Green sheet Online HW assignments Practice Problems Course overview See course website OVERVIEW C 2012 J. Becker

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Chapter 21A Electric Field and Coulomb’s Law Electric charge (sec. 21.1) Conductors, insulators, and induced charge (sec. 21.2) Coulomb’s Law (sec. 21.3) Electric field lines (sec. 21.6) C 2012 J. Becker

Learning Goals - we will learn: The nature of electric charge. How objects become electrically charged. How to use Coulomb’s Law to calculate the electric force between charges. How to calculate the electric field caused by electric charges. How to use the idea of electric field lines to visualize electric fields.

Electric charge Protons have positive charge Electrons have negative charge Opposite signs attract Similar signs repel Electric field – used to calculate force between charges C 2012 J. Becker

Photocopiers are amazing devices. They use electric charge to hold fine dust (toner) in patterns until the pattern may be transferred to paper and made permanent with heat.

LITHIUM (Li) ELEMENT Atom: electrically neutral 3 protons and 3 elec. Positive ion: missing one electron so net charge is positive Negative ion: has added electron so net charge is negative

A positive charge and a negative charge attract each other. Two positive charges (or two negative charges) repel each other.

Figure 21.5 Electric forces in action

Figure 21.1a

Figure 21.1b

Figure 21.1c

When you run 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 run 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 Q21.2

Figure 21.6a Copper is a good conductor of electricity; Glass and nylon are good insulators

Figure 21.6b

Figure 21.6c

CHARGING A METAL SPHERE BY INDUCTION Charges are free to move in a conductor but are tightly bound in an insulator. The earth (“ground”) is a large conductor having many free charges.

POLARIZED insulator

In an insulator the charges can move slightly (called polarization of the insulator). A piece of paper is attracted to a charged comb because the positive charges are closer to the negatively charged comb (in the upper figure). CHARGED COMB ATTRACTS A PIECE OF PAPER

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. Q21.3 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. Q21.4 A positively-charged piece of plastic exerts an attractive force on an electrically neutral piece of paper. This is because

An uncharged conductor can attract the charge imparted to paint droplets. Excess charges can flow to or from “ground” -e Car door

Figure 21.2 The imaging drum is aluminum coated with selenium, which changes from an insulator to a conductor when illuminated with light. LASER PRINTER USES CHARGED TONER

FORCE between two charges is given by Coulomb’s Law: | F | = k | Q q o | / r 2

We can use our notion of the gravitational field to form the concept of an ELECTRIC FIELD (E) Recall force between two masses: F = m g g is the gravitational field (9.8 m/sec 2 ) | F | = G | M m | / r 2 The force between two charges Q and q o is given by: F = q o E | F | = k | Q q o | / r 2

Coulomb’s Law: | F | = k | Q q o | / r 2 Rearranged: | F | = | q o [k Q/r 2 ] | Gives us: F = q o E where the electric field E is: | E | = | k Q / r 2 |

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 Q21.5 Charge #1 Charge #2 Charge #3 +q+q +q+q –q–q x 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 Q21.6 Charge #1 Charge #2 Charge #3 +q+q +q+q –q–q x 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 Q21.7 Charge #1 Charge #2 Charge #3 +q+q –q–q –q–q x 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 Q21.8 Charge #1 Charge #2 Charge #3 +q+q –q–q –q–q x y

ELECTRIC FIELD LINES START AND END AT ELECTRIC CHARGES An electric charge is surrounded by an electric field just as a mass is surrounded by a gravitational field.

Electric field and equipotential lines are perpendicular to each other In Lab #2 a voltmeter is used to measure the equipotential lines (in Volts) in order to determine the magnitude and direction of the electric field lines.

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 Q21.9 Charge #1 Charge #2 –q–q +q+q x 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 Q21.10 Charge #1 Charge #2 –q–q +q+q x 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 Q21.11 Charge #1 Charge #2 –q–q –q–q x 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 Q21.12 Charge #1 Charge #2 –q–q –q–q x y P A. the +x-direction.B. the –x-direction. C. the +y-direction. D. the –y-direction. E. none of the above

A PROTON (+e) is released from rest in an electric field. At ANY later time, the velocity of the proton A. is in the direction of the electric field at the position of the proton. B. is directly opposite the direction of the electric field at the position of the proton. C. is perpendicular to the direction of the electric field at the position of the proton. D. is zero. E. not enough information given to decide Q21.13

A PROTON (+e) is released from rest in an electric field. At ANY later time, the velocity of the proton A. is in the direction of the electric field at the position of the proton. B. is directly opposite the direction of the electric field at the position of the proton. C. is perpendicular to the direction of the electric field at the position of the proton. D. is zero. E. not enough information given to decide Q21.14

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 Q21.15

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 Q21.16

Forces on electron beam in a TV tube (CRT) F = Q E and F = m g (vector equations)

TV tube with electron-deflecting charged plates (orange) F = Q E

see Review

Vectors Review (See Chapter 1) Used extensively throughout course INTRODUCTION: see Ch. 1 C 2012 J. Becker

Vectors are quantities that have both magnitude and direction. An example of a vector quantity is velocity. A velocity has both magnitude (speed) and direction, say 60 miles per hour in a DIRECTION due west. (A scalar quantity is different; it has only magnitude – mass, time, temperature, etc.)

A vector may be composed of its x- and y- components as shown.

Note: The dot product of two vectors is a scalar quantity. The scalar (or dot) product of two vectors is defined as

The vector (or cross) product of two vectors is a vector where the direction of the vector product is given by the right-hand rule. The MAGNITUDE of the vector product is given by:

Figure 21.14

PROFESSIONAL FORMAT