Chapter 15 Coulomb’s Law Electrical Force Superposition

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# Chapter 15 Coulomb’s Law Electrical Force Superposition - PowerPoint PPT Presentation

Chapter 15 Coulomb’s Law Electrical Force Superposition. Lecture 14. 10 February 1999 Wednesday. Physics 112. Last time: two types of charge Positive and Negative Quantized in units of +/- 1, +/- 2, etc.

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Chapter 15

Coulomb’s Law

Electrical Force

Superposition

Lecture 14

10 February 1999

Wednesday

Physics 112

Last time: two types of charge

Positive and Negative

Quantized in units of +/- 1, +/- 2, etc.

The mks unit for charge is the Coulomb, named after Charles Coulomb in honor of his many contributions in the field of electricity.

1e = 1.6 X 10-19 C

Remember to Convert!

Quantifying the

Electrical Force

Joseph Priestly and Charles Coulomb set out to

quantify the electrical force in the late 1700’s.

Let’s see if we can figure it out an expression for

the electric force ourselves...

On What Does the

Electric Force Depend?

F = ???

Distance

Charge on Body 1

Charge on Body 2

Anything else?

1/r2 (units of 1/m2)

q1 (units of C)

q2 (units of C)

Just a constant of

proportionality…

Let’s call it k.

Note on Notation!

The arrow indicates that F is a vector quantity (i.e., to specify F, you need both magnitude and direction)!

Indicates a direction radially away from

the center of our coordinate system. It could

be the x-direction, the y-direction, or

something in between.

Units!

[F] = [k] [q1] [q2] / [r]2

N = [k] C2/m2

[k] = N m2/C2!

The value of k is determined

experimentally to be…

k = 8.99 X 109 N m2/C2

Let’s call it 9 X 109 N m2/C2

We have now determined a quantitative

expression for the electrostatic force!

Notes on Coulomb’s Law:

Applies only to point charges, particles

or spherical charge distributions.

Obeys Newton’s 3rd Law.

The electrical force, like gravity, is a

“field” force…that is, a force is exerted

at a distance despite lack of physical

contact.

The electrostatic force obeys the

Superposition Principle

This implies that to solve problems with

multiple charges, we may consider each

two charge system separately and combine

the results at the end.

Remember, force is a vector quantity,

so you must use vector addition!

Points in the direction

from 1 to 2!

= (9 X 109 N m2 / C2)(-1 mC)(-2 mC)/(1 m)2

= +1.8 X 10-2 N(i.e.,in the +x direction).

Example:

y

q1

q2

q3

r1

r2

x

q1 = -1 mC q2 = -2 mC q3 = +1 mC

r1 = 1 m r2=2 m

What is the Total Force on q2?

1) Start by examining the force exerted by q1 on q2.

Points in the direction

from 3 to 2!

= (9 X 109 N m2 / C2)(+1 mC)(-2 mC)/(2 m)2

= -4.5 X 10-3 NThe minus sign indicates the force is attractive…..therefore it’sin the +x direction.

Example (con’t):

y

q1

q2

q3

r1

r2

x

q1 = -1 mC q2 = -2 mC q3 = +1 mC

r1 = 1 m r2=2 m

What is the Total Force on q2?

2) Then examine the force exerted by q3 on q2.

Example (con’t):

y

q1

q2

q3

r1

r2

x

q1 = -1 mC q2 = -2 mC q3 = +1 mC

r1 = 1 m r2=2 m

What is the Total Force on q2?

3) Finally, carefully add together the results.

F2 = F12 + F32 = 1.8 X 10-2 N + 4.5 X 10-3 N

= 2.25 X 10-2 N (i.e., in the +x direction).

Concept Quiz!

Electrical Forces