Mapping Electric Fields

Lesson 17Electric Fields and PotentialEleanor Roosevelt High SchoolChin-Sung LinElectric FieldsLaw of Universal GravitationCoulombs LawGravitational & Electric Forces

What are the formulas for the following physics laws?Gravitational & Electric ForcesFg = G m1 m2r2Law of Universal GravitationCoulombs LawFe = k q1 q2r2Gravitational Field

Whats the definition of gravitational field?

Gravitational FieldFg: gravitational force (N)m:mass (kg)g: gravitational field strength (N/kg, or m/s2)g = FgmGravitational Field: Force per unit mass Electric FieldFe: electric force (N)q:charge (C)E: electric field strength (N/C)E = FeqElectric Field: Force per unit chargeGravitational & Electric FieldsE = FeqElectric Fieldg = FgmGravitational FieldElectric FieldSource Charge+q+QFeFeqE =Test ChargerElectric FieldSource ChargeFeqE =Test Charge+qQFerElectric FieldElectric field is a vectorA vector includes ___________ and ____________Source ChargeTest Charge+q+QFer

FeqE =+qQFerElectric FieldElectric field is a vectorA vector includes direction and magnitude Source ChargeTest Charge+q+QFerFeqE =+qQFerElectric FieldE = FeqCan you apply Coulombs law to this formula and then simplify it?Source Charge+q+QFeTest Charger = ???

Electric FieldFe: electric force (N)q:test charge (C)Q:source charge (C)E: electric field strength (N/C)r:distance between charges (m)k:electrostatic constant (N m2/C2)Electric Field: Force per unit chargeE = Feq=k q Qr2 q= k Qr2Electric Field ExampleWhat is the magnitude of the electric field strength when an electron experiences a 5.0N force?

Electric Field ExampleE = Fe / qE = 5 N / (1.6 x 10-19 C)= 3.13 x 1019 N/CWhat is the magnitude of the electric field strength when an electron experiences a 5.0N force?Electric Field ExampleWhat are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C?

Electric Field ExampleWhat 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 / r2E = (8.99 x 109 N m2/C2) (2.1 x 10-9 C) / (1.5 m)2= 8.4 N/CDirection: away from the positive chargeElectric Field ExerciseThere 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 mElectric Field ExerciseThere 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 / r2E = (8.99 x 109 N m2/C2) (0.32 C) / (2.0 m)2= 7.2 x 108 N/CDirection: toward the negative chargeElectric Field ExerciseThere 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 / qFe = q EFe = (1.6 x 10-19 C) (7.2 x 108 N/C)= 1.15 x 10-10 NAim: Electric FieldDoNow: (4 minutes)Write down the definition of Electric Field in wordsWrite down the formulas of Electric Field in two different formsDefine every symbol in the formula and identify their unitsIdentify the relationships between Electric Field and other variables

Aim: Electric FieldElectric FieldFe: electric force (N)q:test charge (C)Q:source charge (C)E: electric field strength (N/C)r:distance between charges (m)k:electrostatic constant (N m2/C2)Electric Field: Force per unit chargeE = Feq= k Qr2Electric FieldFe: electric force (N)q:test charge (C)Q:source charge (C)E: electric field strength (N/C)r:distance between charges (m)k:electrostatic constant (N m2/C2)Electric Field: Force per unit chargeE ~ FeE ~ 1r2E ~ Q 1qE ~Electric FieldSource Charge+q+QFeFeqE =Test ChargeIf you shift the test charge around, where can you find the electric field with the same magnitude?

Electric FieldSource Charge+q+QFeTest ChargeFeFeFeEEEEElectric FieldSource ChargeTest Charge+q+QFeFeFeFeEEEE

What will happen if you move the test charges away from the source charge?Electric FieldSource Charge+q+QFeTest ChargeFeFeFeEEEEElectric FieldFeFeSource Charge+q+QFeTest ChargeFeFeEEEEFeFeFeEEEEElectric FieldSource Charge+q+QFeTest ChargeFeFeFeEEEEFeFeFeFeEEEETest Charge+qVector representation-+Electric Field RepresentationLine-of-Force representation-+Electric Field RepresentationElectric Field Representation-+

How do you decide the strength of electric field?Electric Field Representation-+When the field lines are denser, the field is strongerElectric Field Representation

Where can you find the the strongest electric field?ABCDEElectric Field: Point ChargeLine-of-Force representation-+Electric Field: Pair of ChargesLine-of-Force representation

Electric Field: Pair of Charges

Sketch the electric field for like charges?++Electric Field: Pair of ChargesLine-of-Force representation

++Electric Field: Pair of ChargesLine-of-Force representation

Electric Field: Parallel PlatesLine-of-Force representation

Electric Field: Parallel Plates

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

Electric Field: Parallel PlatesThe electric field between the parallel plates is uniform except at both ends

Electric Field ExampleA 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?

Electric Field ExampleA 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 / qFe = E qFg = m gFe = 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 CElectric Field ExampleA 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 Cnumber of e - = 4.80 x 10-16 C / 1.6 x 10-19 C = 3000 e

Electric Field ExerciseA 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?

Electric Field ExerciseA 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 / qFe = E qFg = m gFe = 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 CElectric Field ExerciseA 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

Electric Fields

Electric Fields

Electric Fields

Electric Fields and Shielding

E = 0Electric ShieldingElectric Fields and ShieldingCancellation of electric forceThe electric forces of area A and area B on P completely cancel out ABPElectric PotentialAim: Electric PotentialDoNow: (3 minutes)Write down the formulas of Electric FieldDraw the electric field surrounding a pair of opposite chargeDraw the electric field surrounding a pair of charged parallel platesAim: Electric PotentialElectric FieldFe: electric force (N)q:test charge (C)Q:source charge (C)E: electric field strength (N/C)r:distance between charges (m)k:electrostatic constant (N m2/C2)Electric Field: Force per unit chargeE = Feq= k Qr2Electric Field: Pair of ChargesLine-of-Force representation

Electric Field: Parallel PlatesLine-of-Force representation

Gravitational Potential Energy (GPE)

Gravitational Potential Energy (GPE)

What happens in the picture?What types of energy have been converted?

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?

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?

Electric Potential Energy (EPE)

++++-----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 different?

Electric Potential Energy

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

Electric Potential Energy

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 EnergyThe work performed in taking a mass from height A to height B does not depend on the pathBAAElectric Potential EnergyBAThe work performed in taking a charge from A to B does not depend on the pathGravitational Potential Energy

x2What happens when we double the mass?Gravitational Potential Energy

x2When mass is doubled, the gravitational potential energy is also doubled PE2 = (2m)gh = 2(mgh) = 2PE1Electric Potential Energy

++++---x2What did we double here?Electric Potential Energy

++++---x2We doubled the charge.What happens when we double the charge?Electric Potential Energy

++++---x2When charge is doubled, the electric potential energy is also doubled Electric Potential EnergyW:electric (potential) energy aka work (J)

W is a scalar (not a vector)W or EPEPotential Energy Capability to Do Work Electric Potential EnergyW:electric (potential) energy aka work (J)W or EPEPotential Energy Capability to Do WorkW ~ qElectric PotentialW:electric (potential) energy, aka work (J)V:electric potential, aka potential difference,aka voltage (V, Volts)q:charge (C)V = WqElectric potential energy per unit chargeElectric PotentialElectric potential (V) is based on a zero reference pointOnly the potential difference mattersElectric potential (Voltage, V) is the work (W) required to bring a unit charge (1 C) from the zero reference pointV is a scalar (not a vector)W= qVElectric Potential ExampleHow 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?

Electric Potential ExampleHow 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 / qW = q VW = (3.0 C) (12.0 V) = 36.0 JElectric Potential ExampleIf 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?

Electric Potential ExampleIf 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 / qW = q VV = (1.4 x 10-15 J) / (1.6 x 10-19 C) = 8750 VElectric Potential ExampleCan you make up a question using the definition of electric potential?

Electric Potential Example

Electric Potential: Parallel PlatesWhat is special about the electric field between the charged parallel plates?

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

EEEElectric Potential: Parallel PlatesIf we place test charge at different locations between the charged parallel plates, compare the forces experienced by these test charges

EEE

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

EEEFeFeFeElectric Potential: Parallel PlatesCompare the work required to move test charges from the negative plate to the positive plate

EEEFed

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

EEEFedFeFeElectric Potential: Parallel PlatesV = WqGiven the following formulas, can you derive the formula for Electric potential (V) and Electric field (E)?W = Fe dFe = q E

Electric Potential: Parallel PlatesV = WqElectric potential (V):W = Fe dand Fe = q EW = q E dV = E d or= q E dq= E d E = VdElectric Potential: Parallel PlatesElectric potential (V): relative to the negative plate is proportional to the distance to it due to V = E d

EEEdVEquipotential Lines: Parallel PlatesEquipotential Lines: on an equipotential line, voltages are all the sameEquipotential lines are perpendicular to field lines

EEEd1V1V1V1V2V2V2d2Equipotential Lines: Point ChargeEquipotential Lines for a point charge:

Equipotential Lines: Point ChargeEquipotential Lines for a charge pair

Electric Field and PotentialE: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 fieldE = Vd= FeqElectric Potential ExampleA 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 / qFe = E qFg = m gFe = Fg E q = m gE = V / dV 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 CElectric Potential ExampleThe End