Chapter 21

Copyright © 2012 Pearson Education Inc.

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

Lectures by Wayne Anderson

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



• Calculate the electric field due to many charges

• Visualize and interpret electric fields

• Calculate the properties of electric dipoles



• 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” page 127

• Exercises/Problems 4.2, 37, 54, 62* (calculus)

• Chapter 5: Applications

• Section 5.2 (Dynamics) ex 5.10, 5.11, 5.12

• Section 5.4 (Circular) ex 5.20, 5.21

• “In a Rotating Cone” page 162


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




Electric charge



• 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.




• All observable charge is quantized in this unit.

“½ e”




• 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.



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

• An insulator does not.



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


Charging by induction

• Start with UNCHARGED conducting ball…



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



• 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.



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

• What happens?



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

• Conductor allows electrons to flow from ball to ground…



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



• 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.


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.


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



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



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

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

Fe/Fg = ?

Remember Fg = Gm1m2/r2 & G= 6.67 x 10-11 Nm2/kg2



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?





What is the Force of q1 on q2?

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

x




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?




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?





What is the Force of q1 on q2?

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

F12

x



• Example 21.3 for three charges:

Two point charges, q1 = +1.0nC at x = +2.0 cm, and q2 = -3.0 nC at x = +4.0 cm. What is the Force of q1

and q2 on q3 = + 5.0 nC at x = 0?



• Example 21.3 for three charges:

Two point charges, q1 = +1.0nC at x = +2.0 cm, and q2 = -3.0 nC at x = +4.0 cm. What is the Force of q1

and q2 on q3 = + 5.0 nC at x = 0?


Vector addition of electric forces

• Example 21.4 shows that we must use vector addition when adding electric forces.

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










© 2012 Pearson Education, Inc.

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.


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).





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).





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).





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.







– 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)?



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

• E = -11N/C x + 14 N/C y

• |E| = 17.8 N/C ~ 18 N/C

in direction = arctan Ey/Ex

= -52 degrees



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

• E = -11N/C x + 14 N/C y

• |E| = 17.8 N/C ~ 18 N/C

in direction = arctan Ey/Ex

= -52 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 requires the force on a charge that is in a known 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

• Electron released from rest; what is a? 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.

• Example 21.8 for an electric dipole.


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





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.


A21.6




Charge #1

Charge #2

–q

+qx

y

P


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.


Q21.7

Charge #1

Charge #2

–q

–qx

y

P





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.


A21.7




Charge #1

Charge #2

–q

–qx

y

P


Q21.9

A. the +x-direction.

B. the –x-direction.

C. the +y-direction.

D. the –y-direction.


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.



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 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


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…


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



• Find dQ = dA where dA = (2r)dr

• dEx = [kdQ/(x2 + r2)]cos()



• 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


Electric dipoles

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

• Figure 21.30 at the right illustrates the water molecule, which forms an electric dipole.


Force and torque on a dipole

• Figure 21.31 below left shows the force on a dipole in an electric field.

• Follow the discussion of force, torque, and potential energy in the text.

• Follow Example 21.13 using Figure 21.32 below right.


Electric field of a dipole

• Follow Example 21.14 using Figure 21.33.


Laser printer

• A laser printer makes use of forces between charged bodies.

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Chapter 21

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