#$&* course PHY 202 026. 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: Current is defined as the number of charges passing a point divided by the time needed to pass the point. Or: The number of charges per unit length divided by the drift velocity.
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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 interval `dL of the conductor and the drift velocity is v, all of the N charges will pass the end of the length interval in time interval `dt = `dL / 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 passing the end of the length interval, is • current = N / `dt = N / (`dL / v) = (N / `dL) * v. N / `dL is the number of charges per unit length, and v is the drift velocity, so we can also say that • current = number of charges per unit length * drift velocity &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 will have a greater current because of the larger cross sectional area. A greater current implies a lesser resistance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Greater cross-sectional area implies greater number of available charge carriers. For a given voltage and a given length of wire the electric field (equal to `dV / `dL) will be the same. Since it is the electric field that accelerates the charge carriers each charge will experience the same acceleration, independent of the cross-sectional area. The average drift velocity of the charge carriers will therefore be the same, regardless of the cross-sectional area. The result will be greater current for a given voltage. Greater current for a given voltage implies lesser electrical resistance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qWill a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The wire will have a greater electrical resistance because a longer wire corresponds to less current flow and thus, more resistance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The electric field is E = `dV / `dL, so greater length implies lesser electrical field for a given voltage, 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: The direction of the field is south because that is the direction of the force on the proton. The magnitude can be found by: Force/charge 3.75*10^-14N/1.6*10^-19 = 2.36*10^5 N/C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The direction of the electric field is the same as that as the force on a positive charge. The proton has a positive charge, so the direction of the field is to the south. The magnitude of the field is equal to the magnitude of the force divided by the charge. The charge on a proton is 1.6 * 10^-19 Coulombs. So the magnitude of the field is E = F / q = 3.75 * 10^-14 N / (1.6 * 10^-19 C) = 2.36* 10^5 N / C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery gen phy problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The magnitude of the charge is q. The contribution to the field from each point is: k q / r^2 r = .08 meters. Both charges contribute equally to the field so the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. E = k q / r^2. Now, we just solve for q to get the point charge: = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * We make the following conceptual observations: • At a point halfway between two opposite charges, a positive test charge will be repelled from one and attracted to the other. The repelling force on the test charge will be in the direction from the positive charge toward the negative charge, and the attracting force will also be toward the negative charge, so the two forces will reinforce one another. • • Thus the electric field at the halfway will be directed from the positive charge toward the negative, and will be double the field produced by either of the charges. • • The halfway point is 8 cm from each of the two charges. If the magnitude of the charge is q then the field contribution of each 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 the field of one of the charges is E = 373 N/C. Another expression for this field is E = k q / r^2. We solve for q to obtain q = E * r^2 / k. Substituting our values for k, E and r we 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 field from either charge = k Q / r^2 So, sense both charges are the same, the total field will be: E = 2 k Q / r^2, When r=.08 So, E= 2(375)(2.6*10^-10)/.08^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This calls for a symbolic expression in terms of the symbol Q. The field from either charge is k Q / r^2, directed toward the negative charge. The field of both charges together is therefore E_total = 2 k Q / r^2, where r=.08 meters. ** STUDENT COMMENT: That is a tough one. I will have to read up on this one. I guess you just added the 2 because they are two charges? INSTRUCTOR RESPONSE: There are two charges and you are asked for the field at their midpoint. We find the field due to each of the two charges, then we add the two fields. Had the charges been of the same sign, rather than equal and opposite, the two fields would have been equal, but opposite, and would therefore have added up to zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If we are supposed to find the magnitude of the E field, then: E = kqQ/r^2 So, I substituted values in: E = 9 * 10^9 N m^2 / C^2 * 33. * 10^-6 C / (.200 m)^2 = 7.43 * 10^6 N / C. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A positive test charge Q 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. The field is the force per unit test charge, in this case (k q Q / r^2) / Q = k q / r^2. Substituting our given values 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): Not really sure if I answered the correct question. ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!