Physics 102: Lecture 2, Slide 1

Coulomb’s Lawand Electric Fields

Physics 102: Lecture 02

Today we will … • get some practice using Coulomb’s Law• learn the concept of an Electric Field

Physics 102: Lecture 2, Slide 2

Recall Coulomb’s Law• Magnitude of the force between charges q1 and q2

separated a distance r:F = k q1q2/r2 k = 9x109 Nm2/C2

• Force is – attractive if q1 and q2 have opposite sign

– repulsive if q1 and q2 have same sign

• Units:– q’s have units of Coulombs (C)

• charge on proton is 1.6 x 10-19 C

– r has units of m

– F has units of N

Physics 102: Lecture 2, Slide 3

Coulomb Law practice: Three Charges

Q=-7.0 C

Q=+2.0C

6 m

4 m

• Calculate force on +2C charge due to other two charges– Draw forces

– Calculate force from +7C charge

– Calculate force from –7C charge

– Add (VECTORS!)

1

2

3

4

5

6

7

8

9

10

11

1

2

F+7

Q=+7.0C

F-7

Physics 102: Lecture 2, Slide 4

Three Charges – Calculate forces

Q=-7.0 C

Q=+2.0C

6 m

4 m

• Calculate force on +2C charge due to other two charges– Draw forces

– Calculate force from +7C charge

– Calculate force from –7C charge

– Add (VECTORS!)

• Calculate magnitudes

F+7

Q=+7.0C

F-7

𝐹= 𝑘𝑞1𝑞2𝑟2

Physics 102: Lecture 2, Slide 5

Q=-7.0 C

Q=+2.0C

6 m

4 m

F+7

Q=+7.0C

F-75 m

Three charges – Adding Vectors F+7+F-7

• Calculate components of vectors F+7 and F-7:

𝐹+7,𝑥 = 𝐹+7 cos𝜃 = 5× 10−3𝑁35

𝐹−7,𝑥 = 𝐹−7 cos𝜃 = 5× 10−3𝑁35

𝐹+7,𝑦 = 𝐹+7 sin𝜃 = 5× 10−3𝑁45

𝐹−7,𝑦 = −𝐹−7 sin𝜃 = −5× 10−3𝑁45

Physics 102: Lecture 2, Slide 6

Q=-7.0 C

Q=+2.0C

6 m

4 m

F+7

Q=+7.0C

F-75 m

Three charges – Adding Vectors F+7+F-7

• Add like components of vectors F+7 and F-7:

F

𝐹𝑥 = 𝐹+7,𝑥 + 𝐹−7,𝑥 = 6× 10−3𝑁 𝐹𝑦 = 𝐹+7,𝑦 + 𝐹−7,𝑦 = 0

𝐹=ට𝐹𝑥2 + 𝐹𝑦2 = 6× 10−3𝑁

𝜑 = tan−1 𝐹𝑦𝐹𝑥 = 0

• Final vector F has magnitude and direction

• Double-check with drawing

Physics 102: Lecture 2, Slide 7

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 = kq/r2

+

r = 1x10-10 m

Qp=1.6x10-19 C

E

E = (9109)(1.610-19)/(10-10)2 N = 1.41011 N/C (to the right)

Physics 102: Lecture 2, Slide 8

Preflight 2.3What is the direction of the electric field at point B?

1) Left

2) Right

3) Zero

x

yA

B

Physics 102: Lecture 2, Slide 9

Preflight 2.2

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

1) Up

2) Down

3) Left

4) Right

5) Zero

x

yA

B

Physics 102: Lecture 2, Slide 10

ACT: E FieldWhat is the direction of the electric field at point C?

A. Left

B. Right

C. Zero

x

y

C

A

B

Physics 102: Lecture 2, Slide 11

E Field from 2 Charges

Q = –3.5 C

A

6 m

4 m

• Calculate electric field at point A due to two unequal charges– Draw electric fields

– Calculate E from +7C charge

– Calculate E from –3.5C charge

– Add (VECTORS!)

Note: this is similar to (but a bitharder than) my earlier example.

Q = +7.0C

Physics 102: Lecture 2, Slide 12

Comparison:Electric Force vs. Electric Field

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

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

F = Eq• Both are vectors, with magnitude and direction.

Physics 102: Lecture 2, Slide 13

x

y

C

A

B

• Electric field defined at any location (we did three: A, B, C)

Electric Field Map

Physics 102: Lecture 2, Slide 14

-5 -4 -3 -2 -1 0 1 2 3 4 5-5

-4

-3

-2

-1

0

1

2

3

4

5

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

Electric Field Lines• Closeness of lines shows field strength (lines never cross)• Number of lines at surface Q• Arrow gives direction of E (Start on +, end on –)

This is becoming a mess!!!

Physics 102: Lecture 2, Slide 15

A B

X

Y

Preflight 2.5

Charge A is

1) positive 2) negative 3) unknown

Physics 102: Lecture 2, Slide 16

A B

X

Y

Preflight 2.6

Compare the ratio of charges QA/ QB

1) QA= 0.5QB 2) QA= QB 3) QA= 2 QB

A B

X

Y

Physics 102: Lecture 2, Slide 17

A B

X

Y

Preflight 2.8

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

1) True 2) False

Physics 102: Lecture 2, Slide 18

ACT: E Field Lines

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

1) EA>EB 2) EA=EB 3) EA<EB

A

B

Physics 102: Lecture 2, Slide 19

E inside of conductor• Conductor electrons free to move

– Electrons feels electric force - will move until they feel no more force (F=0)

– F=Eq: if F=0 then E=0

• E=0 inside a conductor (Always!)

Physics 102: Lecture 2, Slide 20

A B

X

Y

Preflight 2.10

"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

Physics 102: Lecture 2, Slide 21

Demo: E-field from dipole

x

y

C

A

B

Physics 102: Lecture 2, Slide 22

Recap• E Field has magnitude and direction:

– E F/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

Physics 102: Lecture 2, Slide 23

To Do

• Do your preflight by 6:00 AM Wednesday.