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Transcript
Think, Answer; Talk to your neighbors, Answer Ch19: Electric Potential A positively charged particle pushed up at constant speed by thin wire a distance d as shown in the animation.. While moving up, the sign of the net work done is ? Or is the work done zero? A: + B: C: zero D: Not enough info d Work done transfers energy Fyou x x Fgravity You do the work against the gravitational field. The work gets transferred into change in Gravitational potential energy. ++++++++++++++++++++ Q E q ----------------- The wire does work against the Electric field. The work gets transferred into change in Electric potential energy. Work by E: WE FE x cos U WExtF WFieldF qE x cos180o qE x U E ? Think, Answer; Talk to your neighbors, Answer An external force that is not shown is making the charge move with constant velocity. The work done by the agent, done by the electric field, and done by the net force on the charge +q are: Agent E-Field Net Force A + - + C - + 0 B D - + + - - 0 E: None of these/I wish I knew E +q Recall, work = Force x displacement x cos(angle between them) ++++++++++++++++++++ ----------------- Let’s work together Q 1. Draw the direction of electric field +++++++++++++ between two parallel charged plate. x 2. What is the direction of force on the E e negative charge? Draw it. ----------- 3. What is the work done by electric force on the charge? 4. What is the change in electric potential energy? From last chapter, E due to charge on parallel plate: E Q o A 5. What is the change in the PE of a proton when it moves 2.0 cm? Assume that each square shaped plate has side length 1.0 cm and charge on the plate is 8.85 µC. PE of point charges Any two charges separated by r. Qq FE k 2 r . Q r . q If r is very-very large (), FE = 0. No work is required to move the charge around over there means UE (at )= 0 When r is finite FE ≠ 0. Work done to bring there turns out to be Qq Qq r k r r2 Since, U WE it becomes: Qq what happened to the sign? U f Ui k r WE FE r k It turns out, using the sign of charges takes care of the sign of the PE. Qq Since Ui = 0, U E (orU f ) k r 1 . Problem: What is the potential energy if Q = 2C , q = 1 C and r = 9 cm? r Q . Electric Potential (or voltage) Electric Potential energy per unit charge is voltage: V q Problem: How much work does it take for a external force to set up the arrangement of charged objects in the diagram on the corners of a right triangle when the three objects are + q1 5.5 C initially very far away from each other. The work done on the charges is equal to their potential energy. qq q q qq UE k 1 2 2 3 1 3 r23 r13 r12 r12 = 12 cm - r23 = 16 cm + q2 6.5 C q3 2.5 C ans : 3.0 J UE q Just like Field is force per unit charge. Unit of Potential: J/C called Volts and written as V. Notice both the potential and its unit are denoted by V. Just like a field tells us about the force on a test charge around a source charge, the potential tells us about the potential energy of the test charge. U E qV k Qq qV r V kQ r Change in potential between two points is called Potential difference. U E V q or, U E qV . r V ++ ++ Source charge, Q Think, Answer; Talk to your neighbors, Answer ECG is display of change in Potential with time Two identical charges, +Q and +Q, are fixed in space. What is the magnitude of the E field, and the value of the voltage, at the midpoint between them? (Assume the potential is zero at infinity.) +Q E=? V=? (Midpoint) +Q A) E=0 , V nonzero B) E nonzero , V=0 C) Both are 0 D) Both are nonzero E) Not enough information Making sense of a battery Label Think, Answer; Talk to your neighbors, Answer Problem: If the 9V battery lights 12 W bulb, how many electrons are flowing through the bulb each second? Did the pot. energy (PE) increase or decrease? Did the voltage (V) at the position of the test charge increase or decrease? What does it mean when it says 9V on this battery? A negative charge -q is released from position i to position f between the charged plate. A: PE , V B: PE , V C: PE , V D: PE , V E: None of these. Problem: What is the potential difference between the two parallel plates separated by distance d? Assume charge and area of each plate is Q and A. V U E qE x Ed q q Q d o A +++++++++++++ . . ----------- +++++++++++++ E f i q d ----------- 2 Where is higher PE? Potential To decide higher PE, let +q move. +++++++++++++ And think about how rocks roll down hill. High hill, high gPE. E q ----------- Similarly, +q moves from higher PE to lower PE. How about -q ? Now how to find where is the Potential high or low? E points from high V to low V (always!). Is ∆UE of the particle positive, negative, or zero as the particle moves from i to f. We know: U U U E f i e e + i f Just like one might choose sea level, or the tabletop, or the ground as zero gPE, you can pick any spot you want and call the electrical potential energy 0 there. Same is true for potential. Potential and field due to a spherical conductor E E0 inside 1 r 2 outside This is field due to sphere. V constant Potential due to the sphere. Notice the potential difference between two consecutive equipotentials is same. V or, E V d q q e b c Q Q r a g At which labeled point is voltage highest? B A D E Ed means, E and d are inversely proportional. Also remember closer to the charge, field is stronger. V k Think, Answer; Talk to your neighbors, Answer Equipotential and Field But the equipotentials are not uniformly distributed. Recall: U E qE x Are points a, b, c, d at same distance from the charge Q? Is potential, V, same at these points? It is equipotential circle for 2D and surface for f 3D (sphere). d Which circle has larger potential? Do we know how to draw E filed lines due to Q? Geometrically, how are they related? draw E field lines before you answer. 1 r outside inside Equipotential Surfaces represents V field Ek Q r2 So E is stronger in the region where equpotentials are crowded. Equipotential Field lines 3 Electric Potential and Conservation of energy Ei E f or, K i U i K f U f Problem: A proton is released from rest at point a. It then travels past b. What is its speed at b? KE is our good old PE is qV 1 2 mv 2 Problem: A proton is fired toward a small hole in the negative plate with the speed shown. What is its kinetic energy when it emerges through the hole in the positive plate? (Hint: The electric field outside of a parallel-plate capacitor is zero). Note: Small energy is expressed as electron volt, eV. 1eV 1.6 1019 J Does this number look familiar? Think, Answer; Talk to your neighbors, Answer Capacitors Q C V A where,C o d If we double L (both of them!) and halve d, by what factor have we changed capacitance? When two metal plates separated by air or insulator makes a capacitor. Simplified sketch: L Q U E qE x Q +++++++++++++ V d Ed d E q o A q o A ----------or, Q V C V A d where,C o It is called capacitance. d Charging a Capacitor When you throw a ball higher, it gains gPE. Ball at height has energy. ---+++++ Battery separates the charges on the plates. +++++++++++++ L d A: no change D: up by 8 B: up by 2. C: up by 4. E: none of these Work, Answer; Talk to your neighbors, Answer A capacitor charged to 1.5 V stores 2.0 mJ of energy. If the capacitor is charged to 3.0 V, it will store A. B. C. D. E. 1.0 mJ 2.0 mJ 4.0 mJ 6.0 mJ 8.0 mJ 1 U Q V 2 Q2 2C 1 C V 2 2 ----------2 1 Energy supplied by the battery is stored U Q V Q as electric energy in the capacitor. 2C 2 4 Problem: A 12 V car battery is hooked up to a capacitor with plates of area 0.1 m2, a distance 8.85 mm apart. (a) How much charge builds up on a plate? Dielectrics in a Capacitor Vacuum or air (b)How much energy is stored in the capacitor now? The electric field due to polarized dielectric reduces net electric field and hence lowers the V . Recall V =Ed. Since, C Q , small V means larger C. V So capacitance increase by a factor called dielectric A constant, . Cdiel o C d Problem: A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are 0.3 m by 0.4 m and the waxed paper, of slightly larger dimensions, is of thickness 0.030 mm and has = 2.5, what is the capacitance of this capacitor? ans 8.85*10^8 F Problem: A parallel-plate capacitor is charged using a 100 V battery; then the battery is removed. If a dielectric slab is slid between the plates, filling the space inside, the capacitor voltage drops to 30 V. What is the dielectric constant of the dielectric? ans 3.3 5