Experimental course (Physics 901), to be

offered next semester as part of an exchange

program with the Bavarian College of

Mad Science. Prerequisites:

Physics 24, Calc. III.

Special Announcement!

Today’s agenda:Announcements and 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.

Remember to check the Physics 24 web page for useful information, handouts, and changes to syllabus.

Announcements

You can help yourself immensely by being able to say in words what each starting equation means.

For Thursday — 21: 55, 59 (explain your answers), 63, 96, 100 (sketch the electric field lines in the three regions in parts a, b, and c) (reminder: all solutions must start with OSE’s).

For Thursday — 21: 55, 59 (explain your answers), 63, 96, 100 (sketch the electric field lines in the three regions in parts a, b, and c) (reminder: all solutions must start with OSE’s).

If you need help with Physics 24:Go to the PLC.Go to the tutor, 7-9 pm Monday-Thursday, 204 Norwood (check here for updated schedule of tutors).See your recitation instructor (he/she will tell you to start by doing the above).Try the counseling center!

Announcements

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.

The Big Picture, Part II

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.”

The Big Picture, Part II

Example: electric field of a point

charge.

2

q ˆE=k rr

+

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

2

qE=k

r

I strongly recommend you start with this and do

x and y components separately:

The Big Picture, Part III

After I show you the dipole, we’ll calculate the electric field due to a collection of many charges.

http://www.amasci.com/emotor/vdg.html

Today’s agenda: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.

A Dipole

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

+ -+q -q

dThe 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.

The Electric Field of a Dipole

i2

i i

qˆE=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

qdE4 r

+ -+q -q

d

P

L

3o

qdE4 r

Caution! The above equation for E applies only to points along the perpendicular bisector of the dipole.It is not a starting equation.

Today’s agenda: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.

The Electric FieldDue to a Collection of Point Charges

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

If necessary, please download and review this lecture.

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 23 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 = r4πε r

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

iii 2

i i0 i

1 qE = E = r4πε 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

xdq

P

2 20 0

1 dq 1 dxdE = r' = r'4πε r' 4πε r'

r’r'

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

I would start a homework or test problem with this:

2

dqdE = k

r

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.

xdq

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

yP

r’r'

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

x

yP

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'

2

0 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 = r4πε r

Just start with or .20

dq1dE =4πε r

“Quiz” time (maybe for points, maybe just for practice!)

The Electric FieldDue to a Continuous Charge

Distribution(worked examples)

In today’s lecture I will work only selected portions of the examples that follow. These lecture notes are provided so that you may review those problems I don’t work in lecture.I will skip the example starting on the next slide, because I already set it up in lecture last week, and you already worked a finite line of charge problem in your homework.Nevertheless, you should study the example because it might help you with future homework or test problems.

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 = LL

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 kx 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 ˆE = dE = -k i = -k i = -k ix x x

1 1 d d L L kQˆ ˆ ˆ ˆE = -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.

Homework hint: you must provide this derivation in your solution to any problems about rings of charge (e.g.

21.59, if assigned).

P

a

dQ

r

dE

x

2

dQdE=kr

x 2

dQdE =k cosr

2 20r = x a 0xcos

r

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 Qr 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.

I’ll show you another way to do the integral in lecture (often needed for homework; e.g. a problem like 21.96, if

assigned).

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 kxx r x r

R1/ 22 20 0 0

x 0 1/ 22 20 00

x r x xE kx 2k1/ 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

1k4

sheet0

E .2

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

I’ve been Really Nice and put this on your starting equation sheet. You don’t have to derive it for your homework!

Example: calculate the electric field due to an infinite line of positive charge. I will skip ahead. Please review on your own.There are two approaches to the mathematics of this problem.One approach (the recommended one) is that of example 21.11, where an equation for the electric field for an infinite line of charge is derived. The resulting equation is not on your OSE sheet. You may not use it as a starting equation!

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; if needed in recitation, your recitation 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.

I’ll set this up at the blackboard. A solution using an alternative mathematical approach is posted here.

2d tan sec d

If you were assigned homework 21.52,* you could do it like the book does it, or the way I do it in the link above. If you do the problem the way it is done in the link above, your recitation instructor would give you these trig facts, which I don’t expect you to have memorized:

1seccos

*Not assigned this semester. Sorry.

Today’s agenda: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.

Electric Field LinesElectric 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. Easier to use this link!

http://www.its.caltech.edu/~phys1/java/phys1/EField/EField.html This applet has some issues with calculating the correct number of field lines, but the “idea” is OK.

Today’s agenda: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.

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 (this equation gives the magnitude only).

3o

qdE .4 r

The electric field depends on the product qd.

Caution! This is not the general expression for the electric field of a dipole!

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 -qp

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

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

Noooooooo! No torques!

Apologies to the ladies in the audience: I do not mean to imply that you are any less competent to deal with torques than the males are.

There are worse things on this earth than torques…

from http://www.nearingzero.net (nz262.jpg)

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

d sin d sinqE qE qdE sin ,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.

What is the maximum torque?

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 , which you learned in Physics 23.zW d

Does that awaken vague memories of Physics 23?

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 23?

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 as

dipoleU 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!