chapter 21: electric force and fields

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson; modified Scott Hildreth 2016

Chapter 21

Electric Charge and Electric Field

Introduction

• Water makes life possible as a solvent for biological molecules. What electrical properties allow it to do this?

• We now begin our study of electromagnetism, one of the four fundamental forces in Nature.

• We start with electric charge and electric fields.

Goals for Chapter 21• Study electric charge & charge conservation

• Learn how objects become charged

• Calculate electric force between objects using Coulomb’s law

F = k|q1q2|/r2 = (1/4π0)|q1q2|/r2

• Learn distinction between electric force and electric field

Goals for Chapter 21

• Calculate the electric field due to many charges

• Visualize and interpret electric fields

• Calculate the properties of electric dipoles

Goals for Chapter 21• Be able to solve this kind of problem: (page 715)

y

x

a

Charge Q

Charge Q is distributed uniformly around a semicircle of radius a.

What is the magnitude and direction of the resulting E field at point P, at the center of curvature of the semicircle?P

Physics from 4A you will need to know!• Forces as vectors

• Establish coordinate frame

• Break into components Fx, Fy, Fz

• Add like components!

• Resolve net vector

• Answers must have three things!

1. Magnitude

2. Direction

3. UNITS

Physics from 4A you will need to know!

• Chapter 4: Forces

• “Links in a Chain” Bridging problem (122)

• Chapter 5: Applications

• “In a Rotating Cone” (162)

• Chapter 6: Energy

• “A Spring That Disobeys Hooke’s Law (193)

Math from 4A you will need to know!

• Integration of continuous variables

Math from 4A you will need to know!

• Integration of continuous variables

Electric charge

• Two positive or two negative charges repel each other.

A positive charge and a negative charge attract each other.

• Check out:

http://www.youtube.com/watch?v=45AAIl9_lsc

Electric charge

• Two positive or two negative charges repel each other.

A positive charge and a negative charge attract each other.

Electric charge

• Two positive or two negative charges repel each other.

A positive charge and a negative charge attract each other.

• Check out Balloons in PhET simulations

Electric charge and the structure of matter

• The particles of the atom are the negative electron, the positive proton, and the uncharged neutron.

You should know this already: Atoms and ions

• A neutral atom has the same number of protons as electrons.

• A positive ion is an atom with one or more electrons removed. A negative ion has gained one or more electrons.

You should know this already: Atoms and ions

Conservation of charge

• Proton & electron have same magnitude charge.

Conservation of charge

• Proton & electron have same magnitude charge.

• All observable charge is quantized in this unit.

“½ e”

Conservation of charge

• Proton & electron have same magnitude charge.

• Universal principle of charge conservation states algebraic sum of all electric charges in any closed system is constant.

+3e – 5e +12e – 43e = -33 e

Conductors and insulators

• A conductor permits the easy movement of charge through it. An insulator does not.

• Most metals are good conductors, while most nonmetals are insulators.

Conductors and insulators

• A conductor permits the easy movement of charge through it.

• An insulator does not.

Conductors and insulators

•Semiconductors are intermediate in their properties between good conductors and good insulators.

Charging by induction

• Start with UNCHARGED conducting ball…

Charging by induction

• Bring a negatively charged rod near – but not touching.

Charging by induction

• Bring a negatively charged rod near – but not touching.

• The negative rod is able to charge the metal ball without losing any of its own charge.

Electric forces on uncharged objects• The charge within an insulator can shift slightly. As a result, an

electric force *can* be exerted upon a neutral object.

Charging by induction

• Bring a negatively charged rod near – but not touching.

• The negative rod is able to charge the metal ball without losing any of its own charge.

Charging by induction• Now connect the conductor to the ground (or neutral “sink”)

• What happens?

Charging by induction• Now connect the conductor to the ground (or neutral “sink”)

• Conductor allows electrons to flow from ball to ground…

Charging by induction• Connect the conductor to the ground (or neutral “sink”)

Charging by induction• The negative rod is able to charge the metal ball without losing

any of its own charge.

Electrostatic painting• Induced positive charge on the metal object attracts the

negatively charged paint droplets. Check out http://www.youtube.com/watch?feature=endscreen&v=zTwkJBtCcBA&NR=1

When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a consequence of rubbing the rod with the fur,

A. the rod and fur both gain mass.

B. the rod and fur both lose mass.

C. the rod gains mass and the fur loses mass.

D. the rod loses mass and the fur gains mass.

E. none of the above

Q21.1

When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a consequence of rubbing the rod with the fur,

A. the rod and fur both gain mass.

B. the rod and fur both lose mass.

C. the rod gains mass and the fur loses mass.

D. the rod loses mass and the fur gains mass.

E. none of the above

A21.1

A. electrons are less massive than atomic nuclei.

B. the electric force between charged particles decreases with increasing distance.

C. an atomic nucleus occupies only a small part of the volume of an atom.

D. a typical atom has many electrons but only one nucleus.

Q21.2

A positively charged piece of plastic exerts an attractive force on an electrically neutral piece of paper. This is because

A positively charged piece of plastic exerts an attractive force on an electrically neutral piece of paper. This is because

A. electrons are less massive than atomic nuclei.

B. the electric force between charged particles decreases with increasing distance.

C. an atomic nucleus occupies only a small part of the volume of an atom.

D. a typical atom has many electrons but only one nucleus.

A21.2

Coulomb’s law – Electric FORCE

• The magnitude of electric force between two point charges is directly proportional to the product of their charges

and

inversely proportional to the square of the distance between them.

Coulomb’s law

• Mathematically:

F = k|q1q2|/r2

= (1/4π0)|q1q2|/r2

• A VECTOR

• Magnitude

• Direction

• Units

Coulomb’s law

• Mathematically:

|F| = k|q1q2|/r2

• “k” = 9 x 109 Newton

meter2/Coulomb2

• “k” = 9 x 109 Nm2/C2

Coulomb’s law

• Mathematically:

|F| = k|q1q2|/r2

= (1/4π0)|q1q2|/r2

• x 10 – 12 C2/Nm2

“Electric Permittivity of Free Space”

Measuring the electric force between point charges

Example 21.1 compares the electric and gravitational forces.

An alpha particle has mass m = 6.64 x 10-27 kg and charge q = +2e = 3.2 x 10-19 C.

Compare the magnitude of the electric repulsion between two alpha particles and their

gravitational attraction

Measuring the electric force between point charges

DRAW the VECTORS!!

An alpha particle has mass m = 6.64 x 10-27 kg and charge q = +2e = 3.2 x 10-19 C.

Compare the magnitude of the electric repulsion between two alpha particles and their

gravitational attraction

Measuring the electric force between point charges

An alpha particle has mass m = 6.64 x 10-27 kg and

charge q = +2e = 3.2 x 10-19 C.

Fe/Fg = ?

Note: Force of Gravity = Gm1m2/r2

where

G= 6.67 x 10-11 Nm2/kg2

Measuring the electric force between point charges

An alpha particle has mass m = 6.64 x 10-27 kg and charge

q = +2e = 3.2 x 10-19 C.

Fe/Fg = 3.1 x 1035!!!!

Force between charges along a line

• Example 21.2 for two charges:

Two point charges, q1 = +25nC, and q2 = -75 nC, separated by r = 3.0 cm.

What is the Force of q1 on q2? What is the Force of q2 on q1?

Force between charges along a line

• Example 21.2 for two charges:

Two point charges, q1 = +25nC, and q2 = -75 nC, separated by r = 3.0 cm.

What is the Force of q1 on q2?

Step 1: Force is a vector – create a coordinate system FIRST!

x

Force between charges along a line

• Example 21.2 for two charges:

Two point charges, q1 = +25nC, and q2 = -75 nC, separated by r = 3.0 cm. What is the Force of q1 on

q2? What is the force of q2 on q1?

Force between charges along a line

• Example 21.2 for two charges:

Two point charges, q1 = +25nC, and q2 = -75 nC, separated by r = 3.0 cm. What is the Force of q1 on

q2? What is the force of q2 on q1?

Force between charges along a line

• Example 21.2 for two charges:

Two point charges, q1 = +25nC, and q2 = -75 nC, separated by r = 3.0 cm.

What is the Force of q1 on q2?

F of q1 on q2 = F12 = 0.019N <-x>

F12

x

Force between charges along a line (Example 21.3)

• Two point charges

– q1 = +1.0nC at x = +2.0 cm, & q2 = -3.0 nC at x = +4.0 cm.

What is the Force of q1 & q2 on

q3 = + 5.0 nC at x = 0?

Force between charges along a line (Example 21.3)

• Two point charges

– q1 = +1.0nC at x = +2.0 cm, & q2 = -3.0 nC at x = +4.0 cm.

What is the Force of q1 & q2 on

q3 = + 5.0 nC at x = 0?

Vector addition of electric forces- Example 21.4

Two equal positive charges, q1 = q2 = +2.0C are located at x=0, y = + 0.30 m

& x=0, y = -.30 m respectively.

What is the Force of q1 and q2 on Q = + 5.0 C at x = 0.40 m, y = 0?

• You must use COMPONENTS!

• No Fruit Salad!! Break vectors into x & y COMPONENTS!!!

Vector addition of electric forces – Example 21.4

Vector addition of electric forces• Example 21.4 shows that we must use vector addition when

adding electric forces.

Vector addition of electric forces• Example 21.4 shows that we must use vector addition when

adding electric forces.

Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charges #1 and #2 are positive (+q) and charge #3 is negative (–q).

The net electric force that charges #2 and #3 exert on charge #1 is in

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

Q21.3

Charge #1

Charge #2

Charge #3

+q

+q

–qx

y

Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charges #1 and #2 are positive (+q) and charge #3 is negative (–q).

The net electric force that charges #2 and #3 exert on charge #1 is in

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

A21.3

Charge #1

Charge #2

Charge #3

+q

+q

–qx

y

Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charge #1 is positive (+q) and charges #2 and #3 are negative (–q).

The net electric force that charges #2 and #3 exert on charge #1 is in

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

Q21.4

Charge #1

Charge #2

Charge #3

+q

–q

–qx

y

Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charge #1 is positive (+q) and charges #2 and #3 are negative (–q).

The net electric force that charges #2 and #3 exert on charge #1 is in

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

A21.4

Charge #1

Charge #2

Charge #3

+q

–q

–qx

y

Electric field• A charged body produces an electric field in the space around it

Electric field• We use a small test charge q0 to find out if an electric field is

present.

Electric field• We use a small test charge q0 to find out if an electric field is

present.

Definition of the electric field

• E fields are VECTOR fields – and solutions to problems require magnitude, direction, and units.

Definition of the electric field

• E fields are VECTOR fields – and solutions to problems require magnitude, direction, and units.

Definition of the electric field

• E fields are VECTOR fields – and solutions to problems require magnitude, direction, and units.

– Need Coordinate System for direction!

– Need Units Force/Charge = Newtons/Coulomb = N/C

– E = (kq0q/r2)/q0 = (kq/r2) (-r direction!) N/C

Electric field of a point charge

• E fields from positive charges point AWAY from the charge

Electric field of a point charge

• E fields point TOWARDS a negative charge:

Electric-field vector of a point charge

• Example 21.6 - the vector nature of the electric field.

• Charge of -8.0 nC at origin.

What is E field at P = (Px,Py) = (1.2, -1.6)?

Electric-field vector of a point charge

• Example 21.6 - the vector nature of the electric field.

• Charge of -8.0 nC at origin.

What is E field at P = (Px,Py) = (1.2, -1.6)?

If you see BOLD, it is a vector!Magnitude, Units, & Direction

Required!

Electric-field vector of a point charge

• What is E field at P = (Px,Py) = (1.2, -1.6)?

• E = -11N/C x + 14 N/C y(2 sig figs)

• |E| = 18 N/C

in direction = arctan Ey/Ex

= 127 degrees from +x or 53˚ from –x

axis

Electric-field vector of a point charge

• What is E field at P = (Px,Py) = (1.2, -1.6)?

• E = -11N/C x + 14 N/C y

• |E| = 18 N/C

in direction = arctan Ey/Ex

= -53 degrees

A positive point charge +Q is released from rest in an electric field. At any later time, the velocity of the point charge

A. is in the direction of the electric field at the position of the point charge.

B. is directly opposite the direction of the electric field at the position of the point charge.

C. is perpendicular to the direction of the electric field at the position of the point charge.

D. is zero.

E. not enough information given to decide

Q21.5

A positive point charge +Q is released from rest in an electric field. At any later time, the velocity of the point charge

A. is in the direction of the electric field at the position of the point charge.

B. is directly opposite the direction of the electric field at the position of the point charge.

C. is perpendicular to the direction of the electric field at the position of the point charge.

D. is zero.

E. none of the above are true.

A21.5

Electron in a uniform field

• Example 21.7: Determine force on a charge in a known uniform electric field.

Electron in a uniform field

• Plates 1.0 cm apart, connected to 100 V battery creating a uniform E field of 100V/0.01 m = 10,000 N/C

• What’s a VOLT?

Electron in a uniform field

• Plates 1.0 cm apart, connected to 100 V battery creating a uniform E field of 100V/0.01 m = 10,000 N/C

• What’s a VOLT?

– A unit of potential energy/charge (Joules/Coulomb)

– Like “electrical water pressure”

– 1 Volt/meter = 1 Newton/Coulomb (E field)

Electron in a uniform field

• Plates 1.0 cm apart, connected to 100 V battery creating a uniform E field of 100V/0.01 m = 10,000 N/C

• Electron released from rest; what is acceleration? Final velocity? Total KE? Time to travel 1.0 cm?

Superposition of electric fields• The total electric field at a point is the vector sum of the fields due

to all the charges present.

• Electric DIPOLE fields are important!!

Two point charges and a point P lie at the vertices of an equilateral triangle as shown. Both point charges have the same magnitude q but opposite signs. There is nothing at point P.

The net electric field that charges #1 and #2 produce at point P is in

Q21.6

Charge #1

Charge #2

–q

+qx

y

P

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

Two point charges and a point P lie at the vertices of an equilateral triangle as shown. Both point charges have the same magnitude q but opposite signs. There is nothing at point P.

The net electric field that charges #1 and #2 produce at point P is in

A21.6

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

Charge #1

Charge #2

–q

+qx

y

P

Dipole E fields

Dipole E fields

Dipole E fields

Dipole E fields

Two point charges and a point P lie at the vertices of an equilateral triangle as shown. Both point charges have the same negative charge (–q). There is nothing at point P.

The net electric field that charges #1 and #2 produce at point P is in

Q21.7

Charge #1

Charge #2

–q

–qx

y

P

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

Two point charges and a point P lie at the vertices of an equilateral triangle as shown. Both point charges have the same negative charge (–q). There is nothing at point P.

The net electric field that charges #1 and #2 produce at point P is in

A21.7

A. the +x-direction. B. the –x-direction.

C. the +y-direction. D. the –y-direction.

E. none of the above

Charge #1

Charge #2

–q

–qx

y

P

Field of a charged line segment Example 21.10• Line of charge => use as Charge/Meter = Q/2a

• Set up dQ = dy; find field dEx and dEy at P (in x and y separately) from kdQ/r2

• Integrate from y = -a to +a (Check out Youtube)

Field of a charged line segment Example 21.10• Find E = Ex only = kQ/x(x2+a2)½ (+x direction)

• For a >>x, E = k(Q/a)/x[(x/a)2+1)]½ ~ k(Q/a)/x ~ 2k x

• So for LONG wire, field nearby goes as 1/r…

Q21.9

A. the +x-direction.

B. the –x-direction.

C. the +y-direction.

D. the –y-direction.

E. none of the above

Positive charge is uniformly distributed around a semicircle.

The electric field that this charge produces at the center

of curvature P is in

A21.9

Positive charge is uniformly distributed around a semicircle.

The electric field that this charge produces at the center

of curvature P is in

A. the +x-direction.

B. the –x-direction.

C. the +y-direction.

D. the –y-direction.

E. none of the above

Field of a ring of charge

• Example 21.9 - a uniform ring of charge.

• Any continuous charge distribution – INTEGRALS!

• Always start with a small “dQ”, and calculate dF or dE created from that dQ.

• Remember dF & dE are still VECTORS!

Field of a ring of charge

• Example 21.9 - a uniform ring of charge.

• Use as Charge/Meter for “charge density” [C/m]

• dQ = charge density) x ds (length of segment)

• dQ = (Coulombs/meter) x ds (meters) = Coulombs

Field of a uniformly charged disk

• Example 21.11 – Superposition of multiple rings!

• Surface of charge – use = Charge/Area = Q/R2

• Find dQ = dA where dA = (2r)dr

Field of a uniformly charged disk

• Find dQ = dA where dA = (2r)dr

• dEx = [kdQ/(x2 + r2)]cos()

Field of a uniformly charged disk

• Ex = k2x direction)

• At R >>x, Ex = k2

R/x

Field of two oppositely charged infinite sheets

• Example 21.12- Superposition of two sheets

Electric field lines

• An electric field line is an imaginary line or curve whose tangent at any point is the direction of the electric field vector at that point.

Electric field lines of point charges

• Figure 21.28 below shows the electric field lines of a single point charge and for two charges of opposite sign and of equal sign.

The illustration shows the electric field lines due to three point charges. The electric field is strongest

A. where the field lines are closest together.

B. where the field lines are farthest apart.

C. where adjacent field lines are parallel.

D. none of the above

Q21.8

The illustration shows the electric field lines due to three point charges. The electric field is strongest

A. where the field lines are closest together.

B. where the field lines are farthest apart.

C. where adjacent field lines are parallel.

D. none of the above

A21.8

Dipole Moments (Torques!)

• An electric dipole is a pair of point charges having equal but opposite sign and separated by a distance.

Electric field of a dipole creates a dipole moment

• What is the field ABOVE the positive charge??

Force and torque on a dipole

• Figure below left shows the force on a dipole in an electric field.