Physics 2112 Unit 2: Electric Fields - College of …Physics 2112 Unit 2: Electric Fields Today’s...

Physics 2112

Unit 2: Electric Fields

Today’s Concepts:

A) The Electric Field

B) Continuous Charge Distributions

Unit 2, Slide 1

Fields

Unit 2, Slide 2

122

12

21 r̂r

qkq

MATH:

12F

122

12

1

2

12 r̂r

kq

q

F

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

142

14

41132

13

31122

12

21 ˆˆˆ rr

qkqr

r

qkqr

r

qkq++

MATH:

1F

142

14

4132

13

3122

12

2

1

1 ˆˆˆ rr

kqr

r

kqr

r

kq

q

F++

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 and

Away 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

q

FE

rr

QkE ˆ

2

i

i

i

i rr

QkE ˆ

2

Electricity & Magnetism Lecture 2, Slide 5

We never talk about the “gravitational field” in 2111.

We just talked about a “gravitational force”.

Let’s say we had. What is the magnitude of the

gravitational field close to the surface of the earth?

A) g

B) mg

C) m/g

D) g/m

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. Up

B. Down

C. Left

D. Right

E. Zero

CheckPoint: Electric Fields1

Unit 2, Slide 6

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. Up

B. Down

C. Left

D. Right

E. Zero

CheckPoint: Electric Fields2


A

Bx+Q -Q

Question


P

d

What is the direction of the

electric field at point P, the

unoccupied corner of the

square? -q +q

+q

d

A)

B)

C)

D)Need to know d

Need to know

sign of charge

placed at point P

E)

0E

Example 2.1 (Field from three charges)


Calculate E at

point P. d

P

d

-q +q

+q1

2

3

CheckPoint: Magnitude of Field (2 Charges)


In which of the two cases shown below is the magnitude of the electric

field at the point labeled A the largest?

A. Case 1

B. Case 2

C. Equal

+Q

+Q

+Q

-Q

A

A

Case 1 Case 2

Two charges q1 and q2 are fixed at points (-a,0) and (a,0) as

shown. Together they produce an electric field at point (0,d) which

is directed along the negative y-axis.

x

y

q1 q2

(-a,0) (a,0)

(0,d)

Which of the following statements is true:

A) Both charges are negative

B) Both charges are positive

C) The charges are opposite

D) There is not enough information to tell how the charges are

related

Question: Two Charges

Unit 2, Slide 11

E

_ _+ +

_+


A 6uC charge is placed in

an electric field and feels a

0.18N force. If the charge

is replaced by a 9uC

charge, what force will the

new charge feel?

Question

Unit 2, Slide 13

E q1

A) 0N

B) 0.09N

C) 0.18N

D) 0.27N

E) 0.36N

CheckPoint Results: Motion of Test Charge


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 left

B. To the right

C. Stay still

D. 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 15

(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


Linear Example:

charges

pt for E

dq l dx

i

i

i

i rr

QkE ˆ

2

r

r

dqkE ˆ

2

“Please go over infinite line

charge.”

Example 2.3 (line of charge)


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.

x

y

a

h

P

x

r

dq l dx

Question: Calculation


2x

dxA) B) C) D) E)

What is ? 2r

dq

22 ha

dx

+22 ha

dx

+

l22 hx

dx

+

l2x

dxl


x

y

a

h

P

x

r

dq l dx

We know:

rr

dqkE ˆ

2



We know:

222 hx

dx

r

dq

+

l xx dEE

What is ?

A) B) C) D)

xdE

2cosdE2cosdE-

1sindE 2sindE


rr

dqkE ˆ

2

xa

P

x

r

1

2

dq l dx

xdE

dEy

2sindE-E)

h

We know:

sin2 DEPENDS ON x!



- 2sindEdEE xx222 hx

dx

r

dq

+

l

What is ?

A) B)

C) neither of the above

xE

-+

-222sin

hx

dxak l +

-

a

hx

dxk

0

222sinl


rr

dqkE ˆ

2

xa

P

x

r

2

dq l dx

xdE

dEy

We know:

Clicker Question: Calculation


2sindEdEE xx222 hx

dx

r

dq

+

l

rr

dqkE ˆ

2

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

2

dq l dx

xdE

dEy

PROBLEM: We have two variables Q and x

Observation


xa

P

x

r2

dq l dx

xdE

dEy

+1

22 ah

h

-

+ 1)(

22 ah

h

h

kPEx

l

Note:

Ex < 0

make sense?

For an infinite line of

charge, we had:

xdE

Back to the pre-lecture


)(PEx

xa

P

x

r

dq l dx

xdE

dEy

How would this integral change we wanted the y component

instead of the x component?

-

f

dh

k

l

0

*sin

A) The limits would be Q1 to –Q1

B) The limits would be +/- infinity

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

D) sinQ would turn to tanQ

E) sinQ would turn to -cosQ

We had:

xdE

Back to the pre-lecture


( )22

2

0)tan*(

)cos(*sec*)(

hh

hdkPE

f

y+

--

l

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

xa

P

x

r

dq l dx

xdE

dEy

How would this integral change if the line of charge were

infinite in both directions?

f

dh

k

l

0

*cos

A) The limits would be Q1 to –Q1

B) The limits would be + to -

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

D) sinQ would turn to tanQ

E) sinQ would turn to -cosQ8 8

k in terms of fundamental constants

Unit 2, Slide 25

Note

o

kp4

1

o

liner

Ep

l

2

r

kEline

l2

CheckPoint: Two Lines of Charge


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 27

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

Question

Unit 2, Slide 28

In the previous question,

what should the electric

field be if h>>>>a?

a) k Q / h2

b) k Q / a2

c) k Q / (a*h)

d) k Q a*h

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

Unit 2, Slide 29

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

A total charge Q is uniformly

distributed over a half ring

with radius R. The total

charge inside a small

element dθ is given by:

p dR

Q

2

A.

p dR

Q

B.

R

Q

p

C.

p dQ

D.

p dQ

2

E.

θ

dθ

Q

R

Question



with radius R. The Y

component of electric field

created by a short element

dθ is given by:

p

sin2R

kQd

A.

p

cos2R

kQd

B.

p

sin3R

kQdC.

p

cos3R

kQdD.

y

x

Question



with radius R. The total

electric field at point C is:

2R

kQC. D.

y

x

Question

2R

kQ

p

A.

2

2

R

Qk

p

B.

0

dipoles

Unit 2, Slide 33

q q

d

- +

Dipole moment = p = qd

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

q qCl Na

Water molecule has dipole moment of 620X10-30 Cm

= 2 * 1.6X10-19C * 190pm

Torque on dipole

Unit 2, Slide 34

q

q

-

+ = 2*(qE X d/2)

= p X E

dU= -dW = -dQ = pE*sinQ*dQ

DU - pEcosQ

Define U = when Q = p/2

U -p E

Q

Example 2.6 (Salt Dipole)

Unit 2, Slide 35

-

+

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+x

A. (a)

B. (b)

C. (c)

D. (d)

E. (c) and (d)

An electrically neutral dipole

is placed in an external field.

In which situation(s) does

the potential energy of the

dipole have the smallest

magnitude using the

standard definition of U?

(a) (b)

(c) (d)

+-

+ -

+- +

-

A. (a)

B. (b)

C. (c)

D. (d)

E. (c) and (d)

An electrically neutral dipole

is placed in an external field.

In which situation(s) does

the potential energy of the

dipole have the lowest value

using the standard definition

of U?

(a) (b)

(c) (d)

+-

+ -

+- +

-

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