college physics b - hadronhadron.physics.fsu.edu/~crede/files/fall2014/electric_forces_aug... ·...

CollegePhysics B

ElectricForces

Coulomb’s Law

Motion ofElectricCharge

Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

College Physics B - PHY2054C

Electric Forces and Fields

08/27/2014

My Office Hours:

Tuesday 10:00 - 12:00 Noon

206 Keen Building

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Outline

1 Electric Forces

Coulomb’s Law

2 Motion of Electric Charge

Conductors

Insulators

Metals

Static Electricity

3 Electric Field

4 Electric Flux

Gauss’ Law

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Discovery of electricity is generally credited to the Greeks.

• About 2500 years ago:

They observed electric charges and the forces between

them in many situations.

• Greeks used amber (a type of dried, hardened tree sap)

and rubbed it with a piece of animal fur.

➜ Greek word for amber was electron.

• The observed force even occurs when the amber (or

plastic in the picture) and the paper are not in contact.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Forces

Electric force can be attractive or repulsive.

• It is extremely large.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Forces



• Assume two charged particles can be modeledas point particles.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Forces



• The magnitude of the electric force between thetwo particles is given by Coulomb’s Law.

Direction of the force is along theline that connects the 2 charges.

• A repulsive force will give apositive value for F .

• An attractive force will give anegative value for F .

• k = 8.99 × 109 N ·m2/C2

“Coulomb constant”

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Features of Coulomb’s Law

Direction of the force

• The product q1 · q2 determines the sign.

• For like charges, the product is positive.➜ The force tends to push the charges farther

apart.

• For unlike charges, the product is negative.➜ The charges are attracted to each other.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Features of Coulomb’s Law

Direction of the force

• The product q1 · q2 determines the sign.

• For like charges, the product is positive.➜ The force tends to push the charges farther

apart.

• For unlike charges, the product is negative.➜ The charges are attracted to each other.

The form of Coulomb’s Law is similar to Newton’s Lawof Universal Gravitation:

F = k q1·q2

r2 F = G m1·m2

r2

r = distance between q1 and q2 or m1 and m2

attractive or repulsive always attractive!

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Gravity

“Inverse-Square Law”

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Coulomb’s Law

Every charged particle in the universe attracts (orrepels) every other charged particle with a forcethat is directly proportional to the product of thecharges of the particles and inversely proportionalto the square of the distance between them.

Electric Field Hockey:

phet.colorado.edu/en/simulation/electric-hockey

phet.colorado.edu/en/simulation/electric-hockey

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Coulomb’s Law

Charles-Augustin de Coulomb

(14 June 1736 - 23 August 1806)

F = kq1 q2

r2

k =1

4πǫ0

= 8.99 × 109 N · m2/C2

ǫ0 is called permittivity of free space

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Example: Hydrogen Atom

F = k ·q proton · q electron

r 2= 8.99 × 109

·−(1.6 × 10−19)2

(0.53 · 10−10)2N

≈ − 8.2 × 10−8 N

a = F/m ≈ 9.0 × 1022 m/s2

with r = 0.53−10 m.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Example: Helium Nucleus

F = k ·q proton · q proton

r 2= 8.99 × 109

·(1.6 × 10−19)2

(10−15)2N

≈ + 230 N

with r = 10−15 m.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Size of the Electric Force

Assume you have two boxes each containing 1 g ofelectrons:

• There would be 1.1 × 1027 electrons in each box.

• The force between the two boxes would be about3 × 1026 N with the boxes separated by r = 1 m.

➜ This is almost a million times larger than the forcebetween the Sun and the Earth.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Superposition of Forces

When there are more than two charges in a problem, the

“Superposition Principle” must be used.

1 Find the forces on the charge of interest due to all the

other forces.

2 Add the forces as vectors. ~F = ~F1 + ~F2

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Question

Two uniformly charged spheres are firmly fastened toand electrically insulated from frictionless pucks on anair table. The charge on sphere 2 is three times thecharge on sphere 1. Which force diagram correctlyshows the magnitude and direction of the electrostaticforces?

A

B

C

D

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

1 Recognize the Principle

• The electric force can be found using Coulomb’s Law.• The principle of superposition may also be needed.

2 Sketch the Problem

• Construct a drawing and show the location and charge of

each object (include a coordinate system).• Include the directions of all electric forces acting on the

particle of interest.

3 Identify the Relationships

• Use Coulomb’s Law to find the magnitude of the forces

acting on the particle of interest.

4 Solve

• The total force on a particle is the sum of all the forces.• Add the forces as vectors.• It is usually easier to work in terms of the components

along the coordinate system.

5 Check

• Consider what the answer means.• Check if the answer makes sense.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Outline

1 Electric Forces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

3 Electric Field

4 Electric Flux

Gauss’ Law

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Structure of Matter

... from an “electrical” point of view:

• Conductors (copper, silver, gold, iron, lead, etc.)

are materials, in which electric charge can move freely.

• Insulators (glass, plastic, rubber, wood, etc.)

do not allow charges to move.

• Semiconductors (silicon, germanium, etc.)

are materials, in which charges can move under the

influence of small electrical forces.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Conductors

• Each atom by itself is

electrically neutral.

➜ Equal numbers of

protons and electrons

• When these atoms come

together to form the metal

piece, electrons are freed.

• The electrons move freely

through the entire piece of

metal leaving behind protons.

➜ Conduction Electrons

A piece of metal can also accept extra electrons or release

some of its conduction electrons so that the entire object can

acquire a net positive or negative charge.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Insulators

• In insulators, the electrons

are not able to move freely

through the material.

• Examples include quartz,

plastic, and amber.

• Electrons cannot escape

from these ions.

• There are no conduction

electrons available to carry

charge through the solid.

• If extra electrons are placed

on an insulator, they tend to

stay in that same place.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Excess Charge on a Metal

• Electrons can move easily

through a metal.

• Excess electrons will be

distributed on the surface of

the metal.

Why?

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Excess Charge on a Metal

• Electrons can move easily

through a metal.

• Excess electrons will be

distributed on the surface of

the metal.

• Eventually, all the charge

carriers will come to rest and

be in static equilibrium.

• For a metal in equilibrium:

1 Any excess electrons must

be at the surface.

2 The electric field is zero

inside metal in equilibrium.

➜ Faraday’s Cage

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Faraday’s Cage

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Charging an Object by Rubbing

The act of rubbing causes some

charge to be transferred from one

material to another

Example:

• When rubbing amber with

fur, electrons are moved

from the fur to the amber.

• The amber acquires a net

negative charge.

• The fur is left with a net

positive charge.

Applies to many combinations of

materials.

phet.colorado.edu/en/simulation/travoltage

phet.colorado.edu/en/simulation/travoltage

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Polarization

Initially, the rod and the paper are both neutral.

• The rod is rubbed by the fur, obtaining a negative charge.

• The presence of the rod causes the electrons in the

paper to be repelled and the positive ions are attracted.

➜ The paper is said to be polarized.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Polarization

The paper is electrically neutral.

• The negative side of the paper is repelled.

• The positive side of the paper is closer to the rod so it will

experience a greater electric force.

• The net effect is that the positive side of the paper is

attracted to the negative side.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Polarization

Balloons and Static Electricity:

phet.colorado.edu/en/simulation/balloons

phet.colorado.edu/en/simulation/balloons

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Outline

1 Electric Forces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

3 Electric Field

4 Electric Flux

Gauss’ Law

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Field

An electric field gives another explanation of electric forces.

The presence of a charge produces an electric field:

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Field

An electric field gives another explanation of electric forces.

The presence of a charge produces an electric field:

• A positive charge produces field

lines that radiate outward.

• For a negative charge, the field

lines are directed inward, toward

the charge.

• The electric field is a vector and

denoted by ~E .

• The density of the field lines is

proportional to the magnitude of E .

➜ The field lines are most dense

in the vicinity of charges.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Field

The Electric Field is defined as the force exerted on a tiny

positive test charge at that point divided by the magnitude

of the test charge:

• Consider a point in space where

the electric field is E .

• If a charge q is placed at the point,

the force is given by F = q E .

• The charge q is called a test

charge.

• By measuring the force on

the test charge, the magnitude

and direction of the electric field

can be inferred.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Field

The Electric Field is defined as the force exerted on a tiny

positive test charge at that point divided by the magnitude

of the test charge:

• Consider a point in space where

the electric field is E .

• If a charge q is placed at the point,

the force is given by F = q E .

• Unit of the electric field is N/C:

F = kQ q

r 2= q E

E = kQ

r 2

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Fields are Vectors

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Rules for Drawing Field Lines

The lines must begin on a positive charge and terminate on a

negative charge (or at infinity).

• The number of lines drawn leaving a positive charge or

approaching a negative charge is proportional to the

magnitude of the charge.

• No lines can cross.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Outline

1 Electric Forces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

3 Electric Field

4 Electric Flux

Gauss’ Law

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Flux

Gauss’ Law can be used to find the electric field of a com-

plex charge distribution.

• Easier than treating it as a collection of point charges

and using superposition.

To use Gauss’ Law, a quantity called the Electric Flux is

needed.

• The electric flux is equal to the number of electric field

lines that pass through a particular surface multiplied

by the area of the surface.

• The electric flux is denoted by ΦE = E A.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Electric Flux: Examples

A The electric field is perpendicular to the surface of A:

➜ ΦE = E A

B The electric field is parallel to the surface of A:

➜ ΦE = E A = 0

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law



➜ ΦE = E A


➜ ΦE = E A = 0

C The electric field forms an angle with the surface A:

➜ ΦE = E A cos Θ

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law



➜ ΦE = E A


➜ ΦE = E A = 0

C The electric field forms an angle with the surface A:

➜ ΦE = E A cos Θ

D The flux is positive if the field is directed out of the region

surrounded by the surface and negative if going into the

region: ΦE = 0

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Gauss’ Law

Carl Friedrich Gauss

(30 April 1777 - 23 February 1855)

Gauss’ Law says that the electric flux through any closed

surface is proportional to the charge Q inside the surface:

ΦE =Q

ǫ0

ǫ0 is called permittivity of free space

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Field of a Point Charge

Choose a Gaussian surface

• Choose a surface that will make the calculation easy.

• Surface should match the symmetry of the problem.

➜ For a point charge, the field lines have a spherical

symmetry.

• The spherical symmetry means that

the magnitude of the electric field

depends only on the distance

from the charge.

• The magnitude of the field is the

same at all points on the surface

(due to the symmetry).

• The field is perpendicular to

the sphere at all points.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Field of a Point Charge

Choose a Gaussian surface

• Choose a surface that will make the calculation easy.

• Surface should match the symmetry of the problem.

➜ For a point charge, the field lines have a spherical

symmetry.

ΦE = E A sphere

= E 4πr2

= q/ǫ0

E =q

4πǫ0 r2→ Coulomb′s Law

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Problem Solving Strategy

1 Recognize the Principle

• Calculate the electric flux through a Gaussian surface.• The choice of the Gaussian surface is key!

2 Sketch the Problem

• Draw the charge distribution.• Use symmetry of distribution to sketch the electric field.• The Gaussian surface should match the symmetry of the

electric field.

3 Identify the Relationships• Your Gaussian surface should satisfy one or more of the

following conditions:

• The field has constant magnitude over all or much of the

surface and makes a constant angle with the surface.• The field may be zero over a portion of the surface.

➜ The flux through that part of the surface is zero.• The field may be parallel to some part of the surface.

➜ The flux through that part of the surface is zero.

CollegePhysics B

ElectricForces

Coulomb’s Law


Conductors

Insulators

Metals

Static Electricity

Electric Field

Electric Flux

Gauss’ Law

Problem Solving Strategy

Solve

• Calculate the total electric flux through the entire

Gaussian surface.• Find the total electric charge inside the surface.• Apply Gauss’ Law to solve for the electric field.

Check

• Consider what the answer means.• Check if the answer makes sense.

college physics b - hadronhadron.physics.fsu.edu/~crede/files/fall2014/electric_forces_aug... ·...

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