MTE1 results Mean 75% = 90/120 Scores available at Learn@UW, your TAs have exams If your score is an F or a D, talk to us and your TAs for suggestions on how to improve
From last times Electric charges and force Electric Field and ways to calculate it Motion of charges in E-field Gauss Law Today: More on Gauss law and conductors in electrostatic equilibrium Work, potential energy Electric potential
What is the Electric Flux? (Component of E-field to surface) x (surface area) So you need to understand how the E-field is directed respect to each piece of surface through which you have to calculate it Why do we need to calculate it? to use Gauss law to determine E PHYS208, SPRING 2008
How to calculate Electric Flux? Φ E Electric flux through a surface: (component of E-field to surface) x (surface area) Ecosθ is component of E-field perpendicular to surface ˆ s PHYS208, SPRING 2008
GAUSS LAW for any closed surface Flux thru closed surface depends ONLY on the charge enclosed by surface!! But you need to calculate the Flux to determine E The Flux depends on the normal component of E that crosses the surface because of the scalar product in the surface integral PHYS208, SPRING 2008
Quiz 1 +Q +Q PHYS208, SPRING 2008 +Q +2Q 3) 1) 2) Rank fluxes through the blue dashed spherical surfaces for the 3 cases: Qr 3 A) 1)<2)=3) < Q < 2Q a 3 ε B) 3)<1)<2) 0 ε 0 ε 0 C) 1) = 3) < 2) Flux thru spheres: 4πr 2 E(r) D) 1) < 2) < 3)
Conductors in equilibrium electrostatic equilibrium In a conductor in electrostatic equilibrium there is no net motion of charge Property 1: E=0 everywhere inside the conductor E in Conductor slab in an external field E: if E-field not null inside the conductor, free electrons would be accelerated These electrons would not be in equilibrium. E tot =0 E tot = E+E in = 0 When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field The total field inside the conductor is zero PHYS208, SPRING 2008
Property 2: Charge on only conductor surface chose a gaussian surface inside the conductor (as close to the surface as desired) There is no net flux through the gaussian surface (since E=0) Any net charge must reside on the surface (cannot be inside!) E=0 PHYS208, SPRING 2008
Magnitude and direction of E-Field E-field perpendicular to the surface: parallel component to E would put force on charges charges would accelerate along the surface NO equilibrium Applying Gauss s law Field lines PHYS208, SPRING 2008
Remember from P207 1st Newton s law: An object at rest will remain at rest and a moving object will continue to move at constant velocity on a straight line if F=0. r F net = ma r 2nd Newton s law(superposition principle applies): A moving particle has kinetic energy Work: F ds θ Total work and work-energy theorem: W = B A F ds dw = F r d r Infinitesimal displacement s = Fdscosθ >0: Force is in direction of motion <0: Force is opposite to direction of motion =0: Force is perpendicular to direction of motion = K B K A K= 1 2 mv 2
Potential energy Work to lift the block done by external agent (F): W = mgh = mg(yf-yi) > 0 (the agent has to make an effort to move the block) Work done by gravitational field: Wg = -mgh y Conservative Forces: work done by the force is independent on the path and depends only on the starting and ending locations (if the initial and final point coincide => W = 0) yf yi h F Fg = mg It is possible to define the potential energy U Wconservative = -Δ U = Uinitial - Ufinal = ΔK ΔUg = Ug - U0 = -mgh Ug = U0 + mgy for gravitational force and U0 =U(y=0) = 0
Electric force work Consider bringing two positive charges together They repel each other Force is conservative Pushing them together requires work Stop after some distance How much work was done? + +
Calculating the work E.g. Keep fixed, push Q 1 at constant velocity Net force on Q 1? Zero Force from hand on Q 1? 1 Q 1 4πε o R 2 r F Coulomb Q 1 R + r + + F x initial rinitial x x final rfinal Total work done by hand Force in direction of motion r F dx r x final Q = k 1 Q e k 1 e R x R x xinitial x final x initial x initial ( ) 2 dx = x final = = k e Q 1 R x final k e Q 1 R x initial == k e Q 1 r final k e Q 1 r initial > 0 r final < r initial 1 r final > 1 r initial for like charges
Potential energy of 2 point charges If we initially put Q1 very far away (rinitial = ) and move Q1 at distance rfinal =r12 from Q2: W ext = k e Q 1 r final k e Q 1 r initial = k e Q 1 r 12 Q2 Q1 The external agent makes positive work (because charges repel) and changes the potential energy of the system Wext = ΔU = Ufinal - Uinitial >0 => U increases. The energy is stored in the electric field as electric potential energy. We can set: Uinitial = U( ) = 0 (at infinite distance force becomes null). Hence, electric potential energy of 2 charges at distance r: U elec = k e Q 1 r
More about U of 2 charges Like charges U > 0 and work must be done to bring the charges together since they repel (W>0) Unlike charges U < 0 and work is done to keep the charges apart since the attract one the other (W<0)
Quick Quiz Two balls of equal mass and equal charge are held fixed a distance r12 apart, then suddenly released. They fly away from each other, each ending up moving at some constant speed. If the initial distance between them is reduced by a factor of four, their final speeds are A. Two times bigger B. Four times bigger C. Two times smaller D. Four times smaller E. None of the above W el = ΔU = U initial U final = k e Q 1 =0 initially fixed =0 very far away W el = ΔK = K final K initial = K final = 1 2 mv 2 = k e Q 1 r 12 if r 12 = 1/4 r12 => K final = 4 Kfinal and v final = 2 vfinal r 12
U with multiple charges If there are more than two charges, then find U for each pair of charges and add them For three charges:
Quick Quiz How much work would it take YOU to assemble 3 negative charges? A. W = +19.8 mj B. W = 0 mj C. W = -19.8 mj 3µC 5 m 5 m 1µC 5 m 2µC
Electric Potential Work done by E-field produced by some source charge on a test charge q to move it along a path: W el = ΔU = r F Coulomb d r s = qe r d r s = q r E d r s Define: ΔU /q V = Electric potential V scalar quantity and units volts 1 V = 1 J/C V created by some source charge or charge distribution and independent on test charge V can be used to determine potential energy of charge q and source charge system V connected to E due to source charge (more in this week laboratory)
Equipotential lines Lines of constant potential In 3D, surfaces of constant potential (your Lab)
Quick Quiz In the Figure, q1 is a negative source charge and q2 is a test charge. If q2 is initially positive and then is changed to a negative charge of the same magnitude, the potential at the position of q2 due to q1 (A.) increases (B.) decreases (C.) remains the same + - Answer: (c). The potential is established only by the source charge and is independent of the test charge.
Electric potential at point charge Consider one charge as creating electric potential, the other charge as experiencing it Q q V q U Qq ( r) = U Qq r q Qq ( r) = k e ( ) r = k e Q r
Electric Potential of point charge Every point in space has a numerical value for the electric potential V = kq r Distance from source charge +Q Potential energy: U = q0v (q0 test charge) VB > VA Means that work must be done to move the test charge q0 from A to B to overcome the Coulomb repulsive force. B Work done = q o V B -q o V A = ( F r Coulomb ) d r l Differential form: q o dv = r F Coulomb d r l A Electric potential energy=q o V B A q o > 0 y +Q x
Quick Quiz Two points in space A and B have electric potential VA=20 volts and VB=100 volts. How much work does it take to move a +100µC charge from A to B? A. +2 mj B. -20 mj C. +8 mj D. +100 mj E. -100 mj
Superposition: electric potential of dipole y x=-a Superposition of potential from +Q potential from -Q x=+a +Q -Q + = Eg:V(y=0) = kq/a-kq/a=0 In general: for a group of point charges qi V = ke i ri
Electric Potential of a continuous charge distribution Consider a small charge element dq. The potential at some point due to this charge element is Total potential: This value of V used the reference that V = 0 at infinite distance
Van de Graaff generator The high-voltage electrode is a hollow metal dome mounted on an insulated column Charge is delivered continuously to dome by a moving belt of insulating material Large potentials can be developed by repeated trips of the belt Protons can be accelerated through such large potentials ΔU = qδv = ΔK = K f Sparks: electrons from rod to dome excite air molecules that emit light ΔV