What is Electricity?

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What is Electricity?. Flow of electrons. Electromagnetic force. BASIC MODEL. Charge: Modern view. charge comes in two flavors (positive and negative) normally balanced (neutral) the amount of positive charge in an object is fixed! imbalance causes “charge”. Problem:. - PowerPoint PPT Presentation

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What is Electricity?

Flow of electrons.Electromagnetic force

BASIC MODEL

Charge: Modern viewCharge: Modern view

•charge comes in two flavors (positive and negative)

•normally balanced (neutral)•the amount of positive

charge in an object is fixed!•imbalance causes “charge”

Problem:Problem:

A bushel basket contains

50 apples. How much of a

bushel is just one apple?

Measuring chargeMeasuring charge

•we measure in units called Coulombs (C)

•6.25x1018 bare protons is required to make 1C.

•What is the charge on a single proton?

elementary chargeelementary charge

•we define the charge on one proton e=1.6 x10-19C

•the charge on the electron is then -e

ProblemProblem

A penny contains about 2 x1023

protons and an equal number of electrons. If the penny is charged to +1C, what fraction of the electrons have been added or removed?

fraction removed = #removedoriginal #

ProblemProblem

A penny contains about 2 x1023

protons and an equal number of electrons. If the penny is charged to +1C, what fraction of the electrons have been added or removed?

fraction removed = #removed

2 x 1023

ProblemProblem

A penny contains about 2 x1023

protons and an equal number of electrons. If the penny is charged to +1C, what fraction of the electrons have been added or removed?

fraction removed = 6.25 x 1018

2 x 1023

ProblemProblem

A penny contains about 2 x1023

protons and an equal number of electrons. If the penny is charged to +1C, what fraction of the electrons have been added or removed?

fraction removed = .00003!

ConductivityConductivity

•conductors: loosely bound electrons; charge flows freely

•insulators: tightly bound electrons; charge hard to move

Induction in a conductor

EBONITE

How will the +’ve charge move?

How will the -’ve charge move?

Induction in a conductor

EBONITEI see a problem; do you see a problem?

Induction in a conductor

EBONITE

Induction in a conductor

EBONITE

Where is the +’ve charge?

Where is the -’ve charge?

Induction in an insulator

EBONITE

Induction in an insulator

EBONITE

Induction in an insulator

EBONITE

Look at the surface.

Fnet

Induction

• conductors: charge flows• insulators: charge shifts

221

r

MMGF

Newton’s Law of Newton’s Law of GravityGravity

M1 M2

r

Constant of proportionality

Mearth = 5.98 x 1024 kg

r = 6.38 x 106 m

What is the force between the Earth and a book? 2

21

r

MMGF

Wearth,book = Mbook (9.8 N/kg)

Wearth,book = (6.67 x 10-11Nm2)(5.98 x 1024 kg) Mbook

(6.38 x 106 m)2kg2

Gravitational Field

consider a 1kg block

F = ?10 N

F = ?5 N

Gravitational Field

consider a 3kg block

F = ?10 N

F = ?3 kg x 5 N/kg= 15 N

gravitational field strength

What is the gravitational field strength at the Earth’s surface?

Gravitational Field

What is the gravitational field strength at this point?

m = 5 kg

F = 35 N

7 N/kg

Gravitational Field

10 N/kg

7 N/kg

5 N/kgF = m g

Electric Force

• Definition: Coulomb’s Law

• Electric field is the force that +1C would feel if it were placed at this location.

What units?

Newtons (N)

Coulombs (C)q is in units of:

k : 9 x 109 N.m2/C2

F is in units of:

E is in units of: N/C

Example #1

Anthea rubs two latex balloons against her hair, causing the balloons to become charged negatively with 2.0 x 10-6 C. She holds them a distance of 0.70 m apart. A) what is the electrical force between the two balloons? B) Is it one of attraction or repulsion?

Example 2

Anthea rubs two latex balloons against her hair, causing the balloons to become charged negatively with 2.0 x 10-6 C. She holds them a distance of 0.70 m apart. A) what is the electrical force between the two balloons? B) Is it one of attraction or repulsion? C) What is the electrical field of the 1st balloon?

Example #3

At the location marked with an x, the electric field is 2000 N/C and points right. What is the electric force (size and direction) on a 6 x 10-6 C charge that is placed at the x?

EF

F = qE = (6x10-6)(2000) = 1.2 x 10-2 N (to the right)

+q

203

3 20 60

Example #3 cont.

What if the charge were the same size but negative?

EF

Same size F = 60 N (to the left)

F = Who Knows?

-q

What if a charge were placed somewhere else?

E

Electric FieldsElectric Fields(again)(again)

Electric Field

• Operational definition:

qEF • Electric field is the force that +1C would

feel if it were placed at this location.

Example #2

A charge of -5 x 10-8 C feels a force of 0.2 N to the right. What is the electric field (magnitude and direction) at the charge’s location?

FEF = qE-q

E = F/q

E = 4 x 106 N/C (to the left)

Where does E come from?

qEF Force on q due to the field, E, at q’s location

Field at q’s location due to other charges at other locations(source charges) The charge that is

feeling the force(test charge)

What is the direction of the electric field at x?

E

The direction of E

E

What is the direction of the electric field at x?

HOW BIG?

Coulomb’s Law

Er

k = 9 x 109 Nm2/C2

qs

E =k qs

r2

Example #last

A 2.5 x 10-6 C charge is placed as shown below. What is the electric field at a point 5 cm to the right?

qs

5 cm

E = kqs /r2 = (9 x 109)(2.5 x 10-6) /(5 x 10-2)2

= +9 x 106 N/C

Which way does E point?

Coulomb’s Law

Er

k = 9 x 109 Nm2/C2

qs

E =k qs

r2

Q1 Q2

What is the electric field strength at the location of Q1 due to Q2?

Which charge do we care about?

0.20 m -6x10-9C3x10-9C

Q2

0.20 m

What is the electric field strength at the location of Q1 due to Q2?

Which charge do we care about?

How does this change the problem?

-6x10-9C

Q2

E = kqs /r2

= (9 x 109)(6 x 10-9) /(0.20)2

= 1350 N/C

1350 N/C -6x10-9C

Q2

1350 N/C

What force does Q1 experience?

F = qE

= (3 x 10-9 C)(1350 N/C)

= 4 x 10-6 N (right)

4 x 10-6 N

Q1 Q2

0.20 m

What is the electric field strength at the location of Q2 due to Q1?

Which charge do we care about?

3x10-9C -6x10-9C

0.20 m

What is the electric field strength at the location of Q2 due to Q1?

Which charge do we care about?

How does this change the problem?

Q1

3x10-9C

E = kqs /r2

= (9 x 109)(3 x 10-9) /(0.20)2

= 675 N/C

675 N/C

Q1

3x10-9C

Q1

What force does Q2 experience?

4 x 10-6 N

F = qE

= (6 x 10-9 C)(675 N/C)

= 4 x 10-6 N (left)

675 N/C

4 x 10-6 N

675 N/C1350 N/C

4 x 10-6 N

due to Q1due to Q2

F2,1 F1,2

Q1 Q2

0.20 m

What is the electric field at the midpoint between the charges?

Which charge do we care about?

Q1 Q2

What is the electric field at the midpoint between the charges?

E2,X

E1,X

Q1 Q2

What is the electric field at the midpoint between the charges?

EnetEX,2

EX,2

EX,1 = kqs /r2

= (9 x 109)(3 x 10-9) /(0.10)2

= 2700 N/C

EX,2 = kqs /r2

= (9 x 109)(6 x 10-9) /(0.10)2

= 5400 N/C

Enet= 8100 N/C

Q1 Q2

What force would a charge Q3 = -2 x 10-6 C experience if placed at the midpoint?

Enet= 8100 N/C

F = qE

= (2 x 10-6 C) (8100 N/C)

= 0.016 N

F

4 x 10-6 N4 x 10-6 N

F2,1 F1,2

Q1 Q2

Two point charges +Q and -Q are fixed in place a distance 2d apart as shown. What direction is the electric field at the midpoint between the charges?

+Q -Q

Student 2: “The electric field is given by E=kqs/r2 so if I do the calculation I get:

Enet = k(+Q)/d2 + k(-Q)/d2 = 0

So, the electric field is zero and has no direction.”

dd

What do you think?

Two identical positive charges, A and B, are arranged as shown. The distance from point C to A is twice the distance from point C to B. Which of the following best represents the electric field at point C?

e

d

c

b

aA

B C

Two identical positive charges, A and B, are arranged as shown. The distance from point C to A is twice the distance from point C to B. Which of the following best represents the electric field at point C?

e

d

c

b

aA

B C

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

What if all the charge were moved to the bottom?

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

Distributed Charge:or what if charge were peanut butter?

+Q

+Q/9

+Q/9

+Q/9

+Q/9+Q/9 +Q/9

+Q/9

+Q/9

+Q/9

What is the electric field in the middle of the circle?

Boston Museum

of Science

a (excess charge is all on

outside surfaces)

+0-

0

b (excess charge is evenly

distributed on both sides)

c (excess charge is all on

inside surfaces)

+0-

0+0

+0

0

–– 0

From Tutorial

L = -0/2

L = +0/2

R = -0/2

R = +0/2

L = -0/2

L = +0/2

R = -0/2

R = +0/2

E = 0

L = -0/2

L = +0/2

R = -0/2

R = +0/2

L = -0/2

L = +0/2

R = -0/2

R = +0/2

E 0

L = +0

R = -0

E-

E+

L = +0

R = -0

E = 0

L = +0

R = -0

E-

E+

L = +0

R = -0

E = 0

L = +0

R = -0

E-

E+

= 0/20

= 0/20

L = +0

R = -0

E = 0/0

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