course Phy 232 ???x???x???????assignment #005
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15:50:18 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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15:50:22 Introductory Problem Set 2
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RESPONSE --> done
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15:50:37 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 --> Greater the length, the more dispersed the currect carrying electrons so the current is less. The smaller the cross sectional area, the faster electrons have to pass through that area and therefore have a greater current.
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15:50:48 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 drift velocity tells you that all the electrons of a certain length of the wire pass through a given point. If you find that percentage and mulitply it by the number of charge carriers - that will tell you how many electrons pass through a point in 1 second. multiply that result by the 1.6 x10^-19 (amount of colombs an electron holds) and you have the current.
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15:50:56 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 thinner wire will have more resistance so fewer electrons will flow through, which will make the generator easier to crank - a greater cross sectional area will offer less resistance allowing more electrons to flow making it harder to crank.
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15:51:10 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 --> A longer wire has a lower current so it will have fewer electrons and the crank will be easier (less elec. resistance).
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16:04:10 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 --> Direction is towards the proton (south) b/c they are opposite charges.
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16:04:24 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 --> yesss
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16:04:49 ** 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 --> is this a question?
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16:40:44 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> at the midpoint, two equal and opposite charges would cancel each other out so the electric field would be 0
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16:41:42 ** 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'm not sure where the 8cm came from
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16:42:20 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 --> I think the previous answer goes with this one...right?
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16:47:16 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 Q / r^2) / Q = (k * q)/r^2 E = (9 x 10^9 * 33.0 x 10^-6)/ .2^2 = 7.43 x 10^6
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16:47:22 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 --> yesssss
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20:00:26 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 --> flux = perpendicular component of mag field * area of coil area of coil in m is .3^2 = .09 for each side, it's E dot perp component of mag field * area. sides that dot with +/- j = 0 b/c there is only an i and k component to E, sides that dot with +/- k equal +/- the z component of E and sides that dot with +/- i equal +/- the x component There are 0 y contributions from S1 and S3, so S2 yields z=3m, S4 gives z=0 so the flux on the z axis is 3 * .09 = .27 S5 gives us x=3 and S6 gives x=0, so the x axis yields -5 * .09 = -.45 total flux is: .27 - .45 = -.18 4 pi k Q = -.18 Q = -.18 / (4 pi k) = -1.6 * 10^-12 C, approx. **
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20:01:00 **** 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 --> I messed up my decimal places somewhere - I'll be doing this one again!!
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