Physics 1161 Lecture 2Vectors

&Electric Fields

Three Charges

Q=-3.5 mCQ=+7.0mC

Q=+2.0mC

6 m

4 m

• Calculate force on +2mC charge due to other two charges– Calculate force from +7mC charge

– Calculate force from –3.5mC charge

– Add (VECTORS!)

6 6

1,3 22 2

7 10 2 10

3 4

k C CF

m m

35.04 10 N

35.04 10 N

6 6

2,3 22 2

3.5 10 2 10

3 4

k C CF

m m

32.52 10 N

32.52 10 N

Three Charges

Q=-3.5 mCQ=+7.0mC

Q=+2.0mC

6 m

4 m

• Resolve each force into x and y components

• Add the x-components & the y-comp.

• Use Pyth. Theorem & Trigonometry to express in R,θ notation

3 31,3 5.04 10 cos53 3.03 10oxF N N

35.04 10 N

3 32,3 2.52 10 cos307 1.51 10oxF N N

32.52 10 N

53o

53o

3 31,3 5.04 10 sin 53 4.03 10oyF N N

3 32,3 2.52 10 sin 307 1.997 10oyF N N

3 3 33.03 10 1.51 10 4.54 10xF N N N

3 3 34.03 10 1.997 10 2.03 10yF N N N

Three Charges

• Use Pyth. Theorem & Trigonometry to express in R,θ notation

34.54 10 N

32.03 10 N

2 23 34.54 10 2.03 10F N N

34.9 10F N

φ

3

3

2.03 10arctan arctan 244.54 10

y o

x

F NF N

Since resultant is in first quadrant, θ = φ 24o

F

Electric Force on Electron by Proton

• What are the magnitude and direction of the force on the electron by the proton?

+

r = 1x10-10 m

q=1.6x10-19 C

Toward the left

22

1 22

9 19 19

210

8

9 10 1.6 10 1.6 10

10

2.30 10

N mC

kq qFr

C CF

m

F N

e-

Comparison:Electric Force vs. Electric Field

• Electric Force (F) - the actual force felt by a charge at some location.

• Electric Field (E) - found for a location only – tells what the electric force would be if a charge were located there:

F = qE• Both are vectors, with magnitude and

direction

Electric Field• Charged particles create electric fields. – Direction is the same as for the force that a + charge

would feel at that location.– Magnitude given by: E F/q

• Field at A due to proton?

+

r = 1x10-10 m

q=1.6x10-19 C

Toward the right

22

2

9 19

210

11

9 10 1.6 10

10

1.44 10

N mC

kqEr

CE

m

NE C

A

What is the direction of the electric field at point A, if the two positive charges have equal magnitude?

1 2 3 4 5

0% 0% 0%0%0%

1. Up2. Down3. Right4. Left5. Zero

x

yA

B+ +

What is the direction of the electric field at point A, if the two positive charges have equal magnitude?

1 2 3 4 5

0% 0% 0%0%0%

1. Up2. Down3. Right4. Left5. Zero

x

yA

B+ +

Two ChargesCheckpoint1What is the direction of the electric field at point A?

1) Up

2) Down

3) Left

4) Right

5) Zerox

yA

B+

Two ChargesCheckpoint 2

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

1) up

2) down

3) Left

4) Right

5) Zero

x

yA

B+

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

1 2 3

0% 0%0%

1. Left2. Right3. zero

x

y

C+ -

Electric Field Applet

• http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

A B

X

Y

Checkpoint

Charge A is

1) positive 2) negative 3) unknown

Field lines start on positive charge, end on negative.

Checkpoint

Compare the ratio of charges QA/ QB

1) QA= 0.5QB 2) QA= QB 3) QA= 2 QB 4) can’t say

# lines proportional to |Q|

A B

X

Y

A B

X

Y

Checkpoint

The magnitude of the electric field at point X is greater than at point Y

1) True 2) False Density of field lines gives E

The electric field is stronger when the lines are located closer to one another.

Compare the magnitude of the electric field at point A and B

A

B

1. EA> EB

2. EA= EB

3. EA< EB

E inside of conductor• Conductor electrons free to move– Electrons feels electric force - will move until they

feel no more force (F=0)– F=qE: if F=0 then E=0

• E=0 inside a conductor (in electrostatics)

A B

X

Y

Checkpoint

"Charge A" is actually a small, charged metal ball (a conductor). The magnitude of the electric field inside the ball is:

(1) Negative (2) Zero (3) Positive

In a static situation

E inside of conductor in electrostatics

• Conductor electrons free to move– Electrons feel electric force - will move until

they feel no more force (F=0)– F=qE: if F=0 then E=0

• E=0 inside a conductor (in a static situation)

Recap

• E Field has magnitude and direction:– EF/q– Calculate just like Coulomb’s law– Careful when adding vectors

• Electric Field Lines– Density gives strength (# proportional to charge.)– Arrow gives direction (Start + end on -)

• Conductors– Electrons free to move E=0

To Do

• Read sections 19-6 & 19-7• Watch Prelecture 3 by 6am 1/15• Complete Checkpoint 3 by 6 am 1/15• Complete Homework 1 by 11pm 1/15• Complete Homework 2 by 11pm 1/17