Electric Potential Energy. Voltage and Capacitance. Electric Work. Electric Potential: Voltage 6/1/2016

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1 Electric Potential Energy Voltage and Capacitance Chapter 17 Potential Energy of a charge Wants to move when it has high PE Point b U = max K = min Point a U = min K = max Electric Work Charge moving between plates. Work = FDs cos0 o = qes i - qes f DU = qes i qes f = qeds W = -DU The electric field of a capacitor is 2.82 X 10 5 N/C. The spacing between the plates is 2.00 mm. a. A proton is released from rest at the positive plate. Calculate the change in potential energy. (-9.02 X J) b. Calculate the final speed of the proton. (3.29 X 10 5 m/s) c. An electron is released from the halfway point between the plates. Calculate the change in PE (DU) and the final speed of the electron. (9.95 X 10 6 m/s, left) A 2.0 cm diameter disk capacitor has a 2.5 mm spacing and an electric field of 2.70 X 10 5 N/C. a. An electron is released from rest at the negative plate. Calculate the change in potential energy (DU). (-1.08 X J) b. Calculate the final speed of the electron. (1.54 X 10 7 m/s) Electric Potential: Voltage Voltage 1 Volt = 1 Joule/Coulomb DV = DU q DV = -W ba q Work done by the electric field to accelerate the charge 1

2 The higher the rock, the greater the PE The greater the Voltage difference, the greater the PE (DU = qv) Great Q, greater DU An electron is accelerated in a TV tube through a potential difference of 5000 V. a. Calculate the change in PE of the electron (-8.0 X J) b. Calculate the final speed of the electron (m = 9.1 X kg) (4.2 X 10 7 m/s (1/7 th speed of light) c. Calculate the final speed of a proton (mass = 1.67 X kg) (9.8 X 10 5 m/s (0.3% speed of light) Equipotential Lines Equipotential lines are perpendicular to electric field lines Voltage is the same along equipotential lines Like contour (elevation) lines on a map Equipotential lines for point charges DU = qeds DV = DU/q DV = EDs Electric Field and Voltage Voltage increases as you move between plates Greater the distance between plates, the greater the voltage The greater the E field, the greater the voltage 2

3 Calculate the electric field between two plates separated by 5.0 cm with a voltage of 50V. (1000 V/m) An electron in a television set is accelerated through a 2.86 X 10 4 V/m electric field. The screen is 35 mm from the cathode. a. Calculate the net change in the potential energy of the electron during the acceleration process (- 1.6 X J) b. How much work is done by the electric field in accelerating the electron? (1.6 X J) c. What is the speed of the electron when it strikes the screen? (1.87 X 10 7 m/s) A capacitor is constructed of 2.0 cm diameter disks separated by a 2.0 mm gap, and charged to 500 V. a. Calculate the electric field strength (V = Es) [2.5 X 10 5 N/C] b. A proton is shot through a hole in the negative plate towards the positive plate. It has an initial speed of 2.0 X 10 5 m/s. Does it have enough energy to reach the other side? (V = DU/q) [DU = 8 X J, K = 3.34 X J] Electron Volt Energy an electron gains moving through a potential difference of 1 V 1 ev = 1.6 X J Ex: An electron moving through 1000 V would gain 1000 ev of energy Potential Energy of point charges U elec = k q 1 q 2 r k = 9.0 X 10 9 Nm 2 /C 2 A proton is fired from far away at a 1.0 cm diameter glass sphere of charge +100 nc. a. Calculate the potential energy of the system (just as it touches the sphere). (2.88 X J) b. Since PE = KE, calculate the needed initial speed of the proton. (5.87 X 10 6 m/s) (derive from Force expression) 3

4 In Rutherford s gold foil experiment, he fired alpha particles (+2 charge, 6.64 X kg) at 1.61 X 10 7 m/s at a gold nucleus (+79 charge). How close could the alpha particle get to the nucleus? An electron and a positron are created in the CERN collider. a. Calculate the potential energy they have when they are 1.00 X m apart. (2.30 X J) b. Calculate the velocity they need to escape from one another. Remember that PE = KE, but you will need to consider the KE of both particles added together. (1.59 X 10 6 m/s) 4.2 X m Voltage due to a Point Charge Voltage is not directional (scalar) Charged particles (i.e.: electrons, protons) have a voltage V = kq r Example 1 Consider a +1.0 nc charge. a. Calculate the electric potential (voltage) at a point 1.0 cm from the charge b. Calculate the electric potential at a point 3.0 cm from the charge. Point Charges: Example 2 Calculate voltage (electric potential) at point A as shown below: A Use Pythagorean theorem to calculate the distance from A to Q 2 : A 30 cm 30 cm 52 cm 52 cm Q 1 = +50 mc Q 2 = -50 mc Q 1 = +50 mc Q 2 = -50 mc 4

5 V A = V 1 + V 2 V 1 = kq = (9.00X 10 9 Nm 2 /C 2 )(5.00X10-5 C) r (0.30 m) V 1 = 1.50 X 10 6 V V 1 = kq = (9.00X 10 9 Nm 2 /C 2 )(-5.00X10-5 C) r (0.60 m) V 2 = -7.5 X 10 5 V V A = 1.50 X 10 6 V -7.5 X 10 5 V V A = 7.5 X 10 5 V Suppose two helium nuclei (+2 each) are brought to within 5.00 X m of each other. a. Calculate the potential energy of the system. (1.8 X J) b. Calculate the work needed to bring them together from very far away (infinity). (1.8 X J) c. Calculate the speed of the nuclei if they are released and allowed to move far away from each other. The mass of one nuclei is 6.64 X kg. (1.65 X 10 6 m/s) d. Calculate the voltage produced by just one nuclei at the distance of 5.00 X m. (57.6 kv) A positive mc charge is placed at the origin. A mc charge is placed at (0.200, 0.000)m. a. Calculate the voltage (potential) at (0.100, 0.000) m. [-4.5 X 10 5 V] b. Calculate the point in between the particles where the voltage is zero. [0.067 m, 6.67 cm] c. Calculate the magnitude of the potential energy between the two particles. [2.25 J] d. Suppose the mc charge is fixed in place. Calculate the speed of mc (m = g) particle when it collides with it. [94.9 m/s] Point Charges: Example 3 Calculate voltage (electric potential) at point B as shown below: 30 cm B 26 cm 26 cm Q 1 = +50 mc Q 2 = -50 mc Use Pythagoream theorem to calculate the distance from B to Q 1 and to Q 2 : 30 cm B V A = V 1 + V 2 V 1 = kq = (9.00X 10 9 Nm 2 /C 2 )(5.00X10-5 C) r (0.40 m) V 1 = X 10 6 V V 1 = kq = (9.00X 10 9 Nm 2 /C 2 )(-5.00X10-5 C) r (0.40 m) V 2 = X 10 5 V Q 1 = +50 mc 26 cm 26 cm Q 2 = -50 mc V A = X 10 6 V X 10 5 V V A = 0 V 5

6 Calculate voltage (electric potential) at point B as shown below (-188 kv or X 10 5 V) B Point Charges: Example 4 How much work is required to bring a charge of q = 3.00 mc to a point m from a charge Q = 20.0 mc? 30 cm V Q = kq = (9.00X 10 9 Nm 2 /C 2 )(2.00X10-5 C) r (0.500 m) Q 1 = +50 mc 30 cm 20 cm Q 2 = -50 mc V Q = 3.6 X 10 5 V (This is the voltage caused by the stationary charge) W = DU (like the work to lift a book to a shelf) V = DU q V = W q W = Vq W = (3.6 X 10 5 V )(3 X 10-6 C) W = 1.08 J Point Charges: Example 5 Which of three sets of charges has the most: positive potential energy? The most negative potential energy? Would require the most work to separate? + - (i) Capacitors (Condensers) + - (ii) Store electric charge for later use Camera flash Energy backup in computers Block surges of charge Stores 0 s and 1 s in RAM + + (iii) 6

7 Anatomy of a capacitor Charge is stored on plates Charges do not jump the gap Often rolled to increase surface area Q = DVC Capacitance Q = charge stored on the plate DV = Voltage C = Capacitance [Farad (F)] Most capacitors between F and 10-6 F (picofarad to microfarad) C = e o A d A = area (larger, more storage) d = distance e o = 8.85 X C 2 /Nm 2 (permittivity of free space) Calculate the capacitance of a capacitor whose plates are 20 cm X 3.0 cm and are separated by a 1.0 mm air gap. a. Calculate the capacitance using: (53 pf) C = e o A d b. If the capacitor is attached to a 12-Volt battery, what is the charge in each plate? (6.4 X C) c) Calculate the electric field between the plates. (1.2 X 10 4 V/m) 7

8 The spacing between the plates of a 1.00 mf capacitor is mm. a. Calculate the surface area of the plates (5.65 m 2 ) b. Calculate the charge if the plate is attached to a 1.5 V battery. (1.5 mc) Dielectrics Insulating paper or plastic Prevents charge from jumping the gap Increases capacitance by a factor of K C = KC i C = Ke o A d Dielectric Constants (K) How Dielectrics Work Molecules orient themselves and/or their electrons Decreases electric field Electric field and capacitance are inversely related Capacitance and Studfinders Stud finder registers a change in capacitance. Wood acts as a dielectric Capacitance and Keyboards Pushing down decreases d C = Ke o A d Change in capacitance detected 8

9 A parallel plate capacitor has plates 2.0 cm by 3.0 cm. The plates are separated by 1.0 mm thickness of paper (K=3.7). Calculate the capacitance. Calculate the charge that can be stored on the capacitor at a voltage of 240 V. Q = 4.8 X 10-9 C or 4.8 nc C = 2.0 X F or 20 pf A 50.0 mf capacitor is charged to V, It is then disconnected from the battery and submerged in water. a. Calculate the charge stored on the plates [8.00 C] b. Calculate the new capacitance [4.00 F] c. Calculate the new voltage [2.00 V] d. Calculate the energy stored before and after plunging it into the water. [640J, 8.00 J] A parallel plate capacitor is filled with a dielectric of K = 3.4. The plates are square and have a side length of 2.0 m. It is connected to a 100 V source. The plates are separated by 4.02 mm. a. Calculate the capacitance (3.0 X 10-8 F) b. Calculate the charge on the plates (3.0 X 10-6 C) c. Calculate the electric field between the plates. (2.5 X 10 4 V/m) d. Calculate the energy stored (1.50 X 10-4 J) e. The battery is disconnected and the dielectric removed. Calculate the new capacitance, voltage and energy stored. (8.8 X 10-9 F, 341 V, 5.1 X 10-4 J) Storage of Electric Energy A camera flash stores energy in a 150 mf capacitor at 200 V. How much energy can be stored? U = 3.0 J 9

10 An electric device must hold 0.45 J of energy while operating at 110 V. What size capacitor should be chosen? A 2.0 mf capacitor is charged to 5000 V a. Calculate the charge on one of the plates. [0.010C] b. Calculate the energy stored [25 J] c. Calculate the power if the capacitor is discharged in 10 ms [2.5 MW] (ANS: 7.4 X 10-5 F, 74 mf) X 10 6 m/s 4. 25,000 m/s X 10-7 J X 10 7 m/s X 10 4 V V 16. a) 1833 b) a) 1.5 V b) 8.3 X C 20. a) 200 V b) 400 V m, m kv V V, 10.0 nc cm 32. a) Both positive b) (draw graph) nc, +40 nc 36. a) zero at + b) (same as (a)) J b) 14.4 N c) 21.9 m/s & 11.0 m/s X 105 m/s 44. b) 9.6 X x 2 J b) 1.92 X N/m 46. a) 2.1 X 10 6 V/m b) 9.4 X 10 7 m/s X 10 7 m/s 10

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13 One Charge Two Charges Python Challenge: #We will model a charged particle released from one plate of a capacitor. We will assume it races toward the other plate. Allow the user to input: Voltage between plates Charge and mass of the particle Output: Speed when it hits the other plate Can you modify the program to run multiple times? Can you build error checking into the program? W = -DU DV = DU q DV = EDs U elec = k q 1 q 2 r V = kq r Q = DVC m e = 9.11 X kg k = 9.0 X 10 9 Nm 2 /C 2 13