dx

Consider charge uniformly distributed along a line (e.g., electrons on a thread).

The electric field at point P due to the charge dq is

x

dq

P

2 20 0

1 dq 1 dxdE = r' = r'

4πε r' 4πε r'

r’r'

dE

I’m assuming positively charged objects in these “distribution of charges” slides.

The electric field at P due to the entire line of charge is

20

1 λ(x) dxE = r' .

4πε r'

The integration is carried out over the entire length of the line, which need not be straight. Also, could be a function of position, and can be taken outside the integral only if the charge distribution is uniform.

x

dq

P

r’r'

E

Example: calculate the electric field due to an infinite line of positive charge.

There are two approaches to the mathematics of this problem.One approach is that of example 21.11, where an equation for the electric field for an infinite line of charge is derived.

If you use the text’s approach, you must evaluate this indefinite integral, which is in appendix B (page A4) of your text:

3

2 2 2

dx

x a

Thus, if this were given for homework, you would need to repeat this derivation!

If you need it, you can look this integral up; your instructor will give it to or let you look it up in the text.

Example: calculate the electric field due to an infinite line of positive charge.

Today’s agendum:

Review.

The electric field of a dipole.You must be able to calculate the electric field of a dipole.

The electric field due to a collection of point charges (continued).You must be able to calculate the electric field of a collection of point charges.

Electric field lines.You must be able to draw electric field lines, and interpret diagrams that show electric field lines.

A dipole in an external electric field.You must be able to calculate the moment of an electric dipole, the torque on a dipole in an external electric field, and the energy of a dipole in an external electric field.

In Lecture 1 you learned Coulomb's Law:

+ -

The Big Picture, Part I

Q1 Q2

r121 2

20 12

q q1F = ,12 4πε r

Coulomb’s Law quantifies the force between charged particles.

In Lecture 2 you learned about the electric field.

The Big Picture, Part II

There were two kinds of problems you had to solve:

1. Given an electric field, calculate the force on a charged particle.

2. Given one or more charged particles, calculate the electric field they produce.


FE =

q

1. Given an electric field, calculate the force on a charged particle.

-F E

F= qE

You may not be given any information about where this electric field “comes

from.”


Example: electric field of a point

charge.

2

qÊ=k r

r

+

source point

field point

r̂

2. Given one or more charged particles, calculate the electric field they produce. 2 slides from now we’ll do this for a dipole.

E

Today’s agendum:

Review.





A Dipole

A combination of two electric charges with equal magnitude and opposite sign is called a dipole.

+ -+q -q

d

The charge on this dipole is q (not zero, not +q, not –q, not 2q). The distance between the charges is d. Dipoles are “everywhere” in nature.

This is an electric dipole. Later in the course we’ll study magnetic dipoles, which, as you might guess, have a north and a south magnetic pole.

The Electric Field of a Dipole

i2

i i

qÊ=k r

r

Example: calculate the electric field at point P, which lies on the perpendicular bisector a distance L from a dipole of charge q.

+ -+q -q

d

P

Lto be worked at the blackboard

3o

qdE

4 r

+ -+q -q

d

P

L

3o

qdE

4 r

Caution! The above equation for E applies only to points along the perpendicular bisector of the dipole.

Today’s agendum:

Review.





The Electric FieldDue to a Collection of Point Charges

Hang on to your seats, because the next 11 slides will go by very fast!

In today's lecture I show how to calculate the electric field due to a distribution of charges.

Matter is made of discrete atoms, but appears "continuous" to us, and in Physics 103 we treated matter as being a continuous entity.

Similarly, a charge distribution is made of individual charged particles, but we can treat it as if the charge were continuous.

The electric field due to a small "chunk" q of charge is

20

1 qE = r

4πε r

The electric field due to collection of "chunks" of charge is

iii 2

i i0 i

1 qE = E = r

4πε r

unit vector from q to wherever you want to calculate E

unit vector from qi to wherever you want to calculate E

As qdq0, the sum becomes an integral.

If charge is distributed along a straight line segment parallel to the x-axis, the amount of charge dq on a segment of length dx is dx.

is the linear density of charge (amount of charge per unit length). may be a function of position.

Think length. times the length of line segment is the total charge on the line segment.

x

dx

dx

The electric field at point P due to the charge dq is

x

dq

P

2 20 0

1 dq 1 dxdE = r' = r'

4πε r' 4πε r'

r’r'

dE

I’m assuming positively charged objects in these “distribution of charges” slides.

The electric field at P due to the entire line of charge is

20

1 λ(x) dxE = r' .

4πε r'

The integration is carried out over the entire length of the line, which need not be straight. Also, could be a function of position, and can be taken outside the integral only if the charge distribution is uniform.

x

dq

P

r’r'

E

If charge is distributed over a two-dimensional surface, the amount of charge dq on an infinitesimal piece of the surface is dS, where is the surface density of charge (amount of charge per unit area).

x

y

area = dS

charge dq = dS

dE

The electric field at P due to the charge dq is

2 20 0

1 dq 1 dSdE = r' = r'

4πε r' 4πε r'

x

y

P

r’r'

The net electric field at P due to the entire surface of charge is

x

y

P

r’r'

20 S

1 (x, y) dSE = r'

4πε r'

E

x

zP

r’r'

20 V

1 (x, y, z) dVE = r' .

4πε r'

After you have seen the above, I hope you believe that the net electric field at P due to a three-dimensional distribution of charge is…

y

E

Summarizing:

20

1 λ dxE = r' .

4πε r'

20 S

1 dSE = r' .

4πε r'

20 V

1 dVE = r' .

4πε r'

Charge distributed along a line:

Charge distributed over a surface:

Charge distributed inside a volume:

If the charge distribution is uniform, then , , and can be taken outside the integrals.

20

1 λ dxE = r'

4πε r'

20 S

1 dSE = r'

4πε r'

2

0 V

1 dVE = r'

4πε r'

“Hold it right there! These equations…

…are not on my starting equation sheet. That’s not fair!”

20

1 qE = r

4πε r

Just start with or .2

0

dq1dE =

4πε r

The Electric FieldDue to a Continuous Charge

Distribution(worked examples)

Example: A rod of length L has a uniform charge per unit length and a total charge Q. Calculate the electric field at a point P along the axis of the rod at a distance d from one end.

P x

y

d L

Let’s put the origin at P. The linear charge density and Q are related by

Q = and Q = L

L

Let’s assume Q is positive.

P x

y

d L

The electric field points away from the rod. By symmetry, the electric field on the axis of the rod has no y-component. dE from the charge on an infinitesimal length dx of rod is

dE x dx dQ = dx

2 2

dq dxdE = k k

x x

Note: dE is in the –x direction. dE is the magnitude of dE. I’ve used the fact that Q>0 (so dq=0) to eliminate the absolute value signs in the starting equation.

P x

y

d L

dE x dx dQ = dx

d Ld+L d+L d+L

x 2 2d d dd

dx dx 1ˆ ˆ Ê = dE = -k i = -k i = -k ix x x

1 1 d d L L kQˆ ˆ ˆ Ê = -k i = -k i= -k i= - i

d L d d d L d d L d d L

Example: A ring of radius a has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at a point P along the axis of the ring at a distance x0 from its center.

P

a

dQ

r

x0

dE

x

By symmetry, the y- and z-components of E are zero, and all points on the ring are a distance r from point P.

P

a

dQ

r

dE

x

2

dQdE=k

r

x 2

dQdE =k cos

r

2 20r = x a 0x

cosr

x0

0 0 0 0

x x 3/ 22 3 3 2 2ring ring ring 0

x x x kx QdQE dE k k dQ k Q

r r r r x a

Or, in general, on the ring axis x,ring 3/ 22 2

kxQE .

x a

For a given x0, r is a constant for points on the ring.

No absolute value signs because Q is positive.

Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axis of the disc at a distance x0 from its center.

P

r

dQ

xx0R

The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr).

We can use the equation on the previous slide for the electric field due to a ring, replace a by r, and integrate from r=0 to r=R.

0

ring 3/ 22 20

kx 2 rdrdE .

x r

Caution! I’ve switched the “meaning” of r!

P

r

dQ

xx0R

R

0x x 03/ 2 3/ 202 2 2 2

disc disc 0 0

kx 2 rdr 2r drE dE kx

x r x r

R1/ 22 20 0 0

x 0 1/ 22 20 0

0

x r x xE kx 2k

1/ 2 x x R

Example: Calculate the electric field at a distance x0 from an infinite plane sheet with a uniform charge density .Treat the infinite sheet as disc of infinite radius.

Let R and use to get0

1k

4

sheet0

E .2

Interesting...does not depend on distance from the sheet.

Today’s agendum:

Review.





Electric Field Lines

Electric field lines help us visualize the electric field and predict how charged particles would respond the field.

Example: electric field lines for isolated +2e and -e charges.

+-

Here’s how electric field lines are related to the field:

The electric field vector E is tangent to the field lines. The number of lines per unit area through a surface perpendicular to the lines is proportional to the electric field strength in that region

The field lines begin on positive charges and end on negative charges.

The number of lines leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.

No two field lines can cross.

Example: draw the electric field lines for charges +2e and -1e, separated by a fixed distance.

Today’s agendum:

Review.





Electric Dipole in anExternal Electric Field

An electric dipole consists of two charges +q and -q, equal in magnitude but opposite in sign, separated by a fixed distance d. q is the “charge on the dipole.”

Earlier, I calculated the electric field along the perpendicular bisector of a dipole.

3o

qdE .

4 r

The electric field depends on the product qd.

q and d are parameters that specify the dipole; we define the "dipole moment" of a dipole to be the vector

p qd,

where the direction of p is from negative to positive (NOT away from +).

+q -q

p

caution: this p is not momentum!

To help you remember the direction of p, this is on the equation sheet:

p q d, from to plus

E

A dipole in a uniform electric field experiences no net force, but probably experiences a torque…

+q

-q

pF+

F-

F F F qE qE 0.

There is no net force on the dipole:

E+q

-q

pF+

F-

½ d sin

If we choose the midpoint of the dipole as the origin for calculating the torque, we find

dsin dsinqE qE qdEsin ,

2 2

and in this case the direction is into the plane of the figure. Expressed as a vector,

p E.

½ d sin

Recall that the unit of torque is N·m, which is not a joule!

E+q

-q

pF+

F-

½ d sin½ d sin

The torque’s magnitude is p E sin and the direction is given by the right-hand rule.

Energy of an Electric Dipole in anExternal Electric Field

If the dipole is free to rotate, the electric field does work* to rotate the dipole.

E+q

-q

pF+

F-

initial finalW pE(cos cos ).

The work depends only on the initial and final coordinates, and not on how you go from initial to final.

*Calculated using , using torque times angular displacement.zW d

Does that awaken vague memories of Physics 103?

If a force is conservative, you can define a potential energy associated with it.

What kinds of potential energies did you learn about in Physics 103?

Because the electric force is conservative, we can define a potential energy for a dipole. The equation for work

initial finalW pE(cos cos )

suggests we should define

dipoleU pE cos .

E+q

-q

pF+

F-

dipoleU pE cos

With the definition on the previous slide, U is zero when =/2.U is maximum when cos=-1, or = (a point of unstable equilibrium).

U is minimum when cos=+1, or =0 (stable equilibrium).

It is “better” to express the dipole potential energy asdipoleU p E.

Recall that the unit of energy is the joule, which is a N·m, but is not the same as the N·m of torque!

Today’s agendum:

Electric flux.You must be able to calculate the electric flux through a surface.

Gauss’ “Law.”You must be able to use Gauss’ “Law” to calculate the electric field of a high-symmetry charge distribution.

Conductors in electrostatic equilibrium.You must be able to use Gauss’ “Law” to draw conclusions about the behavior of charged particles on, and electric fields in, conductors in electrostatic equilibrium.

Gauss’ “Law”

Electric Flux

We have used electric field lines to visualize electric fields and indicate their strength.

We are now going to count* the number of electric field lines passing through a surface, and use this count to determine the electric field.

E

*There are 3 kinds of people in this world: those who can count, and those who can’t.

The electric flux passing through a surface is the number of electric field lines that pass through it.

Because electric field lines are drawn arbitrarily, we quantify electric flux like this: E=EA.

If the surface is tilted, fewer lines cut the surface.

EA

Later we’ll learn about magnetic flux, which is why I will use the subscript E on electric flux.

E

E

A

The “amount of surface” perpendicular to the electric field is A cos .

A = A cos so E = EA = EA cos .

We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface.

Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos .

Remember the dot product from Physics 103?

E E A

If the electric field is not uniform, or the surface is not flat…

divide the surface into infinitesimal surface elements and add the flux through each…

dAE

iE i i

A 0i

lim E A

E E dA

A

If the surface is closed (completely encloses a volume)…

E

…we count* lines going out as positive and lines going in as negative…

dAa surface integral, therefore a double integral

*There are 10 kinds of people in this world: those who can count in binary, and those who can’t.

AdEE

Nothing to panic about!

The circle just reminds you to integrate over a closed surface.

What is this thing?

Question: you gave me five different equations for electric flux. Which one do I need to use?

E E dA

E E A

E EA cos

E EA

Answer: use the simplest (easiest!) one that works.

Flat surface, E A, E constant over surface. Easy!

Flat surface, E not A, E constant over surface.

Flat surface, E not A, E constant over surface.

Surface not flat, E not uniform.

Closed surface. Most general. Most complex.

If the surface is closed, you may be able to “break it up” into simple segments and still use E=E·A for each segment.

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Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.

E

To be worked at the blackboard…

If there were a + charge inside the cylinder, there would be more lines going out than in.

If there were a - charge inside the cylinder, there would be more lines going in than out…

…which leads us to…

E+-

Today’s agendum:




Gauss’ “Law”

Mathematically*, we express the idea two slides back as

Gauss’ “Law”

We will find that Gauss’ “Law” gives a simple way to calculate electric fields for charge distributions that exhibit a high degree of symmetry…

…and save more complex and realistic charge distributions for advanced classes.

*“Mathematics is the Queen of the Sciences.”—Karl Gauss

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To see how this works, let’s do an example.

Example: use Gauss’ “Law” to calculate the electric field from an isolated point charge q.

To apply Gauss’ “Law”, we construct a “Gaussian Surface” enclosing the charge.

The Gaussian surface should mimic the symmetry of the charge distribution.

For this example, choose for our Gaussian surface a sphere of radius r, with the point charge at the center.

I’ll work the rest of the example on the blackboard.

Strategy for Solving Gauss’ “Law” Problems

Select a Gaussian surface with symmetry that matches the charge distribution.

Draw the Gaussian surface so that the electric field is either constant or zero at all points on the Gaussian surface.

Evaluate the surface integral (electric flux).

Determine the charge inside the Gaussian surface.

Solve for E.

Use symmetry to determine the direction of E on the Gaussian surface.

Example: calculate the electric field outside a long cylinder of finite radius R with a uniform volume charge density spread throughout the volume of the cylinder.

The cylinder being “long” and the radius “finite” are “code words” that tell you to neglect end effects from the cylinder (i.e., assume it is infinitely long).

Know how to interpret “code words” when exam time comes!Let’s go through this a step at a time (work to be shown at board).

Select a Gaussian surface with symmetry that matches the charge distribution.

Pick a cylinder of length L and radius r, concentric with the cylinder of the problem.

Draw the Gaussian surface so that the electric field is either constant or zero at all points on the Gaussian surface.Already done!

Use symmetry to determine the direction of E on the Gaussian surface.

Electric field points radially away from cylinder, and magnitude does not depend on direction.

Evaluate the surface integral (electric flux).

Surface integral is just E times the curved area.

Determine the charge inside the Gaussian surface.

The charge inside is just the volume of charged cylinder inside the Gaussian surface, times the charge per volume.

Solve for E.

2

0

RE

2 r

Example: use Gauss’ “Law” to calculate the electric field due to a long line of charge, with linear charge density .

Example: use Gauss’ “Law” to calculate the electric field due to an infinite sheet of charge, with surface charge density .

These are easy using Gauss’ “Law” (remember what a pain they were in the previous chapter). Study these examples and others in your text!

sheet0

E .2

line0

E .2 r

Today’s agendum:




Conductors in Electrostatic Equilibrium

Electrostatic equilibrium means there is no net motion of the charges inside the conductor.

The electric field inside must be zero.

Any excess charge must reside on the outside surface of the conductor.

If this were not the case, charges would move.

Apply Gauss’ “Law” to a Gaussian surface just inside the conductor surface. The electric field is zero, so the net charge inside the Gaussian surface is zero.

The electric field just outside a charged conductor must be perpendicular to the conductor’s surface.

Otherwise, the component of the electric field parallel to the surface would cause charges to accelerate.

The magnitude of the electric field just outside a charged conductor is equal to /0, where is the local surface charge density.

A simple application Gauss’ “Law”, which I will show if time permits.

If there is an empty nonconducting cavity inside the conductor, Gauss’ “Law” tells us there is no net charge on the interior surface of the conductor.

If there is a nonconducting cavity inside the conductor, with a net charge inside the cavity, Gauss’ “Law” tells us there is an equal and opposite induced charge on the interior surface of the conductor.

Construct a Gaussian surface that includes the inner surface of the conductor. The electric field at the Gaussian surface is zero, so no electric flux passes through the Gaussian surface. Gauss’ Law says the charge inside must be zero. The conductor does not have to be symmetric, as shown.

+QConstruct a Gaussian surface that includes the inner surface of the conductor. The electric field at the Gaussian surface is zero, so no electric flux passes through the Gaussian surface. Gauss’ Law says the charge inside must be zero.

Construct a Gaussian surface that includes the inner surface of the conductor. The electric field at the Gaussian surface is zero, so no electric flux passes through the Gaussian surface. Gauss’ Law says the charge inside must be zero. There must be a –Q on the inner surface. The conductor does not have to be symmetric, as shown.

-Q

Example: a conducting spherical shell of inner radius a and outer radius b with a net charge -Q is centered on point charge +2Q. Use Gauss’s “Law” to show that there is a net charge of -2Q on the inner surface of the shell, and a net charge of +Q on the outer surface of the shell.

a

b

-Q

+2Q

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Example (if time permits): an insulating sphere of radius a has a uniform charge density ρ and a total positive charge Q. Calculate the electric field at a point inside the sphere

a

Q

r

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Documents

Welcome… …to Physics 104.. Prologue Some things you will recall from “last”* semester… *or whenever you took your previous physics class Newton’s Laws