course PHY 202 ????t???????assignment #005005. Query 27
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13:03:52 Note that the solutions given here use k notation rather than epsilon0 notation. The k notation is more closely related to the geometry of Gauss' Law and is consistent with the notation used in the Introductory Problem Sets. However you can easily switch between the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k).
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RESPONSE --> ok
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13:15:24 Based on what you learned from Introductory Problem Set 2, how does the current in a wire of a given material, for a given voltage, depend on its length and cross-sectional area?
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RESPONSE --> The current depends on its length and cross sectional area because each play an important role in determining the magnitude of the potential gradient which is V/m. If given the current of one wire and you need to find the current in an identical wire, you would multiply the ratio of the potential gradients by the given current magnitude, to get the mag. of the current in the identical wire.
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13:18:14 How can the current in a wire be determined from the drift velocity of its charge carriers and the number of charge carriers per unit length?
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RESPONSE --> You can use the drift velocity to obtain the amount of charge carriers that pass a given length every second. You can then use multiply this number by the number of charge carriers in the wire to determine the current.
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13:20:39 Will a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why?
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RESPONSE --> Cross sectional area is related to by electrical resistance by the following formula: R=p(L/A). One can see that the resistance is indirecty proportional to its cross sectional area. Thus as the area is greater the electrical resistance will be less.
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13:21:08 Will a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why?
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RESPONSE --> Length is related to by electrical resistance by the following formula: R=p(L/A). One can see that the resistance is diirecty proportional to its cross sectional area. Thus as the length is greater the electrical resistance will be greater.
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13:21:39 Query Principles and General Physics 16.24: Force on proton is 3.75 * 10^-14 N toward south. What is magnitude and direction of the field?
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RESPONSE --> 2.3E5, direction will be to the north.
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13:22:42 The direction of the electric field is the same as that as the force on a positive charge. The proton has a positive charge, so the direction of the field is to the south.{}{}The magnitude of the field is equal to the magnitude of the force divided by the charge. The charge on a proton is 1.6 * 10^-19 Coulombs. So the magnitude of the field is {}{}E = F / q = 3.75 * 10^-14 N / (1.6 * 19^-19 C) = 2.36* 10^5 N / C.
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RESPONSE --> My direction was incorrect. I have to remember the direction of the field is the same as the force on a postive charge.
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13:23:48 ** If the magnitude of the charge is q then the field contribution of each charge is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. Thus E = 373 N/C and E = k q / r^2. We know k, E and r so we solve for q to obtain q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 373 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **
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RESPONSE --> ok, but I think the program skipped this question.
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13:26:43 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> THe electric field at the midpoint is E=k(Q)/(r/2)^2, multiplied by two.
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13:28:09 ** this calls for a symbolic expression in terms of the symbol Q. The field would be 2 k Q / r^2, where r=.08 meters and the factor 2 is because there are two charges of magnitude Q both at the same distance from the point. **
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RESPONSE --> ok. I have to be careful with my distance. I was right about multiplying the field by two, but I must be sure not to miscalculate my distance at the midpt.
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13:28:50 Query gen phy problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge?
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RESPONSE --> 2.65E-10
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13:32:50 Query Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
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RESPONSE --> Use formula: E=K(Q)/r^2 So: 9.0E9(33E-6)/(.20)^2= 7.4E6
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13:33:16 A positive charge at the given point will be repelled by the given positive charge, so will experience a force which is directly upward. The field has magnitude E = (k q Q / r^2) / Q, where q is the given charge and Q an arbitrary test charge introduced at the point in question. Since (k q Q / r^2) / Q = k q / r^2, we obtain E = 9 * 10^9 N m^2 / C^2 * 33.0 * 10^-6 C / (.200 m)^2 = 7.43 * 10^6 N / C.
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RESPONSE --> Remember to use units with all answers, and direction.
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15:31:11 Query Principles and General Physics 17.4: work by field on proton from potential +135 V to potential -55 V.
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RESPONSE --> Using formula: -Wab=Vba(q)= -W=-190(1.60E-19)= 3.04E-17.
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15:33:07 The change in potential is final potential - initial potential = -55 V - (125 V) = -180 V, so the change in the potential energy of the proton is {}{}-180 V * 1.6 * 10^-19 C ={}-180 J / C * 1.6 * 10^-19 C = -2.9 * 10^-17 J. {}{}In the absence of dissipative forces this is equal and opposite to the change in the KE of the proton; i.e., the proton would gain 2.9 * 10^-17 J of kinetic energy.{}{} Change in potential energy is equal and opposite to the work done by the field on the charge, so the field does 2.9 * 10^-17 J of work on the charge.{}{}Since the charge of the proton is equal in magnitude to that of an electron, he work in electron volts would be 180 volts * charge of 1 electron= 180 eV.
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RESPONSE --> ok I calculated my change in potential wrong. I used -190, instead of -180, but my process was correct.
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15:40:01 Query Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE.
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RESPONSE --> Using formula: change in V=-change in KE/q where q=2e 65.0/2(-1.60E-19)=2.03E20
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15:45:00 65.0 keV is 65.0 * 10^3 eV, or 6.50 * 10^4 eV, of energy. {}{}The charge on a He nucleus is +2 e, where e is the charge on an electron. So assuming no dissipative forces, for every volt of potential difference, the He nuclues would gain 2 eV of kinetic energy.{}{}To gain 6.50 * 10^4 eV of energy the voltage difference would therefore be half of 6.50 * 10^4 voles, or 3.35 * 10^4 volts.
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RESPONSE --> The charge on the He nucleus of 2e does not mean 2(1.60E-19). It means 2e. In my previous answer, I should have divided by 2 rather than 3.2E-19.
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17:18:58 Query gen phy text problem 17.18 potential 2.5 * 10^-15 m from proton; PE of two protons at this separation in a nucleus. What is the electrostatic potential at a distance of 2.5 * 10^-15 meters?
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RESPONSE --> Using the formula: PE=K(Q1)(Q2)/r 9.0E-19(1.60E-19/2.5E-15)= 9.2E-14 joules.
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17:24:05 STUDENT SOLUTION: For a part, to determine the electric potential a distance fo 2.5810^-15m away from a proton, I simply used the equation V = k q / r for electric potential for point charge: q = 1.60*10^-19C=charge on proton V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5V. Part B was the more difficult portion of the problem. You have to consider a system that consists of two protons 2.5*10^-5m apart. The work done against the electric field to assemble these charges is W = qV. The potential energy is equal to the work done against the field. PE=(1.60*10^-19C)(5.8*10^5V) = 9.2*10^-14 J.
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RESPONSE --> Part B is the answer I put, that was correct. I also used the same method for Part A of the problem as the solution to the right. V=k(Q)/r 9.0E9(1.60E-19/2.5E-15)=5.74E5
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???J??????^??? assignment #006 006. Physics II 06-17-2007"