course Phy 202 ??K??????·??????assignment #005
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16:32:24 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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16:32:26 Introductory Problem Set 2
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16:33:44 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 length and the cross-sectional area of a wire are partial in determining how many electrons are traveling and carrying a charge. the shorter the wire, the more electrons must pass in a shorter distance, and the thicker the wire, the more electrons are allowed to pass through.
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16:35:23 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 velocity of the carriers determines the distance of the wire they travel. This is them multiplied by the charge carriers to find the charge carriers/second
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16:36:11 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 --> a greater cross-sectional area of a wire allows a greater amperage because there are more charge carriers present
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16:37:00 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 --> with a greater length, the electrical resistance will be less. A shorter distance means more charge carriers crammed into the smaller distance and carrying more charge.
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16:38:20 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 --> the direction of the field is 270 degrees and the magnitude is 3.75x10^-14N
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16:39:37 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 --> ok the magnitude is the force divided by the charge. the charge of a proton is 1.6x10-19 C
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16:40:04 ** 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
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16:40:50 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 center will be 2kQ/r^2
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16:41:50 ** 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
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16:42:26 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 --> (9x10^9 Nm^2/C^2)(x^2)/(.16m)^2=745N/C x^2=2.119x10^-9 x=4.6x10^-5
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16:48:53 Query Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
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RESPONSE --> the force will be directly upward. E=kq/r^2=(9x10^9Nm^2/C^2)/(.2m^2) =7.43x10^6 N/C
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16:51:26 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 --> ok
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16:51:29 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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16:51:31 **** 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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