Physics 1502: Lecture 2 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, solutions etc. Homework #1:Homework #1: –On Masterphysics this Friday Homeworks posted on Masteringphysics –You need to register (included in cost of book) –Go to masteringphysics.com and register –Course ID: MPCOTE33308 Labs: Begin in two weeks

Today’s Topic : End of Chapter 20 –Define Electric Field in terms of force on "test charge" –Electric Field Lines –Example Calculations –Continuous charge distributions => integrate –Moving charges: Use Newton’s law Demonstration of Mastering Physics

Coulomb's Law SI Units: r in meters q in Coulombs F in Newtons  F 12 = 1 4   q1q2q1q2 r2r2 1 = N m 2 /C 2 r q2q2 r F 12 q1q1 F 21 r Charles Coulomb ( )

Electric Fields Introducing the Electric Field: a quantity, which is independent of that charge q, and depends only upon its position relative to the collection of charges. The force, F, on any charge q due to some collection of charges is always proportional to q: A FIELD is something that can be defined anywhere in space it can be a scalar field (e.g., a Temperature Field) it can be a vector field (as we have for the Electric Field)

Lecture 2, ACT 1 Two charges, Q 1 and Q 2, fixed along the x-axis as shown, produce an electric field E at a point (x,y) = (0,d) which is directed along the negative y-axis. –Which of the following statements is true? (a) Both charges Q 1 and Q 2 must be positive. Q2Q2 Q1Q1 x y E d (b) Both charges Q 1 and Q 2 must be negative. (c) The charges Q 1 and Q 2 must have opposite signs.

How Can We Visualize the E Field? Vector Maps: arrow length indicates vector magnitude + chg + O Graphs: E x, E y, E z as a function of (x, y, z) E r, E , E  as a function of (r, ,  ) x ExEx

Example Consider a point charge fixed at the origin of a co-ordinate system as shown. –The following graphs represent the functional dependence of the Electric Field. Q x y r  r 0 ErEr  0 22 ErEr As the distance from the charge increases, the field falls off as 1/r 2. At fixed r, the radial component of the field is a constant, independent of  !! E

Lecture 2, ACT 2 x Consider a point charge fixed at the origin of a co-ordinate system as shown. –Which of the following graphs best represents the functional dependence of the Electric Field at the point (r,  )? Q y r  Fixed r>0  0 22 ExEx  0 22 ExEx  0 22 ExEx

Another Way to Visualize E... The Old Way: Vector Maps Lines leave positive charges and return to negative charges Number of lines leaving/entering charge = amount of charge Tangent of line = direction of E Density of lines = magnitude of E + O A New Way: Electric Field Lines + chg - chg + O  O

 x y a a +Q -Q r Electric Dipole E E Symmetry E x = ?? E y = ?? Calculate for a pt along x-axis: (x,0) What is the Electric Field generated by this charge arrangement?

Electric Dipole: Field Lines Lines leave positive charge and return to negative charge E x (x,0) = 0 What can we observe about E? E x (0,y) = 0 Field largest in space between the two charges We derived:... for r >> a,

Field Lines from 2 Like Charges Note the field lines from 2 like charges are quite different from the field lines of 2 opposite charges (the electric dipole) There is a zero halfway between charges r>>a: looks like field of point charge (+2q) at origin.

Lecture 2, ACT 3 Consider a dipole aligned with the y-axis as shown. –Which of the following statements about E x (2a,a) is true? +Q x y a a -Q a 2a (a) E x (2a,a) < 0 (b) E x (2a,a) = 0(c) E x (2a,a) > 0 q1q1 q2q2

Electric Dipole x y a a +Q -Q Coulomb Force Radial   Ey Q yaya y ,               Ey Qay y a y y ,          E Now calculate for a pt along y-axis: (0,y) E x = ?? E y = ?? What is the Electric Field generated by this charge arrangement?

Electric Dipole x y a +Q -Q Case of special interest: (antennas, molecules) r > > a r For pts along x-axis: For pts along y-axis:  Er Qa r y,   For r >>a,  Er Qa r y ,   a  Er Qar r a r y ,         

Electric Dipole Summary x y a a +Q -Q Case of special interest: (antennas, molecules) r > > a r Along y-axis  Er Qa r y ,   Along x-axis  Er Qa r y,   Along arbitrary angle  dipole moment with

Electric Fields from Continuous Charge Distributions Principles (Coulomb's Law + Law of Superposition) remain the same. Only change:     See Examples in text P r qq EE r

Charge Densities How do we represent the charge “Q” on an extended object? total charge Q small pieces of charge dq Surface of charge:   = charge per unit area dq =  dA Line of charge:  = charge per unit length dq = dx Volume of charge:   = charge per unit volume dq =  dV

r E(r) = ? Example: Infinite line of charge

How do we approach this calculation? r E(r) = ? In words: “add up the electric field contribution from each bit of charge, using superposition of the results to get the final field” In practice: Use Coulomb’s Law to find the E field per segment of charge Plan to integrate along the line… x: from  to  OR  : from  to   Any symmetries ? This may help for easy cancellations.

Infinite Line of Charge x y dx r' r  dE Charge density = We need to add up the E field contributions from all segments dx along the line.

Infinite Line of Charge x y dx r' r  dE We use Coulomb’s Law to find dE: But x and  are not independent! x = r tan  dx = r sec 2  d  And what is r’ in terms of r? But what is dq in terms of dx? Therefore,

Infinite Line of Charge Components: x y dx r' r  dE Integrate: EdE d r xx            si n / / EdE d r yy            cos / / ExEx  EyEy

Infinite Line of Charge Solution: Conclusion: The Electric Field produced by an infinite line of charge is: –everywhere perpendicular to the line – is proportional to the charge density –decreases as 1/r x y dx r' r  dE 0sin 2/ 2/        d 2cos 2/ 2/        d

Lecture 2, ACT 4 Consider a circular ring with a uniform charge distribution ( charge per unit length) as shown. The total charge of this ring is +Q. The electric field at the origin is (a) zero R x y (b) (c)

Summary Electric Field Distibutions Dipole ~ 1 / R 3 Point Charge ~ 1 / R 2 Infinite Line of Charge ~ 1 / R

Motion of Charged Particles in Electric Fields Remember our definition of the Electric Field, And remembering Physics 1501, Now consider particles moving in fields. Note that for a charge moving in a constant field this is just like a particle moving near the earth’s surface. a x = 0 a y = constant v x = v ox v y = v oy + at x = x o + v ox ty = y o + v oy t + ½ at 2

Motion of Charged Particles in Electric Fields Consider the following set up, e-e- For an electron beginning at rest at the bottom plate, what will be its speed when it crashes into the top plate? Spacing = 10 cm, E = 100 N/C, e = 1.6 x C, m = 9.1 x kg

Motion of Charged Particles in Electric Fields e-e- v o = 0, y o = 0 v f 2 – v o 2 = 2a  x Or,

Homework #1 on Mastering Physics –From Chapter 20 Recap of today’s lecture Define Electric Field in terms of force on "test charge" Electric Field Lines Example Calculations Continuous charge distributions => integrate Moving charges: Use Newton’s law