course phy122 I got help here with the text problems. Lots of work to do here with my new tutor before I will pass any exam on the subject matter. I went back and entered textbook problem's answers after cut and pasting to this submit work form--you'll see in time line--I did not consult with answers below though. Hope that is OK.
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11:23:16 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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11:23:23 Introductory Problem Set 2
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RESPONSE --> ok
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11:34:09 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 greater the resistance the lower the current. The longer the wire the greater the resitance. The greater the cross-sectional area the lower the resistance resulting in a higher current.
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11:39:29 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 --> Current = Charges per unit volume * Magnitude of the charge on each moving charge * Drift Velocity * Cross sectional area perpendicular to the flow
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11:44:59 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 --> I = nAvQ Current = charge particles per unit volume * cross sectional area of the conductor * drift velocity * charge on each particle ...is perhaps a better answer for the previous question and for this question. The greater cross sectional area of a given material will have less resistance beucause it makes it easier for current to flow.
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11:50:11 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 --> The longer wire will have more resistance because more push (voltage) is required to flow the current throught he longer distance.
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11:55:31 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> I am not really sure.
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11:56:05 ** 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 --> would this be the same a 8kq/r^2 towards the right?
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11:56:31 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 --> 745 N/C = (8)(9.0*10^9 **m^2/c^2)(Q/.16m^2) Solving for Q =2.65 * 10^-10C?
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11:56:37 Query Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
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RESPONSE --> E = (9.0*10^9 N*m^2/C^2) * [(33.0*10^-6C)/(0.2m)^2] = 7.43*10^6 N/C The direction of the electric field is outward from the charge.
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11:56:48 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 -->
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11:57:00 query 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?
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RESPONSE -->
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