Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Chapter 21 Electric Charge and Electric Field

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. 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|q 1 q 2 |/r 2 = (1/4π  0 )|q 1 q 2 |/r 2 Learn distinction between electric force and electric field

Copyright © 2012 Pearson Education Inc. Goals for Chapter 21 Calculate the electric field due to many charges Visualize and interpret electric fields Calculate the properties of electric dipoles

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. Physics from 4A you will need to know! Forces as vectors Establish coordinate frame Break into components F x, F y, F z Add like components! Resolve net vector Answers must have three things! 1.Magnitude 2.Direction 3.UNITS

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. Math from 4A you will need to know! Integration of continuous variables

Copyright © 2012 Pearson Education Inc. Electric charge Two positive or two negative charges repel each other. A positive charge and a negative charge attract each other. Check out: ch?v=45AAIl9_lsc

Copyright © 2012 Pearson Education Inc. Electric charge Two positive or two negative charges repel each other. A positive charge and a negative charge attract each other.

Copyright © 2012 Pearson Education Inc. 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 simulationsBalloons

Copyright © 2012 Pearson Education Inc. Electric charge and the structure of matter The particles of the atom are the negative electron, the positive proton, and the uncharged neutron.

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. You should know this already: Atoms and ions

Copyright © 2012 Pearson Education Inc. Conservation of charge Proton & electron have same magnitude charge.

Copyright © 2012 Pearson Education Inc. Conservation of charge Proton & electron have same magnitude charge. All observable charge is quantized in this unit. “½ e”

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. Conductors and insulators A conductor permits the easy movement of charge through it. An insulator does not.

Copyright © 2012 Pearson Education Inc. Conductors and insulators Semiconductors are intermediate in their properties between good conductors and good insulators.

Copyright © 2012 Pearson Education Inc. Charging by induction Start with UNCHARGED conducting ball…

Copyright © 2012 Pearson Education Inc. Charging by induction Bring a negatively charged rod near – but not touching.

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. Charging by induction Now connect the conductor to the ground (or neutral “sink”) What happens?

Copyright © 2012 Pearson Education Inc. Charging by induction Now connect the conductor to the ground (or neutral “sink”) Conductor allows electrons to flow from ball to ground…

Copyright © 2012 Pearson Education Inc. Charging by induction Connect the conductor to the ground (or neutral “sink”)

Copyright © 2012 Pearson Education Inc. Charging by induction The negative rod is able to charge the metal ball without losing any of its own charge.

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. Electrostatic painting Induced positive charge on the metal object attracts the negatively charged paint droplets. Check out

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. Coulomb’s law Mathematically: F = k|q 1 q 2 |/r 2 = (1/4π  0 )|q 1 q 2 |/r 2 A VECTOR Magnitude Direction Units

Copyright © 2012 Pearson Education Inc. Coulomb’s law Mathematically: |F| = k|q 1 q 2 |/r 2 “k” = 9 x 10 9 Newton meter 2 /Coulomb 2 “k” = 9 x 10 9 Nm 2 /C 2

Copyright © 2012 Pearson Education Inc. Coulomb’s law Mathematically: |F| = k|q 1 q 2 |/r 2 = (1/4π  0 )|q 1 q 2 |/r 2    x 10 – 12 C 2 /Nm 2     “ Electric Permittivity of Free Space”

Copyright © 2012 Pearson Education Inc. 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 kg and charge q = +2e = 3.2 x C. Compare the magnitude of the electric repulsion between two alpha particles and their gravitational attraction

Copyright © 2012 Pearson Education Inc. Measuring the electric force between point charges DRAW the VECTORS!! An alpha particle has mass m = 6.64 x kg and charge q = +2e = 3.2 x C. Compare the magnitude of the electric repulsion between two alpha particles and their gravitational attraction

Copyright © 2012 Pearson Education Inc. Measuring the electric force between point charges An alpha particle has mass m = 6.64 x kg and charge q = +2e = 3.2 x C. F e /F g = ? Remember F g = Gm 1 m 2 /r 2 & G= 6.67 x Nm 2 /kg 2

Copyright © 2012 Pearson Education Inc. Measuring the electric force between point charges An alpha particle has mass m = 6.64 x kg and charge q = +2e = 3.2 x C. Fe/Fg = 3.1 x !!!!

Copyright © 2012 Pearson Education Inc. 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 q 1 on q 2 ? What is the Force of q 2 on q 1 ?

Copyright © 2012 Pearson Education Inc. 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 q 1 on q 2 ? Step 1: Force is a vector – create a coordinate system FIRST! x

Copyright © 2012 Pearson Education Inc. 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 q 1 on q 2 ? What is the force of q 2 on q 1 ?

Copyright © 2012 Pearson Education Inc. 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 q 1 on q 2 ? What is the force of q 2 on q 1 ?

Copyright © 2012 Pearson Education Inc. 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 q 1 on q 2 ? F of q 1 on q 2 = F 12 = 0.019N F 12 x

Copyright © 2012 Pearson Education Inc. Force between charges along a line 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 = nC at x = 0?

Copyright © 2012 Pearson Education Inc. Force between charges along a line 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 = nC at x = 0?

Copyright © 2012 Pearson Education Inc. 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 = +2.0  C 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 =  C at x = 0.40 m, y = 0?

Copyright © 2012 Pearson Education Inc. Vector addition of electric forces Example 21.4 shows that we must use vector addition when adding electric forces.

Copyright © 2012 Pearson Education Inc. Vector addition of electric forces Example 21.4 shows that we must use vector addition when adding electric forces.

Copyright © 2012 Pearson Education Inc. Vector addition of electric forces Example 21.4 shows that we must use vector addition when adding electric forces.

© 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. E. none of the above Q21.3 Charge #1 Charge #2 Charge #3 +q+q +q+q –q–q x y

© 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. E. none of the above A21.3 Charge #1 Charge #2 Charge #3 +q+q +q+q –q–q x y

© 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 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 –q–q –q–q x y

© 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 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 –q–q –q–q x y

Copyright © 2012 Pearson Education Inc. Electric field A charged body produces an electric field in the space around it

Copyright © 2012 Pearson Education Inc. Electric field We use a small test charge q 0 to find out if an electric field is present.

Copyright © 2012 Pearson Education Inc. Electric field We use a small test charge q 0 to find out if an electric field is present.

Copyright © 2012 Pearson Education Inc. Definition of the electric field E fields are VECTOR fields – and solutions to problems require magnitude, direction, and units.

Copyright © 2012 Pearson Education Inc. Definition of the electric field E fields are VECTOR fields – and solutions to problems require magnitude, direction, and units.

Copyright © 2012 Pearson Education Inc. 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 = (kq 0 q/r 2 )/q 0 = (kq/r 2 ) (-r direction!) N/C

Copyright © 2012 Pearson Education Inc. Electric field of a point charge E fields from positive charges point AWAY from the charge

Copyright © 2012 Pearson Education Inc. Electric field of a point charge E fields point TOWARDS a negative charge:

Copyright © 2012 Pearson Education Inc. Electric-field vector of a point charge Example 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)?

Copyright © 2012 Pearson Education Inc. 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| = 17.8 N/C ~ 18 N/C in direction  = arctan Ey/Ex = -52 degrees

Copyright © 2012 Pearson Education Inc. 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| = 17.8 N/C ~ 18 N/C in direction  = arctan Ey/Ex = -52 degrees

© 2012 Pearson Education, Inc. 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

© 2012 Pearson Education, Inc. 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

Copyright © 2012 Pearson Education Inc. Electron in a uniform field Example 21.7 requires the force on a charge that is in a known electric field.

Copyright © 2012 Pearson Education Inc. 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?

Copyright © 2012 Pearson Education Inc. 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.

© 2012 Pearson Education, Inc. 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–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

© 2012 Pearson Education, Inc. 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–q +q+q x y P

© 2012 Pearson Education, Inc. 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–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

© 2012 Pearson Education, Inc. 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–q –q–q x y P

© 2012 Pearson Education, Inc. 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

© 2012 Pearson Education, Inc. 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

Copyright © 2012 Pearson Education Inc. Field of a ring of charge Example 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!

Copyright © 2012 Pearson Education Inc. Field of a ring of charge Example 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

Copyright © 2012 Pearson Education Inc. Field of a charged line segment Example Line of charge => use as Charge/Meter = Q/2a Set up dQ =  dy; find field dE x and dE y at P (in x and y separately) from kdQ/r 2 Integrate from y = -a to +a

Copyright © 2012 Pearson Education Inc. Field of a charged line segment Example Find E = Ex only = kQ/x(x 2 +a 2 ) 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…

Copyright © 2012 Pearson Education Inc. Field of a uniformly charged disk Example – Superposition of multiple rings! Surface of charge – use  = Charge/Area = Q/  R 2 Find dQ =  dA where dA = (2  r)dr

Copyright © 2012 Pearson Education Inc. Field of a uniformly charged disk Find dQ =  dA where dA = (2  r)dr dEx = [kdQ/(x 2 + r 2 )]cos(  )  

Copyright © 2012 Pearson Education Inc. Field of a uniformly charged disk E x = k2  x direction) At R >>x, E x = k2    R  /x  

Copyright © 2012 Pearson Education Inc. Field of two oppositely charged infinite sheets Example Superposition of two sheets

Copyright © 2012 Pearson Education Inc. 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.

Copyright © 2012 Pearson Education Inc. Electric field lines of point charges Figure below shows the electric field lines of a single point charge and for two charges of opposite sign and of equal sign.

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. 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

Copyright © 2012 Pearson Education Inc. Electric dipoles An electric dipole is a pair of point charges having equal but opposite sign and separated by a distance. Figure at the right illustrates the water molecule, which forms an electric dipole.

Copyright © 2012 Pearson Education Inc. Force and torque on a dipole Figure 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 using Figure below right.

Copyright © 2012 Pearson Education Inc. Electric field of a dipole Follow Example using Figure

Copyright © 2012 Pearson Education Inc. Laser printer A laser printer makes use of forces between charged bodies.