Applications of Coulomb’s Law

Applications of Coulomb’s Law

Physics 12

Joke of the day/clip of the day: Minute physics again!

http://www.youtube.com/watch?v=eTfm03T3VAE&list=PLED25F943F8D6081C&index=17

Review: Coulomb’s Law

Fe is electrical force (N) k is the Coulomb constant q1 is the electrical charge

on object 1 (C) q2 is the electrical charge

on object 2 (C) r is distance between the

objects (m)

221

rqkqFe

Part 1

3 Charges Straight line

Sample Problem 1: Three charges are arranged in a line; if the

three charges are 15μC, -12μC and 18μC respectively.

The distance between the first two charges is 0.20m and the second and third charges is 0.30m.

1. Draw a diagram2. Draw FB diagrams for each charge3. What is the force (Fnet) experienced by the charge

A?4. What is the force (Fnet) experienced by the charge

B?5. What is the force (Fnet) experienced by the charge

C?

Practise in a straight line: Coulomb’s Law WS#2

Questions 1-3

Part 2

3 Charges with angles!

What happens if the charges are NOT in a straight line?

We have to find the components (x and y) for each force and do some vector addition!

Force (Vector) Addition To add forces, resolve each force into its

components and treat the forces in the x-direction and y-direction independently

Once you sum the x and y components, use Pythagorean Theorem and Trigonometry to resolve into a resultant force

Example: Force Addition A point P has forces of 12.0N at 24.3°, 17.6N

at 112°, 6.78N at 241° and 10.2N at 74.4°.

A diagram may be helpful! Determine the resultant vector.

Example cont`d: XAx = 12.0cos24.3= 10.9Bx = 17.6cos112= -6.59Cx = 6.78cos241= -3.29Dx = 10.2cos74.4= 2.74

+3.76

YAy = 12.0sin24.3= 4.94By = 17.6sin112= 16.3Cy = 6.78sin241= -5.93Dy = 10.2sin74.4= 9.82

+25.1

Example cont`d  

θ

Practise vector addition: Add the following vectors head to tail, mark

angle!1. F(x) = -2 , F(y) = 42. F(x) = -0.2 , F(y) = -0.63. F(x) = 100 , F(y) = 504. F(x) = 3 , F(y) = -25. F(x) = -0.25 , F(y) = 0.11

Coulomb’s Law and Vector Addition When we consider an electrostatic system, we

need to use Coulomb’s Law to determine the magnitude and direction of each force

Once the magnitude and direction of each force has been determined, then the vector sum can be completed

Sample problem 2: Three charges are arranged as follows; (A)

+2.2μC is placed 2.3m due north of (B) -3.7μC charge and (C) +1.9μC charge is 3.1m due east of (B).

Draw a diagram What is the force experienced by the (B)?

Coulomb’s WS #2 Try questions 4, 5

Sample problem 3: Three charges are arranged as follows; a -

2.0μC is placed 4.0m due north of a 3.0μC charge and 3.0m due west of a 5.0μC charge.

Draw a diagram What is the force experienced by the 3.0μC

charge?

Concerned with forces on 3.0μC charge:

NxFrqkqF

e

e

312

221

12

104.3

-2.0μC

3.0 μ C

5.0 μ C

4.0m

3.0m

NxFrqkqF

e

e

323

223

23

104.5

1

2

3

5.0m

F1on2 is only in y anyway, need to find angle for F3on2

:

-2.0μC

3.0 μ C

5.0 μ C

4.0m

3.0m1

2

3θ

tan θ = 4.0/3.0 θ = 53°

Total angle from x-axis:

Summary of components:

3.0 μ C2

Use the x and y component data to determine the resultant force vector:

NxF

NxNxF

NxF

NxNF

ey

ey

ex

ex

4

33

3

3

100.9

103.4104.3

102.3

102.30

oe NxF 196,103.3 3

Use Pythagorean to find resultant force:

(-3.2x10-3)2 + (-9.0x10-4)2 = c2

3.3 x10-3N = c

tan θ = (-9.0x10-4) (-3.2x10-3)

θ = 16°

θ

Finish Coulomb’s WS #2 Question 6

Page 640 Questions 6, 7, 8

Part 3

Hanging angles!

Pith Ball Demo:

Sample problem 4: A negatively charged pith ball, with a mass of

1.25g is suspended from a thread from above. It makes a angle of 22° with the vertical when another negatively charged pith ball is brought near. The distance between the two pith balls is 3.35cm. What is the electrostatic force on the first pith ball?

Pith Ball

cos

0

TFTFTF

e

xe

xe

sin

0

TF

TF

TF

g

gg

yg

tan

sincos

sin

mgF

mgF

mgT

e

e

Coulomb’s WS #2 Question 7

Page 641 Questions 9, 10