course Phy 202 Sorry, accidentally hit enter when I was typing the assignment number on the last form. à³øŠ‘ÎzÐ’’æ•ÊÉÇS«©{„¬GÁ¿E«ûassignment #005
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16:51:46 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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16:51:54 Introductory Problem Set 2
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RESPONSE --> Yes?
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17:00:21 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 shorter wires have higher currents because the proportion of potential difference to length is higher. As the cross-section increases, the current increases exponentially.
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17:12:41 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 --> The current in a wire can be determined based on these two variables by dividing the length the charge carriers move in a second by the total length of the wire to find the portion of the wire covered by the electrons in one second. The number of charge carriers can be multiplied by this number to find the number of charge carriers in the increment found before. This result represents the charge carrier/second in the wire. The charge of the carrier can be multiplied by the carrier/second value to find the current in Coulombs/second.
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17:20:49 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 --> Because a greater cross-sectional area increases current, the wire will have less electrical resistance; larger currents yield smaller resistance and vice versa.
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17:23:42 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 --> If the length is greater, the current is proportionally smaller, so the electrical resistance will be greater due to the smaller current.
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17:29:05 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 --> Q = 1.602*10^-19 C F = 3.75*10^-14 N south E = F/Q = 3.75*10^-14 N / 1.602*10^-19 C = 2.34*10^5 N/C south
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17:29:17 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 -->
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17:33:28 ** 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 --> I didn't see the problem to this solution, and I can't find the related one in my practice problems. I will summarize the solution, however. The field of each charge that is .08 meters from the midpoint has a magnitude of half of the given field on the midpoint. The given E, k, and r can be used to solve for q.
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17:37:19 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 0.
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17:39:09 ** 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 --> I don't really understand this. Does r=.08 meters come from the previous problem?
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17:43:31 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 --> E = k(Q/r^2) Q=(E*r^2)/k Q = (745 N/C * (.16 m)^2) / 9.0*10^9 N*m^2/C^2) Q = 2.12*10^-9
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17:45:17 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 = k*(Q/r^2) E = 9.0*10^9 N*m^2/C^2 * (33.0*10^6 C/(.20 m)^2) E = 7.4*10^6 N/C
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17:46:02 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 --> I forgot to specify a direction, but I understand that the two like charges repel.
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17:46:10 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 --> n/a?
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17:46:18 **** Advance correction of a common misconception: Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The vectors normal to S1, S2, ..., S6 are respectively -j, k, j, -k, i and -i. For any face the flux is the dot product of the field with the normal vector, multiplied by the area. The area of each face is (.3 m)^2 = .09 m^2 So we have: For S1 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0. For S2 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( k) * .09 m^2 = 3 z N / (C m) * .09 m^2. For S3 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0. For S4 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-k) * .09 m^2 = -3 z N / (C m) * .09 m^2. For S5 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( i) * .09 m^2 = -5 x N / (C m) * .09 m^2. For S6 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-i) * .09 m^2 = 5 x N / (C m) * .09 m^2. On S2 and S4 we have z = .3 m and z = 0 m, respectively, giving us flux .027 N m^2 / C on S2 and flux 0 on S4. On S5 and S6 we have x = .3 m and x = 0 m, respectively, giving us flux -.045 N m^2 / C on S5 and flux 0 on S6. The total flux is therefore .027 N m^2 / C - .045 N m^2 / C = -.018 N m^2 / C. Since the total flux is 4 pi k Q, where Q is the charge enclosed by the surface, we have 4 pi k Q = -.018 N m^2 / C and Q = -.018 N m^2 / C / (4 pi k) = -.018 N m^2 / C / (4 pi * 9 * 10^9 N m^2 / C^2) = -1.6 * 10^-13 C, approx. **
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RESPONSE --> ok
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