Physics 2112 Unit 2: Electric Fields

Today’s Concepts:A) The Electric FieldB) Continuous Charge Distributions

Unit 2, Slide 1

Fields

Unit 2, Slide 2

12212

21 r̂rqkq

MATH:

12F

12212

1

2

12 r̂rkq

qF

2E

q1

2

What if I remove q2? Is there anything at that point in space?

q1q2

F12

If there are more than two charges present, the total force on any given charge is just the vector sum of the forces due to each of the other charges:

+q1 -> -q1 direction reversed

Vector Field

Unit 2, Slide 3

F2,1

F3,1

F4,1

F1

q1

q2

q3

q4

F2,1

F3,1

F4,1

F1

F2,1

F3,1

F4,1

F1

F2,1

F3,1

F4,1

F1

q1

q2

q3

q4

14214

41132

13

31122

12

21 ˆˆˆ rrqkqr

rqkqr

rqkq

++

MATH:

1F

14214

4132

13

3122

12

2

1

1 ˆˆˆ rrkqr

rkqr

rkq

qF

++

E

The electric field E at a point in space is simply the force per unit charge at that point.

Electric field due to a point charged particle

Superposition

E2

E3

E4

EField points toward negative andAway from positive charges.

Field point in the direction of the force on a positive charge.

“What exactly does the electric field that we calculate mean/represent? ““What is the essence of an electric field? “

Electric Field

Unit 2, Slide 4

q4

q2

q3

qFE

rrQkE ˆ2

i

ii

i rrQkE ˆ2

Two equal, but opposite charges are placed on the x axis. The positive charge is placed to the left of the origin and the negative charge is placed to the right, as shown in the figure above.

What is the direction of the electric field at point A?

A. UpB. DownC. Left D. RightE. Zero

CheckPoint: Electric Fields1

Unit 2, Slide 5

A

Bx+Q -Q

Two equal, but opposite charges are placed on the x axis. The positive charge is placed to the left of the origin and the negative charge is placed to the right, as shown in the figure above.

What is the direction of the electric field at point B?

A. UpB. DownC. Left D. RightE. Zero

CheckPoint: Electric Fields2

Electricity & Magnetism Lecture 2, Slide 6

A

Bx+Q -Q

Example 2.1 (Field from three charges)

Electricity & Magnetism Lecture 2, Slide 7

Calculate E at point P. d

P

d

-q +q

+q1

23

CheckPoint: Magnitude of Field (2 Charges)

Electricity & Magnetism Lecture 2, Slide 8

In which of the two cases shown below is the magnitude of the electric field at the point labeled A the largest?

A. Case 1B. Case 2C. Equal

+Q

+Q

+Q

-Q

AA

Case 1 Case 2

_ _+ +

_+

Electricity & Magnetism Lecture 2, Slide 9

CheckPoint Results: Motion of Test Charge

Electricity & Magnetism Lecture 2, Slide 10

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2Q. How will the test charge move immediately after being released?

A. To the leftB. To the rightC. Stay stillD. Other

xq1 Q2

(0.4m,0)

A charge of q1 = +4uC is placed at the origin and another charge Q2 = +10uC is placed 0.4m away. At what point on the line connected the two charges is the electric field zero?

Example 2.2 (Zero Electric Field)

Unit 2, Slide 11

(0,0)

l Q/L

Summation becomes an integral (be careful with vector nature)

“I don't understand the whole dq thing and lambda.”

WHAT DOES THIS MEAN ?

Integrate over all charges (dq)

r is vector from dq to the point at which E is defined

r

dE

Continuous Charge Distributions

Electricity & Magnetism Lecture 2, Slide 12

Linear Example:

charges

pt for E

dq l dx

i

ii

i rrQkE ˆ2

r

rdqkE ˆ2

Clicker Question: Charge Density

Electricity & Magnetism Lecture 2, Slide 13

Linear (l Q/L) Coulombs/meterSurface (s Q/A) Coulombs/meter2

Volume (r Q/V) Coulombs/meter3

What has more net charge?. A) A sphere w/ radius 2 meters and volume charge density r = 2 C/m3

B) A sphere w/ radius 2 meters and surface charge density s = 2 C/m2

C) Both A) and B) have the same net charge.

“I would like to know more about the charge density.”

Some Geometry

24 RAsphere

334 RVsphere LRVcylinder

2

RLAcylinder 2

334 RVQA rr

24 RAQB ss R

RR

QQ

B

A

sr

sr

31

4 2

334

“Please go over infinite line charge.”

Example 2.3 (line of charge)

Electricity & Magnetism Lecture 2, Slide 14

Charge is uniformly distributed along the x-axis from the origin to x a. The charge density is l C/m. What is the x-component of the electric field at point P: (x,y) (0,h)?

Let’s do one slightly different.

y

a

h

P

x

r

dq l dx

Clicker Question: Calculation

Electricity & Magnetism Lecture 2, Slide 15

2xdxA) B) C) D) E)

What is ? 2rdq

22 hadx+ 22 ha

dx+l

22 hxdx+l

2xdxl

Charge is uniformly distributed along the x-axis from the origin to x a. The charge density is l C/m. What is the x-component of the electric field at point P: (x,y) (0,h)?

x

y

a

h

P

x

r

dq l dx

We know:

rrdqkE ˆ2

Clicker Question: Calculation

Electricity & Magnetism Lecture 2, Slide 16

We know:

222 hxdx

rdq

+

l xx dEE

What is ?

A) B) C) D)

xdE

1cosdE 2cosdE 1sindE 2sindE

Charge is uniformly distributed along the x-axis from the origin to x a. The charge density is l C/m. What is the x-component of the electric field at point P: (x,y) (0,h)?

rrdqkE ˆ2

xa

P

x

r

1

2

dq l dx

xdE

dEy

We know:

sin2 DEPENDS ON x!

Clicker Question: Calculation

Electricity & Magnetism Lecture 2, Slide 17

2sindEdEE xx222 hxdx

rdq

+

l

What is ?

A) B)

C) neither of the above

xE

- + 222sinhx

dxk l +

a

hxdxak

0222sinl

Charge is uniformly distributed along the x-axis from the origin to x a. The charge density is l C/m. What is the x-component of the electric field at point P: (x,y) (0,h)?

rrdqkE ˆ2

x

a

P

x

r

1

2

dq l dx

xdE

dEy

We know:

Clicker Question: Calculation

Electricity & Magnetism Lecture 2, Slide 18

2sindEdEE xx222 hxdx

rdq

+

l

Charge is uniformly distributed along the x-axis from the origin to x a. The charge density is l C/m. What is the x-component of the electric field at point P: (x,y) (a,h)?

rrdqkE ˆ2

What is x in terms of Q ?

A) B) C) 22 ha

a

+ 22)( hxa

a

+- D)x = h*tanQ2 x = h*cosQ2 x = h*sinQ2 x = h / cosQ2

xa

P

x

r

1

2

dq l dx

xdE

dEy

We know:

xdE

Calculation

Electricity & Magnetism Lecture 2, Slide 19

222 hxdx

rdq

+

l 2cosdEdEE xx

What is ?

( )22

2

0 )tan*(sin*sec*)(hh

hdkPEf

x +

l

)(PEx

-

+ 1)(

22 ahhkPEx l

rrdqkE ˆ2

Charge is uniformly distributed along the x-axis from the origin to x a. The charge density is l C/m. What is the x-component of the electric field at point P: (x,y) (a,h)?

x = h*tanQ

xa

P

x

r

1

2

dq l dx

xdE

dEy

dx = h*sec2Q d

Exercise for student:

Change variables: write Q in terms of x

Result: do integral with trig sub

Observation

Electricity & Magnetism Lecture 2, Slide 20

xa

P

x

r

1

2

dq l dx

xdE

dEy

+1

22 ah

h

-

+ 1)(

22 ah

hkPEx l

since

Ex < 0

make since?

We had:

xdE

Back to the pre-lecture

Electricity & Magnetism Lecture 2, Slide 21

( )22

2

0 )tan*(sin*sec*)(hh

hdkPEf

x +

l

“Please go over infinte line charge. How does R get outside the intergral?”

xa

P

x

r

1

2

dq l dx

xdE

dEy

How would this integral change if the line of charge were infinite in both directions?

hdkf l

sin

0

A) The limits would be Q1 to –Q1

B) The limits would be + to -

C) The limits would be -/2 to /2

D) sinQ would turn to tanQ

E) sinQ would turn to cosQ8 8

For an infinite line of charge, we had:

xdE

Back to the pre-lecture

Electricity & Magnetism Lecture 2, Slide 22

)(PEx

xa

P

x

r

1

2

2

dq l dx

xdE

dEy

How would this integral change we wanted the y component instead of the x component?

hdk l

sin2/

2/

-

A) The limits would be Q1 to –Q1

B) The limits would be +/- infinity

C) The limits would be -/2 to /2

D) sinQ would turn to tanQ

E) sinQ would turn to cosQ

k in terms of fundamental constants

Unit 2, Slide 23

Note

o

k41

rk

rE

oline

ll 2

2

CheckPoint: Two Lines of Charge

Electricity & Magnetism Lecture 2, Slide 24

Two infinite lines of charge are shown below.

Both lines have identical charge densities +λ C/m. Point A is equidistant from both lines and Point B is located a above the top line as shown. How does EA, the magnitude of the electric field at point A, compare to EB, the magnitude of the electric field at point B?

A. EA < EB

B. EA = EB

C. EA > EB

Example 2.4 (E-field above a ring of charge)

Unit 2, Slide 25

a

h

P

What is the electric field a distance h above the center of ring of uniform charge Q and radius a?

y

x

Example 2.5 (E-field above a disk)

Unit 2, Slide 26

a

h

P

What is the electric field a distance h above the center of disk of uniform charge Q and radius a?

y

x

dipoles

Unit 2, Slide 27

q qd

- +

Dipole moment = p = qd

Points from negative to positive.(opposite the electric field.)

q qCl Na

Torque on dipole

Unit 2, Slide 28

q

q

-+ t = 2*(qE X d/2)

= p X E

dU= -dW = tdQ = pE*sinQ*dQDU -pEcosQ

Define U = when Q = /2

U -p E

Q

Example 2.6 (Salt Dipole)

Unit 2, Slide 29

-+

The two atoms in a salt (NaCl) molecule are separated by about 500pm. The molecule is placed in an electric field of strength 10N/C at an angle of 20o.

What is the torque on the molecule?What is the potential energy of the molecule?

20o