course Phys 202 8:43 PM 10-24-09 026. Query 27
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Given Solution: The electric field in the wire is equal to the voltage divided by the length of the wire. So a longer wire has a lesser electric field, which results in less acceleration of the free charges (in this case the electrons in the conduction band), and therefore a lower average charge velocity and less current. The greater the cross-sectional area the greater the volume of wire in any given length, so the greater the number of charge carriers (in this case electrons), and the more charges to respond to the electric field. This results in a greater current, in proportion to the cross-sectional area. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qHow 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The ratio of drift velocities will equal the ratio of potential gradients. Drift velocity is how quickly charge carriers will travel the length of the wire. The current in a wire is proportional to the drift velocity and is equivalent to the number of charges passing a specific point per unit of time. To get the current then you can take number of charges and divide by the drift velocity which is L/v.
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Given Solution: The charge carriers in a unit length will travel that length in a time determined by the average drift velocity. The higher the drift velocity the more quickly they will travel the unit length. This will result in a flow of current which is proportional to the drift velocity. Specifically if there are N charges in length L of the conductor and the drift velocity is v, the N charges will pass the end of the length in time interval `dt = L / v. The current can be defined as • current = # of charges passing a point / time required to pass the point Thus the current, in charges / unit of time, is N / (L / v) = N / L * v &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWill a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A wire of a given length and material will have a lesser electrical resistance if its cross-sectional area is greater because there is a greater volume of charge available to give a greater current at the same voltage. I = V/R If I increases R decreases. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Greater cross-sectional area implies greater available charge, which implies greater current for a given voltage. Greater current for a given voltage implies less electrical resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWill a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: A wire of a given material and cross-sectional area will have greater resistance if its length is greater because this causes a decrease in current. If current decreases resistance increases. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Greater length implies lesser electrical field for a given resistance, which implies less current flow. This implies greater electrical resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = F/ q E = 3.75 * 10^-14 N/ 1.6 * 10 ^ -19 C = 2.34 * 10^ 5 N/C towards the South Since it is a positive charge it will travel in the same direction as the electric field. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 * 10^-19 C) = 2.36* 10^5 N / C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = k q/r^2 373 N/C = (9*10^9 N m^2 / C^2) * q / (.08 m )^2 q = 2.65 * 10 ^ -10 C confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIf the charges are represented by Q and -Q, what is the electric field at the midpoint? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The electric field at the midpoint would be equal to E/2 = k Q/r^2 Which is E = 2k Q/r^2 R would be half the distance between the two since the charge is at the midpoint. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: E = k Q/r^2 E = (9 * 10 ^ 9 N m^2 / C^2) (33.0 * 10^-6 C) / (.2 m)^2 E = 7.43 * 10 ^ 6 N/C Since the charge is positive a positive charge at the point would be repelled, upwards. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery univ 22.30 (10th edition 23.30). Cube with faces S1 in xz plane, S2 top, S3 right side, S4 bottom, S5 front, S6 back on yz plane. E = -5 N/(C m) x i + 3 N/(C m) z k. What is the flux through each face of the cube, and what is the total charge enclosed by the cube? ********************************************* Question: `qquery univ 22.37 (23.27 10th edition) Spherical conducting shell inner radius a outer b, concentric with larger conducting shell inner radius c outer d. Total charges +2q, +4q. ********************************************* Question: `qquery univ 23.46 (23.34 10th edition). Long conducting tube inner radius a, outer b. Lin chg density `alpha. Line of charge, same density along axis.