AP Physics Charge Wrap up

Transcription

1 AP Phyic Charge Wrap up Quite a few complicated euation for you to play with in thi unit. Here them babie i: F r Thi i good old Coulomb law. You ue it to calculate the force exerted 1 by two charge on each other. The term i baically a contant. 4 It called Coulomb contant. (Uually mot authoritie aign the thing, Coulomb contant, a ingle ymbol, k, but the College Board in it tremendou widom doe not do that. So we have thi very awkward multi-value term.) 0 F Thi i the definition of the electric field. i the field trength. F i the force exerted on a tet charge divided by the tet charge. An important point - i not the charge that created the field. U r 1 Thi euation give you the electric potential energy. Avg d Thi euation relate the potential difference with the ditance d a charge i moved within an electric field. 1 4 r 0 i i i Thi euation provide u with the potential difference caued at a ditance r from a field etablihed by a charge. Potential difference i a calar, o it i the imple algebraic um of all the potential difference that are acting on a point from difference charge. The main difficulty with doing electric field, force, potential energy, and potential difference i that there are a lot of term and they can be confuing. So it really 38

2 important to know all thi tuff cold. Somebody ay, Potential energy in a field. And you are immediately thinking U. Next, a ha been our cutom, are the thing that the ever-wie College Board folk who produce the AP curriculum reuire you, the proud tudent, to do. A. lectrotatic 1. Charge, Field, and Potential a. You hould undertand the concept of electric field o you can: (1) Define it in term of the force on a tet charge. Thi i jut an application of F. () Calculate the magnitude and direction of the force on a poitive or negative charge placed in a pecified field. F Similar to the one above. The force i found uing. The direction i figured out logically uing the old like charge repel unlike charge attract deal. (3) Given a diagram on which an electric field i repreented by flux line, determine the direction of the field at a given point, identify location where the field i trong and where it i weak, and identify where poitive or negative charge mut be preent. Simple tuff. Baically pie. The direction of the force i the direction of the force on a poitive tet charge and i tangent to the line of force. The arrow point in the direction of the force for a poitive charge. The field i trong where the line of force are cloe together. The force i weak where the line of force are far apart. The line of force come out of a charged body. If the line are pointing away, the body i poitive if the line of force are pointing toward the body it negative. (4) Analyze the motion of a particle of pecified charge and ma in a uniform electric field. Same kind of tuff a the one above, except we throw in Newton law. So you find the force uing Coulomb law, then figure out the acceleration uing the econd law (F = ma), and then the velocity. You can alo ue conervation of energy. You can alo ue energy. The potential energy of the field i going to be eual to the kinetic energy of the particle. We did everal uch problem. b. You hould undertand the concept of electric potential o you can: (1) Calculate the electrical work done on a poitive or negative charge that move through a pecified potential difference. 39

3 The idea here i that the work done when moving a charge i eual to the change in it potential energy. You ue the euation for potential energy. () Given a ketch of euipotential for a charge configuration, determine the direction and approximate magnitude of the electric field at variou poition. Thi would be a et of curve around a charge. Any point on the curve ha the ame potential difference a any other point on the curve. Pretty imple tuff. Ue the line of force to figure out the direction. (3) Apply conervation of energy to determine the peed of a charged particle that ha been accelerated through a pecified potential difference. Find the potential energy, aume it i all turned into kinetic energy and then olve 1 for the velocity. K mv. We did everal of thee problem. (4) Calculate the potential difference between two point in a uniform electric field, and tate which i at the higher potential. Ue the euation for electric field trength, potential difference, and ditance - Avg. Calculate the potential for each point. Then recall that potential d difference i a calar, o you jut add them algebraically to find their difference.. Coulomb Law and Field and Potential of Point Charge a. You hould undertand Coulomb Law and the principle of uperpoition o you can: (1) Determine the force that act between pecified point charge, and decribe the electric field of a ingle point charge. You ue Coulomb law to find the force. The electric field around a ingle point charge i imple arrow point out for a poitive charge and inward for a negative charge. () Ue vector addition to determine the electric field produced by two or more point charge. Thi i where you add up the field between the charge. Thi i a vector addition deal, o you have to reolve the vector into x and y component. Remember the joy of that? b. You hould know the potential function for a point charge o you can: (1) Determine the electric potential in the vicinity of one or more point charge. 330

4 Thi i where you ue the euation for the potential difference in a field, good old 1 i. Anyway, pretty imple tuff. The potential difference i a 4 r 0 i calar, o you jut add em up. i 3. Field and Potential of Other Charge Ditribution a. You hould know the field of highly ymmetric charge ditribution o you can: (1) Decribe the electric field of: (a) Parallel charged plate. Baically the field look like thi: B. Conductor To decribe the field, I would ay that the line of force are perpendicular to the urface of the plate and are all parallel and evenly paced. Therefore the field i uniform (except at the end). The arrow hould point from poitive to negative. 1. lectrotatic with Conductor a. You hould undertand the nature of electric field in and around conductor o you can: (1) xplain the mechanic reponible for the abence of electric field inide a conductor, and why all exce charge mut reide on the urface of the conductor. See the ection on electric field. () xplain why a conductor mut be an euipotential, and apply thi principle in analyzing what happen when conductor are joined by wire. All point in a conductor have the ame potential difference. Thi i becaue the charge i free to move about there i nothing to top their moving about a much a they like. Join wire together and they will till have the ame potential difference. (3) Determine the direction of the force on a charged particle brought near an uncharged or grounded conductor. Thi imply ha to do with the old baic law of tatic electricity like charge repel and unlike charge attract. A poitively charged particle will be attracted to the grounded conductor, which it will polarize. The ame thing will happen to a negatively charged particle. It too will polarize the conductor. Pretty much the ame thing will happen to the uncharged conductor. 331

5 b. You hould be able to decribe and ketch a graph of the electric field and potential inide and outide a charged conducting phere. The electric field urrounding a charged conducting phere (poitive charge let ay) would look like thi: A negative charge on the phere would be the ame except that the arrow would be toward the phere. The electric field and potential within the phere would be zero no field, no line of force. c. You hould undertand induced charge and electrotatic hielding o you can: (1) Decribe ualitatively the proce of charging by induction. See the handout. () Determine the direction of the force on a charged particle brought near an uncharged or grounded conductor. Didn t we do thi one already? Good thing that you don t have to do much in t it? Poitively charged object Naty Old AP Tet Quetion: From 001: 33

6 Four charged particle are held fixed at the corner of a uare of ide. All the charge have the ame magnitude Q, but two are poitive and two are negative. In Arrangement 1, hown to the right, charge of the ame ign are at oppoite corner. xpre your anwer to part (a) and (b) in term of the given uantitie and fundamental contant. (a) For Arrangement 1, determine the following. i. The electrotatic potential at the center of the uare. (Thi mean potential difference.) Potential difference from one charge: 1 4 o r r r Finding r: 1 Q 4 o 4o Potential difference at a point i the algebraic um of all the potential acting at that point; potential difference i a calar: tot Q Q Q Q 0 4o 4o 4o 4o a +Q - Q b The potential difference i zero. The four value add up to zero. d b ii. The magnitude of the electric field at the center of the uare. The electric field from each of the charge i a vector. The vector appear a hown. You can ee that the um of thee vector i zero. Therefore, the electric field in the center of the uare i zero. c - Q c a +Q d The bottom two charged particle are now witched to form Arrangement, hown to the right, in which the poitively charged particle are on the left and the negatively charged particle are on the right. +Q - Q (b) For Arrangement, determine the following. +Q - Q i. The electrotatic potential at the center of the uare. 333

7 tot 1 Q 1 Q 1 Q 1 Q 0 4o 4o 4o 4o Once again, the voltage in the center i zero. ii. Find the magnitude of the electric field at the at the center of the uare. A D B C But A D B C So: B C a +Q - Q b b c To find, we need the voltage: d c +Q - Q d a d We already figured out the voltage in the firt part of the problem, o we can plug that in. Remember that the ditance wa: Q r b 4o 1 Q 1 Q Q B d 4 o 4 o o C will have ame magnitude. Q B C o Q o Q Q Q Q 4 o o o o 334

8 (c) In which of the two arrangement would more work be reuired to remove the particle at the upper right corner from it preent poition to a ditance a long way away from the arrangement? Arrangement 1 Arrangement Jutify your anwer. Arrangement 1 would reuire more work to move the upper right particle a great ditance away. The reaoning? The removal of a negative particle i reited becaue it cloe to a poitive charge and it i helped becaue of it cloe ditance to a negative charge. Arrangement move a poitive charge farther away from the upper right particle and move a negative charge cloer to the upper right particle, both action reducing the work needed to remove the upper right particle. Therefore, the arrangement 1 will reuire more work. From 1999: In a televiion et, electron are firt accelerated from ret through a potential difference in an electron gun. They then pa through deflecting plate before triking the creen. a. Determine the potential difference through which the electron mut be accelerated in the electron gun in order to have a peed of 6.0 x 10 7 m/ when they enter the deflecting plate m kg mv 4 U K mv C The pair of horizontal plate hown below i ued to deflect electron up or down in the televiion et by placing a potential difference acro them. The plate have length m and eparation 0.01 m, and the right edge of the plate i 0.50 m from the creen. A potential difference of 00. i applied acro the plate, and the electron are deflected toward the top creen. Aume that the electron enter horizontally midway between the plate with a peed of 6.0x10 7 m/ and that fringing at the edge of the plate and gravity are negligible. b. Which plate in the pair mut be at the higher potential for the electron to be deflected upward? Check the appropriate box below. Jutify your anwer. Upper plate Lower plate The electron are attracted to the upper poitive plate. The poitive plate mut be at the higher potential and the negative plate i ground. c. Conidering only an electron motion a it move through the pace between the plate, compute the following. i. The time reuired for the electron to move through the plate. 335

9 x x m 10 v t t v 7 m 610 ii. The vertical diplacement of the electron while it i between the F plate. F ma d d C00 15 m a md kg 0.01 m m y at m d. Show why it i a reaonable aumption to neglect gravity in part (c). The force of electricity i billion and billion of time tronger. Gravity i negligible by comparion. e. Still neglecting gravity, decribe the path of the electron from the time they leave the plate until they trike the creen. State a reaon for your anwer. Move in a traight line due to inertia. The electric field no longer act on the particle. Gravity i too weak to caue a meaurable deflection. From 1998: A wall ha a negative charge ditribution producing a uniform horizontal electric field. A mall platic ball of ma kg, carrying a charge of 80.0 μc, i upended by an uncharged, nonconducting thread 0.30 m long. The thread i attached to the wall and the ball hang in euilibrium, a hown below, in the electric and gravitational field. The electric force on the ball ha a magnitude of 0.03 N. t F w a. On the diagram below, draw and label the force acting on the ball. See above right b. Calculate the magnitude of the electric field at the ball location due to the charged wall, and tate it direction relative to the coordinate axi hown. 336

10 F F 0.03N N C C The negative x direction. The electric field direction i the direction a poitive tet charge would go. Since the negative charge move right, the field mut be to the left. c. Determine the perpendicular ditance from the wall to the center of the ball N t m wmg kg N F t F w 0.03 N N N The angle made by the tenion force with the vertical i the ame a the angle formed by the tring with the vertical. Thi i hown in the drawing. We can olve for uing trig. Once we know we can find r ince we know l. F 0.03 N in 18.1 t N o w r l r o in r l co 0.30 m in m l d. The tring i now cut. i. Calculate the magnitude of the reulting acceleration of the ball, and tate it direction relative to the coordinate axi hown. The tenion diappear, but w and F are till preent. The net force will be the um of thee two force. Since they are perpendicular, we can ue the good old Pythagorean theorem to find the net force. F F w 0.03 N N N We now ue the econd law to find the acceleration acting on the ball: F ma a F N 10.3 m m kg 9.8 m tan 3. m o 7, below the x axi ii. Decribe the reulting path of the ball. 337

11 Straight to the right and down. If t i miing from the FBD two force are till acting. So the net force i down and to the right From 1996: Robert Millikan received a Nobel Prize for determining the charge on the electron. To do thi, he et up a potential difference between two horizontal parallel metal plate. He then prayed drop of oil between the plate and adjuted the potential difference until drop of a certain ize remained upended at ret between the plate, a hown to the right. Suppoe that when the potential difference between the plate i adjuted until the electric field i 10,000 N/C downward, a certain drop with a ma of 3.7 x kg remain upended. a. What i the magnitude of the charge on thi drop? F ma F F ma ma 16 m 3.7 x10 kg x10 N x10 C 4 N 4 N C C b. The electric field i downward, but the electric force on the drop i upward. xplain why. The electric field i oriented in reference to a poitive tet charge. A poitive charge would move downward. However, a negative charge would experience the oppoite effect, or a force upward. The drop mut be negatively charged. c. If the ditance between the plate i 0.01 m, what i the potential difference between the plate? Avg ignore the ign d m 100 d d m w F d. The oil in the drop lowly evaporate while the drop i being oberved, but the charge on the drop remain the ame. Indicate whether the drop remain at ret, move upward, or move downward. xplain briefly. Originally the drop did not move ince gravity and the electric force were in euilibrium. But, if the drop looe ma then the force of gravity get maller. Since 338

12 the charge doe not change the upward electric force, i greater than the downward force. Thi unbalanced force will caue the drop to accelerate upward. 339