Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Potential energy wikipedia , lookup
Electron mobility wikipedia , lookup
Conservation of energy wikipedia , lookup
Electromagnetism wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Density of states wikipedia , lookup
Nuclear physics wikipedia , lookup
Electric charge wikipedia , lookup
Electrostatics wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Heimadæmi 3 Due: 11:00pm on Thursday, February 4, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy Exercise 24.5 A parallelplate capacitor with circular plates and a capacitance of 13.8 μF is connected to a battery which provides a voltage of 10.5 V . Part A What is the charge on each plate? ANSWER: = 1.45×10−4 C Q Correct Part B How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? ANSWER: = 7.25×10−5 C Q Correct Part C How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation? ANSWER: = 5.80×10−4 C Q Correct Exercise 24.21 For the system of capacitors shown in the the figure below, a potential difference of 25.0 V is maintained across ab. Part A What is the equivalent capacitance of this system between a and b? ANSWER: C = 19.3 nF Correct Part B How much charge is stored by this system? ANSWER: = 482 nC Q Correct Part C How much charge does the 6.50 nF capacitor store? ANSWER: = 162 nC Q Correct Part D What is the potential difference across the 7.50 nF capacitor? ANSWER: V = 25.0 V Correct Exercise 24.44 A parallelplate capacitor has plates with area 2.30×10−2 m2 separated by 1.80 mm of Teflon. Part A Calculate the charge on the plates when they are charged to a potential difference of 10.0 V . ANSWER: = 2.37×10−9 C Q Correct Part B Use Gauss's law to calculate the electric field inside the Teflon. ANSWER: E = 5560 N/C Correct Part C Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed. ANSWER: E = 1.17×104 N/C Correct Problem 24.45 Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for a time interval of 1.48×10−3 s with an average light power output of 2.20×105 W . Part A If the conversion of electrical energy to light has an efficiency of 95.0 % (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? ANSWER: U = 343 J Correct Part B The capacitor has a potential difference between its plates of 140 V when the stored energy equals the value calculated in part A. What is the capacitance? ANSWER: C = 3.50×10−2 F Correct Problem 24.66 A parallelplate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. (See the figure below.) An 18.0 V battery is connected across the plates. Part A What is the capacitance of this combination? ANSWER: C = 6.23×10−11 F Correct Part B How much energy is stored in the capacitor? ANSWER: U = 1.01×10−8 J Correct Part C If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor? ANSWER: U = 4.59×10−9 J Correct Exercise 22.17 A very long uniform line of charge has charge per unit length 4.64 μC/m and lies along the xaxis. A second long uniform line of charge has charge per unit length 2.24 μC/m and is parallel to the xaxis at y1 = 0.410 m . Part A What is the magnitude of the net electric field at point y2 = 0.180 m on the yaxis? ANSWER: E = 6.39×105 N/C Correct Part B What is the direction of the net electric field at point y2 = 0.180 m on the yaxis? ANSWER: y axis + y axis Correct Part C What is the magnitude of the net electric field at point y3 = 0.612 m on the yaxis? ANSWER: E = 6.31×104 N/C Correct Part D What is the direction of the net electric field at point y3 = 0.612 m on the yaxis? ANSWER: y axis + y axis Correct Prelecture Concept Question 23.07 Part A The electric potential at a certain distance from a point charge can be represented by V. What is the value of the electric potential at twice the distance from the point charge? ANSWER: At twice the distance, the electric potential is V/2. At twice the distance, the electric potential remains V. At twice the distance, the electric potential is V/4. At twice the distance, the electric potential is 2V. At twice the distance, the electric potential is 4V. Correct Problem 23.34 Part A If the electric potential in a region is given by V (x) = 6/x 2 , the x component of the electric field in that region is ANSWER: −6x. 12x . 12x−3 . 6x. −12x−3 . Correct Bouncing Electrons Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other. Part A Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two electrons reach their minimum separation? ANSWER: Electron A is moving faster than electron B. Electron B is moving faster than electron A. Both electrons are moving at the same (nonzero) speed in opposite directions. Both electrons are moving at the same (nonzero) speed in the same direction. Both electrons are momentarily stationary. Correct If at a given moment the electrons are still moving toward each other, then they will be closer in the next instant. If at a given moment the electrons are moving away from each other, then they were closer in the previous instant. The electrons will be traveling in the same direction at the same speed at the moment they reach their minimum separation. Only in a reference frame in which the total momentum is zero (the center of momentum frame) would the electrons be stationary at their minimum separation. Part B What is the minimum separation rmin that the electrons reach? Express your answer in term of q, m, v , and k (where k = 1 4πϵ0 ). Hint 1. How to approach the problem Since no external or nonconservative forces act on the system of the two electrons, both momentum and total energy (kinetic plus potential) are conserved. Find one expression for the energy when the electrons are far apart, and another when they reach their minimum separation rmin . This will give you an equation in which the only unknown is the speed of the electrons at the moment of their minimum separation. Apply conservation of momentum, using the same initial and final states, to obtain a second equation involving the speed of the electrons. Solve the simultaneous energy and momentum equations to obtain rmin . Hint 2. Find the initial energy What is the total energy Einitial of the two electrons when they are initially released? Assume that the electrons are so far apart that their potential energy is zero. Express your answer in terms of m and v . ANSWER: Einitial = 5mv 2 Hint 3. Find the final energy What is the total energy Ef inal of the electrons when they reach their minimum separation rmin ? Assume that the (identical) speed of the two electrons is u. Express your answer in terms of m, u, q, rmin , and k (where k = Hint 1. Find the final kinetic energy What is the final kinetic energy (both electrons)? Express your answer in terms of u, and m. ANSWER: 1 4πϵ0 ). = Kf inal mu 2 Hint 2. Find the final potential energy What is the final potential energy of this 2electron system? Express your answer in terms of k, q, and rmin . ANSWER: Uf inal = kq 2 rmin ANSWER: Ef inal = mu 2 + kq 2 rmin Hint 4. Find the initial momentum ⃗ What is the total momentum p initial of the two electrons when they are initially released? Express your answer as a vector in terms of m, v , and ^i . ANSWER: p ⃗ initial = ^ −2mv i Hint 5. Find the final momentum What is the total momentum p f⃗ inal of the two electrons when they reach their minimum separation rmin ? Assume that the (identical) velocity of the two electrons is u⃗ . Express your answer as a vector in terms of m and u⃗ . ANSWER: p f⃗ inal = 2mu⃗ Hint 6. Some math help From the momentum equations, ∣u⃗ ∣ find rmin . ANSWER: rmin = kq 2 2 4mv ; that is, u = v = v . Substitute for u in the energy conservation equation to Correct An experienced physicist might approach this problem by considering the system of electrons in a reference frame in which the initial momentum is zero. In this frame the initial speed of each electron is 2v. Try solving the problem this way. Make sure that you obtain the same result for rmin , and decide for yourself which approach is easier. Score Summary: Your score on this assignment is 97.8%. You received 8.8 out of a possible total of 9 points.