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Transcript
Announcements I usually answer questions and comment on essays on Web assign. Answers are available immediately after the due date. Tutorials: Olin 107 with Doug Bonessi Monday 6-8pm Tuesday 6-8pm Thursday 7-9pm . Reading Quizzes Several of you had problems remembering Newton’s second Law, F=ma. You may wish to reread Ch. 5 on the Laws of Motion Keep in mind that a force is a vector, as in acceleration, so they have magnitude and direction. Charge Densities Charge can be localized to discrete points (point charges), or it may be spread out over a volume, a surface or a line •Charge density units C/m3 •Surface charge density units C/m2 •Linear charge density units C/m A cube with side 1 cm has a charge density of = 1 C/m3. What is the charge of the cube? A) 1 C B) 0.01 C = 10 mC C) 10-4 C = 100 m C 1 cm D) 10-6C = 1 mC Coulomb’s Law •Like charges repel, unlike charges attract •Force is directly along a line joining the two charges q1 ke q1q2 Fe rˆ 2 r q2 r q1q2 Fe rˆ 2 4 0 r 0 = 8.85410-12 C2/ (N●m2) •Permittivity of free space •An inverse square law, just like gravity •Can be attractive or repulsive – unlike gravity •Constant is enormous compared to gravity Coulomb’s Law: Applied A Helium nucleus (charge +2e) is separated from one of its electrons (charge –e) by about 3.00 10-11m. What is the force the nucleus exerts on the electron? Is it attractive or repulsive? ke q1q2 9 Nm2/C2 k = 8.98810 Fe e 2 r -11m the force on the electron rHow = just 3.00 10 We calculated from the q1 =of3.204 10-19C does the acceleration the nucleus compare nucleus. does this q2 =compare -1.602with 10-19the C force to that ofHow the electron? theacceleration nucleus from the nucleus electron?is larger A) on The of the A) on the nucleus is twiceisassmaller big The acceleration of the Attractive nucleus FB)e=The - force 0.513 mN Force B) The force on the nucleus is half as big C) The accelerations are equal. C) The forces are equal in magnitude Newton’s Laws and Kinematics Newton’s laws and all the kinematics you learned in 113 are still true! Fnet ma F12 F21 A body in motion tends to stay in motion, therefore changing velocity, i.e. acceleration, requires a force! Also the same forces on different particles can lead to different accelerations, depending on the masses. t' t' dx dv v ;a v adt vt0 ; x vdt x0 dt dt t0 t0 If a does not depend on time, then 1 2 v at v0 ; x at v 0 t x 0 2 Coulomb’s Law vs. Gravity A Helium nucleus (charge +2e) is separated from one of its electrons (charge –e) by about 3.00 10-11m, and we just calculated the electrostatic forces involved. ke q1q2 9 Nm2/C2 k = 8.98810 Fe e 2 r Suppose we could adjust the distance between the nucleus (considered as a point particle) and one electron. Can we find a point at which the electric and gravitational forces are equal? A) Yes, move the particles apart. B) Yes, move the particles together. C) No, they will never be equal. Electric Fields •Electric Field is the ability to extert a force at a distance on a charge •It is defined as force on a test charge divided by the charge •Denoted by the letter E •Units N/C + + – –– + + + F Small test charge q E F /q E-Field: Why? •When we have a charge distribution, and we want to know what effect they would have external charges, we can either •Do many sums (or integrations) every time a charge comes in to find the force on that charge •Or calculate the field from the charge distribution, and multiply the field by the external charge to obtain the force •Simplification! Electric Field from a Point Charge keQq F 2 rˆ r E F /q ke Q E 2 rˆ r Point charge Q Small test charge q Note: the field is a vector! Quiz Two test charges are brought separately into the vicinity of a charge +Q. First,• test T charge +q is brought to point A a distance r from charge +Q. Next, the +q charge is removed and a test charge +2q is brought to point B a distance 2r from charge +Q. Compared with the electric field of the charge at A, the electric field of the charge at B is: +Q +q +Q A +2q B A) Greater B) Smaller C) The same. E-Field from two or more charges •Each charge creates its own Electric Field •Electric Fields must be added as a vector sum q2 = -2 mC 10 cm 5 cm q1 = +1 mC ke qi E 2 rˆi ri E2 = 1.797 106 N/C Etot = 4.019106 N/C = 26.6o E1 = 3.595 106 N/C Electric Field Lines •Graphical Illustration of Electrical Fields •Lines start on positive charges and end on negative •Number of lines from/to a charge is proportional to that charge •Density of lines tells strength of field. + - + - Electric Field from a Point Charge ke Q E 2 rˆ r + Positive Charge – Negative Charge Electric Field from two Charges + + + + Electric Field from two Charges + - + - Electric Field Lines Consider the four field patterns below: Assuming that there are no charges in the region of space depicted, which field pattern(s) could represent electrostatic field(s)? Electric Field Lines Which This thin of disk the two has surfaces charge on(top theand top bottom) and on the hasbottom. more charge Whaton type? it? A) Top Positive has more charge on both B) Bottom Negativehas charge moreon both C) They Positive arecharge roughly onequal top, negative on bottom D) Can’t Negative tell charge from the ongiven top, positive information on bottom Electric Field Lines Which of the following is true: A) the electric field is strongest midway between Y and Z B) the magnitude of the electric field is the same everywhere C) a small negatively charged body placed at X would be pushed to the right D) Y and Z must have the same sign E-Field from continuous charges ke qi E 2 rˆi ri ke dq E 2 rˆ r •Volume charge density : dq = dV •Surface charge density : dq = dA •Linear charge density : dq = dl E-Field from continuous charges Length L Distance R Linear charge density E-field here? R L ke dr R r 2 R L ke L ke ke ke ˆ r rˆ rˆ R L R r R R R L ke dq ke dr E 2 rˆ 2 rˆ rˆ r r Electric Fields and Forces E F /q F qE F ma a qE / m A region of space has an electric field of 104 N/C, pointing in the plus x direction. At t = 0, an object of mass 1 g carrying a charge of 1mC is placed at rest at x = 0. Where is the object at t = 4 sec? A) x = 0.2 m C) x = 20 m B) x = 0.8 m D) x = 80 m