Chapter 16

Electric Charge and Electric Field

Demonstration #1

1. Demonstrate how you can pick up the tissue without touching it in any way with your body.

2. What is occurring on the atomic level that lets you do this?

The atom

The atom has positive charge in the nucleus, located in the protons. The positive charge cannot move from the atom unless there is a nuclear reaction.

The atom has negative charge in the electron cloud on the outside of the atom. Electrons can move from atom to atom without all that much difficulty.

• Demonstration #2• 1. Rub the black rod with the fur. Bring the rod

toward the pole of the electroscope. What happens to the vanes?

• - - -• 2. Come up with an atomic level explanation for

your observations.

• Demonstration #3

+++

• 1. Rub the glass rod with the silk. Bring the rod toward the pole of the electroscope. What happens to the vanes?

• 2. Come up with an atomic level explanation for your observations.

• Demonstration #4

• 1. What happens when your touch the electroscope with the glass rod?

Sample Problem

A certain static discharge delivers -0.5 Coulombs of electrical charge.

• How many electrons are in this discharge?

Sample Problem

1. How much positive charge resides in two moles ofhydrogen gas (H2)?

2. How much negative charge?

3. How much net charge?

Sample Problem (end 1)

The total charge of a system composed of 1800 particles, all of which are protons or electrons, is 31x10-18 C.

How many protons are in the system?

How many electrons are in the system?

Coulomb’s Law andElectrical Force

Day 2

Sample Problem

A point charge of positive 12.0 μC experiences an attractive force of 51 mN when it is placed 15 cm from another point charge. What is the othercharge?

Sample Problem

Calculate the mass of ball B, which is suspended in midair.

Sample Problem

• What is the force on the 4 μC charge?

Sample Problem end 2

• What is the force on the 4 μC charge?

The Electric FieldDay 3

Sample Problem end 3

A 400 μg styrofoam bead has 600 excess electrons on its surface. What is the magnitude and direction of the electric field that will suspend the bead in midair?

SuperpositionDay 4

Sample Problem

A proton traveling at 440 m/s in the +x direction enters an electric field of magnitude 5400 N/C directed in the+y direction. Find the acceleration.

Sample Problem

A particle bearing -5.0 μC is placed at -2.0 cm, and aparticle bearing 5.0 μC is placed at 2.0 cm. What is thefield at the origin?

Sample Problem

A particle bearing 10.0 mC is placed at the origin, and aparticle bearing 5.0 mC is placed at 1.0 m. Where is thefield zero?

Sample Problem end 4

What is the charge on the bead? It’s mass is 32 mg.

Electric Potential andPotential Energy

Day 5

Chapter 17

Electric Potential

Sample Problem

A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

Electrical Potential in Uniform Electric Fields

The electric potential is related in a simple way to a uniform electric field.

DV = -Ed DV: change in electrical potential (V) E: Constant electric field strength (N/C or V/m) d: distance moved (m)

Sample Problem

An electric field is parallel to the x-axis. What is itsmagnitude and direction of the electric field if thepotential difference between x =1.0 m and x = 2.5 m isfound to be +900 V?

Sample Problem

What is the voltmeter reading between A and B? Between A and C? Assume that the electric field has a magnitude of 400 N/C.

Sample Problem end 5

How much work would be done BY THE ELECTRIC FIELD in moving a 2 mC charge from A to C? From A to B? from B to C?. How much work would be done by an external force in each case? End 5

Electric Field and Shielding

Day 6

Sample Problem

If a proton is accelerated through a potential difference of -2,000 V, what is its change in potential energy?

How fast will this proton be moving if it started at rest?

Sample Problem

A proton at rest is released in a uniform electric field. How fast is it moving after it travels through a potentialdifference of -1200 V? How far has it moved?

Sample ProblemDraw field lines for the charge configuration below. The field is 600 V/m, and the plates are 2 m apart. Label each plate with its properpotential, and draw and label 3 equipotential surfaces between the plates. You may ignore edge effects.

-- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + +

Sample Problem

Draw a negative point charge of -Q and its associated electric field. Draw 4 equipotential surfaces such that DV is the same between the surfaces, and draw them at the correct relative locations. What do you observe about the spacing between the equipotential surfaces?

Objectives: After finishing this

unit, you should be able to:• Explain and demonstrate the Explain and demonstrate the first law of electro- first law of electro-

staticsstatics and discuss charging by and discuss charging by contact contact and by and by inductioninduction..

• Write and apply Write and apply Coulomb’s Coulomb’s LawLaw and apply it to problems and apply it to problems involving electric forces.involving electric forces.

• Define the Define the electronelectron, the , the coulombcoulomb, and the , and the microcoulombmicrocoulomb as units of electric charge.as units of electric charge.

Objectives: After finishing this unit you should be able to:

• Define the electric field and explain what determines its magnitude and direction.

• Discuss electric field lines and the meaning of permittivity of space.

• Write and apply formulas for the electric field intensity at known distances from point charges.

• Write and apply Gauss's law for fields around surfaces of known charge densities.

16.1 Static Electricity; Electric Charge and Its Conservation

When a rubber rod is rubbed against fur, electrons When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod.are removed from the fur and deposited on the rod.

The rod is said to be The rod is said to be negatively chargednegatively charged because of an because of an excessexcess of electrons. The fur is said to be of electrons. The fur is said to be positively positively chargedcharged because of a because of a deficiencydeficiency of electrons. of electrons.

Electrons Electrons move from move from fur to the fur to the rubber rod.rubber rod.

positive

negative

+ + + +

--

--

11.1

Two Negative Charges RepelTwo Negative Charges Repel1. Charge the rubber rod by rubbing against fur.1. Charge the rubber rod by rubbing against fur.

2. Transfer electrons from rod to each pith ball.2. Transfer electrons from rod to each pith ball.

The two negative charges repel each other.The two negative charges repel each other.

The Two Types of ChargeThe Two Types of Charge

fur

Rubber

fur

RubberRubber

AttractionAttraction

Note that the negatively charged Note that the negatively charged (green)(green) ball is ball is attractedattracted to the positively charged to the positively charged (red)(red) ball.ball.

Opposite Charges Attract!Opposite Charges Attract!

silk

glass

silk

glass

16.2 Electric Charge in the Atom

Atom:

Nucleus (small, massive, positive charge)

Electron cloud (large, very low density, negative charge)

11.2

16.2 Electric Charge in the Atom

Atom is electrically neutral.

Rubbing charges objects by moving electrons from one to the other.

16.2 Electric Charge in the Atom

Polar molecule: neutral overall, but charge not evenly distributed

16.3 Insulators and Conductors

Conductor:

Charge flows freely

Metals

Insulator:

Almost no charge flows

Most other materials

Some materials are semiconductors.

16.4 Induced Charge; the Electroscope

Charging Spheres by InductionCharging Spheres by Induction

--- - ---- - -

Uncharged SpheresUncharged Spheres Separation of Charge--- - ---- - ---- - -

Isolation of Spheres Charged by Induction

----

++++

------------

++++++++++

----

++++

------------

++++++++++ +

++ +

-

-- -+

++ +

+

++ +

-

-- -

-

--

-- -

InductionInduction

Electrons Electrons RepelledRepelledElectrons Electrons RepelledRepelled

11.3

Induction for a Single SphereInduction for a Single Sphere

--- - ---- - -

Uncharged Sphere Separation of Charge

Electrons move to ground.

Charged by Induction

+

++ +

+

++ +

+

++ +

InductionInduction

----

--- - -

++++

------------

--- - -

++++

--- - ---- - -

++++++++++

------------

- - - -

----

++++

----------------

++++++++++

------------

The Quantity of ChargeThe Quantity of ChargeThe The quantity of chargequantity of charge (q)(q) can be defined in can be defined in terms of the number of electrons, but the terms of the number of electrons, but the Coulomb (C)Coulomb (C) is a better unit for later work. A is a better unit for later work. A temporarytemporary definition might be as given below:definition might be as given below:

The Coulomb: 1 C = 6.25 x 1018 electronsThe Coulomb: 1 C = 6.25 x 1018 electrons

Which means that the charge on a single electron is:Which means that the charge on a single electron is:

1 electron: e- = -1.6 x 10-19 C1 electron: e- = -1.6 x 10-19 C

Units of ChargeThe The coulombcoulomb (selected for use with (selected for use with electric currents) is actually a electric currents) is actually a very large very large unitunit for static electricity. Thus, we often for static electricity. Thus, we often encounter a need to use the metric encounter a need to use the metric prefixes.prefixes.

1 C = 1 x 10-6 C1 C = 1 x 10-6 C 1 nC = 1 x 10-9 C1 nC = 1 x 10-9 C

1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C

16.4 Induced Charge; the Electroscope

The electroscope can be used for detecting charge:

16.4 Induced Charge; the ElectroscopeThe electroscope can be charged either by conduction or by induction.

16.4 Induced Charge; the Electroscope

The charged electroscope can then be used to determine the sign of an unknown charge.

16.5 Coulomb’s Law

Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them.

Coulomb’s Law

The force of attraction or repulsion between two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

The force of attraction or repulsion between two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

F

r

FF

q

q q’

q’- +

- -

2

'qqF

r 2

'qqF

r

11.4

16.5 Coulomb’s LawThe force is along the line connecting the

charges, and is attractive if the charges are opposite, and repulsive if they are the same.

16.5 Coulomb’s Law

Unit of charge: coulomb, C

The proportionality constant in Coulomb’s law is then:

Charges produced by rubbing are typically around a microcoulomb:

16.5 Coulomb’s Law

Charge on the electron:

Electric charge is quantized in units of the electron charge.

16.5 Coulomb’s Law

The proportionality constant k can also be written in terms of , the permittivity of free space:

(16-2)

ProblemProblem--Solving StrategiesSolving Strategies

1. Read, draw, and label a sketch showing all 1. Read, draw, and label a sketch showing all given information in appropriate given information in appropriate SI unitsSI units..

2. Do not confuse sign of charge with sign of 2. Do not confuse sign of charge with sign of forces. forces. Attraction/RepulsionAttraction/Repulsion determines the determines the direction (or sign) of the force.direction (or sign) of the force.

3. 3. Resultant forceResultant force is found by considering force is found by considering force due to each charge due to each charge independentlyindependently. Review . Review module on module on vectorsvectors, if necessary., if necessary.

4. For forces in equilibrium: 4. For forces in equilibrium: FFxx = 0 = = 0 = FFyy = 0.= 0.

16.5 Coulomb’s LawCoulomb’s law strictly applies only to point charges.

Superposition: for multiple point charges, the forces on each charge from every other charge can be calculated and then added as vectors.

Summary of Formulas:Summary of Formulas:

Like Charges Repel; Unlike Charges Attract.Like Charges Repel; Unlike Charges Attract.

1 electron: e- = -1.6 x 10-19 C1 electron: e- = -1.6 x 10-19 C

1 C = 1 x 10-6 C1 C = 1 x 10-6 C 1 nC = 1 x 10-9 C1 nC = 1 x 10-9 C

1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C

29

2

N m9 x 10

Ck

2

'kqqF

r

16.6 Solving Problems Involving Coulomb’s Law and Vectors

The net force on a charge is the vector sum of all the forces acting on it.

16.6 Solving Problems Involving Coulomb’s Law and Vectors

Vector addition review:

16.7 The Electric Field

The electric field is the force on a small charge, divided by the charge:

(16-3)

16.7 The Electric Field

For a point charge:

(16-4a)

(16-4b)

11.5

16.7 The Electric Field

Force on a point charge in an electric field:

(16-5)

Superposition principle for electric fields:

16.7 The Electric Field

Problem solving in electrostatics: electric forces and electric fields

1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions

2. Calculate forces using Coulomb’s law

3. Add forces vectorially to get result

16.8 Field Lines

The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

16.8 Field Lines

The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge.

The electric field is stronger where the field lines are closer together.

16.8 Field Lines

Examples of EExamples of E--Field LinesField LinesTwo equal but Two equal but oppositeopposite charges.charges.

Two Two identical identical charges (both +).charges (both +).

Notice that lines Notice that lines leave +leave + charges and charges and enterenter -- charges.charges.

Also, Also, EE is is strongeststrongest where field lines are where field lines are most densemost dense..

16.8 Field Lines

The electric field between two closely spaced, oppositely charged parallel plates is constant.

16.8 Field Lines

Summary of field lines:

1. Field lines indicate the direction of the field; the field is tangent to the line.

2. The magnitude of the field is proportional to the density of the lines.

3. Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

16.9 Electric Fields and Conductors

The static electric field inside a conductor is zero – if it were not, the charges would move.

The net charge on a conductor is on its surface.

16.9 Electric Fields and Conductors

The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.

16.10 Gauss’s Law

Electric flux:

(16-7)

Electric flux through an area is proportional to the total number of field lines crossing the area.

11.6

16.10 Gauss’s Law

Flux through a closed surface:

16.10 Gauss’s Law

The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law:

(16-9)

This can be used to find the electric field in situations with a high degree of symmetry.

The Density of Field LinesThe Density of Field Lines

NGaussian Surface

N

A

Line density

Gauss’s Law: The field E at any point in space is proportional to the line density at that point.

Gauss’s Law: The field E at any point in space is proportional to the line density at that point.

A

Radius r

rr

Radius r

rr

Gauss’s LawGauss’s Law

Gauss’s Law: The net number of electric field lines crossing any closed surface in an outward direction is numerically equal to the net total charge within that surface.

Gauss’s Law:Gauss’s Law: The net number of electric field The net number of electric field lines crossing any closed surface in an outward lines crossing any closed surface in an outward direction is numerically equal to the net total direction is numerically equal to the net total charge within that surface.charge within that surface.

0N EA q

If we represent If we represent q q as as net enclosed net enclosed positive chargepositive charge , we can write , we can write rewrite Gauss’s law as:rewrite Gauss’s law as: 0

qEA