Lesson 17 Electric Fields and Potential Eleanor Roosevelt High School Chin-Sung Lin.

Lesson 17

Electric Fields and Potential

Eleanor Roosevelt High School

Chin-Sung Lin

Electric Fields

Law of Universal Gravitation

Coulomb’s Law

Gravitational & Electric Forces

What are the formulas for the following physics laws?

Gravitational & Electric Forces

Fg = G m1 m2

r2

Law of Universal Gravitation

Coulomb’s Law

Fe = k q1 q2

r2

Gravitational Field What’s the definition of gravitational field?

Gravitational Field

• Fg: gravitational force (N)

• m: mass (kg)

• g: gravitational field strength (N/kg, or m/s2)

g = Fg

m

Gravitational Field: Force per unit mass

Electric Field

• Fe: electric force (N)

• q: charge (C)

• E: electric field strength (N/C)

E = Fe

q

Electric Field: Force per unit charge

Gravitational & Electric Fields

E = Fe

q

Electric Field

g = Fg

m

Gravitational Field

Electric Field

Source Charge

+q+QFe

Fe

qE =

Test Charge

r

Electric Field

Source Charge

Fe

qE =

Test Charge

+q–Q Fer

Electric Field Electric field is a vector

A vector includes ___________ and ____________

Source Charge

Test Charge

+q+QFe

r Fe

qE =

+q–Q Fer

Electric Field Electric field is a vector

A vector includes direction and magnitude

Source Charge

Test Charge

+q+QFe

r Fe

qE =

+q–Q Fer

Electric Field

E = Fe

q

Can you apply Coulomb’s law to this formula and then simplify it?

Source Charge

+q+QFe

Test Charge

r

= ???

Electric Field


• q: test charge (C)

• Q: source charge (C)


• r: distance between charges (m)

• k: electrostatic constant (N m2/C2)


E = Fe

q= k

q Q

r2 q= k

Q

r2

Electric Field Example

What is the magnitude of the electric field strength when an electron experiences a 5.0N force?


E = Fe / q

E = 5 N / (1.6 x 10-19 C)

= 3.13 x 1019 N/C

What is the magnitude of the electric field strength when an electron experiences a 5.0N force?


What are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C?


What are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C?

E = k Q / r2

E = (8.99 x 109 N m2/C2) (2.1 x 10-9 C) / (1.5 m)2

= 8.4 N/C

Direction: away from the positive charge

Electric Field Exercise

There is a negative charged particle of 0.32 C in the free space. (a) What are the magnitude and direction of the electric field 2.0 m away from the particle? (b) What are the magnitude and direction of the electric force when an electron is placed 2.0 m away from this particle?

[3 minutes] e –– 0.32 C2.0 m


There is a negative charged particle of 0.32 C in the free space. (a) What are the magnitude and direction of the electric field 2.0 m away from the particle?

E = k Q / r2

E = (8.99 x 109 N m2/C2) (0.32 C) / (2.0 m)2

= 7.2 x 108 N/C

Direction: toward the negative charge


There is a negative charged particle of 0.32 C in the free space. (b) What are the magnitude and direction of the electric force when an electron is placed 2.0 m away from this particle? E = Fe / q Fe = q E

Fe = (1.6 x 10-19 C) (7.2 x 108 N/C)

= 1.15 x 10-10 N

Aim: Electric Field

DoNow: (4 minutes)

Write down the definition of Electric Field in words

Write down the formulas of Electric Field in two different forms

Define every symbol in the formula and identify their units

Identify the relationships between Electric Field and other variables

Aim: Electric Field

Electric Field








E = Fe

q= k

Q

r2

Electric Field








E ~ Fe E ~ 1r2

E ~ Q 1qE ~

Electric Field

Source Charge

+q+QFe

Fe

qE =

Test Charge

If you shift the test charge around, where can you find the electric field with the same magnitude?

Electric Field

Source Charge

+q+QFe

Test Charge

Fe

Fe

Fe

E

E

E

E

Electric Field

Source Charge

Test Charge

+q+QFe

Fe

Fe

Fe

E

E

E

E

What will happen if you move the test charges away from the source charge?

Electric Field

Source Charge

+q+QFe

Test Charge

Fe

Fe

Fe

E

E

E

E

Electric Field

Fe

Fe

Source Charge

+q+QFe

Test Charge

Fe

Fe

E

E

E

E

Fe

Fe

Fe

E

E

E

E

Electric Field

Source Charge

+q+QFe

Test Charge

Fe

Fe

Fe

E

E

E

E

Fe

Fe

Fe

Fe

E

E

E

E

Test Charge

+q

Vector representation

-+

Electric Field Representation

Line-of-Force representation

-+



-+

How do you decide the strength of electric field?


-+

When the field lines are denser, the field is stronger

Electric Field Representation Where can you find the the strongest

electric field?

A

B

C

D

E

Electric Field: Point Charge


-+

Electric Field: Pair of Charges



Sketch the electric field for like charges?

++



++

Electric Field: Parallel Plates



Anything special for the electric field between the parallel plates charged with opposite charges?


The electric field between the parallel plates is uniform except at both ends


A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (a) What is the charge on the particle? (b) By how many electrons is the particle deficient?


A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (a) What is the charge on the particle? E = Fe / q Fe = E q Fg = m g Fe = Fg

E q = m g

(1.2 x 107 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2)

q = 4.80 x 10-16 C


A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (b) By how many electrons is the particle deficient?e - = 1.6 x 10-19 C

number of e - = 4.80 x 10-16 C / 1.6 x 10-19 C = 3000 e –


A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 9.6 x 106 N/C and are 2.00 mm apart. What is the charge on the particle?


A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 9.6 x 106 N/C and are 2.00 mm apart. What is the charge on the particle? E = Fe / q Fe = E q Fg = m g Fe = Fg

E q = m g

(9.6 x 106 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2)

q = 6.0 x 10-16 C


A positively charged ball with mass 20 g is hanging between two charged parallel plates from the ceiling through an insulating wire with length 0.1 m. The electric field strength of the charged parallel plates is 4.2 x 109 N/C. When the ball is in balance, the wire and the vertical line form an angle of 60o. What is the charge of the ball?

Electric Fields

Electric Fields and Shielding

E = 0Electric Shielding

Electric Fields and Shielding

Cancellation of electric force

The electric forces of area A and area B on P completely cancel out

A BP

Electric Potential

Aim: Electric PotentialDoNow: (3 minutes)

Write down the formulas of Electric Field

Draw the electric field surrounding a pair of opposite charge

Draw the electric field surrounding a pair of charged parallel plates

Aim: Electric Potential

Electric Field








E = Fe

q= k

Q

r2

Gravitational Potential Energy (GPE)


What happens in the picture?

What types of energy have been converted?




If we want to pull the weight up back to its original position, what should we do?





How much work do we need?

Electric Potential Energy (EPE)

++

++

-----

What are they in common?


++

++

-----




How much work do we need?


++

++

-----

What are they different?

Electric Potential Energy

What did the monkey do in order to bring a positively charged ball toward a

positively charged object?


What did the monkey do in order to bring a positively charged ball toward a

positively charged object?

When the ball was released, what happened to the ball? Why?

Gravitational Potential Energy

The work performed in taking a mass from height A to height B does not depend on the path

B

A A


B

A

The work performed in taking a charge from A to B does not depend on the path


x2

What happens when we double the mass?


x2

When mass is doubled, the gravitational potential energy is also doubled

PE2 = (2m)gh = 2(mgh) = 2PE1


++

++

---

x2

What did we double here?


++

++

---

x2

We doubled the charge.

What happens when we double the charge?


++

++

---

x2

When charge is doubled, the electric potential energy is also doubled


• W: electric (potential) energy aka work (J)

W is a scalar (not a vector)

W or EPE

Potential Energy – Capability to Do Work


• W: electric (potential) energy aka work (J)

W or EPE

Potential Energy – Capability to Do Work

W ~ q

Electric Potential

• W: electric (potential) energy, aka work (J)

• V: electric potential, aka potential difference,

aka voltage (V, Volts)

• q: charge (C)

V = Wq

Electric potential energy per unit charge

Electric Potential

Electric potential (V) is based on a zero reference point

Only the potential difference matters Electric potential (Voltage, V) is the work

(W) required to bring a unit charge (1 C) from the zero reference point

V is a scalar (not a vector)

W = qV

Electric Potential Example

How much work is required to move 3.0 C of positive charge from the negative terminal of a 12-volt battery to the positive terminal?


How much work is required to move 3.0 C of positive charge from the negative terminal of a 12-volt battery to the positive terminal?

V = W / q W = q V

W = (3.0 C) (12.0 V) = 36.0 J


If an electron loses 1.4 x 10-15 J of energy in traveling from the cathode to the screen of a computer monitor, across what potential difference must it travel?


If an electron loses 1.4 x 10-15 J of energy in traveling from the cathode to the screen of a computer monitor, across what potential difference must it travel?

V = W / q W = q V

V = (1.4 x 10-15 J) / (1.6 x 10-19 C) = 8750 V


Can you make up a question using the definition of electric potential?

Electric Potential: Parallel Plates

What is special about the electric field between the charged parallel plates?


Electric Field (E): is uniform due to uniform density of the electric field lines

E E E


If we place test charge at different locations between the charged parallel plates, compare the forces experienced by these test charges

E E E


Electric Force (Fe): experienced by a test charge is constant due to Fe = qE

E E E

Fe

FeFe


Compare the work required to move test charges from the negative plate to the positive plate

E E EFe dFeFe


Electric potential energy / work (W): required to move a test charge from negative plate to positive plate is constant due to W = Fe d

E E EFe dFe Fe


V = Wq

Given the following formulas, can you derive the formula for Electric potential (V) and Electric field (E)?

W = Fe d

Fe = q E


V = Wq

Electric potential (V):

W = Fe d and Fe = q E

W = q E d

V = E d or

= q E dq = E d

E = Vd


Electric potential (V): relative to the negative plate is proportional to the distance to it due to V = E d

E E E

dV

Equipotential Lines: Parallel Plates

Equipotential Lines: on an equipotential line, voltages are all the same

Equipotential lines are perpendicular to field lines

E E E

d1

V1V1V1

V2V2V2

d2

Equipotential Lines: Point Charge

Equipotential Lines for a point charge:

Equipotential Lines: Point Charge

Equipotential Lines for a charge pair

Electric Field and Potential

• E: electric field (N/C or V/m)• V: electric potential / potential difference /

voltage (V, Volts)• d: distance between parallel plates (m)• q: charge (C)

Electric field

E = Vd =

Fe

q


A charged droplet of mass 5.87x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a potential difference of 24000 V and are 2.00 mm apart. What is the charge on the particle?

A charged droplet of mass 5.87x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a potential difference of 24000 V and are 2.00 mm apart. What is the charge on the particle?

E = Fe / q Fe = E q Fg = m g Fe = Fg

E q = m g E = V / d V q / d = m g

(24000 V) q / ((0.002 m) = (5.87 x 10-10 kg) (9.81 m/s2)

q = 4.80 x 10-16 C


The End

Lesson 17 Electric Fields and Potential Eleanor Roosevelt High School Chin-Sung Lin.

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Transcript of Lesson 17 Electric Fields and Potential Eleanor Roosevelt High School Chin-Sung Lin.