course phy 232 I am really starting to have trouble with these qa s none of the questions are completely typed and I cannot find them in the newer 12th edition book so I cannot answer some of the question so I tried to give explanations. xvܹ|ysx{assignment #005
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22:35:01 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 --> k
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22:35:08 Introductory Problem Set 2
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RESPONSE --> k
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22:40:35 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 current speed and KE all depend on the crossectional area and length. The velocity of electrons through a wire is similar to the flow rate formula v2=a1/a2 * v1 this means if the first area is greater then the wire is thicker and the electrons will move faster through the smaller one. The KE2=KE1 *[a1/a2]^2 where the same counts for KE as well. The KE is higher when traveling faster.
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22:43:17 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 --> By knowing the drift velocity(m/s) and the number of the charge carriers per unit length(C/m) then by multipling them together you get c/s which are amperes.
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22:45:27 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 --> Greater because there will be more electrons to move in a certain direction meaning more energy needed like in the burning of light bulb filaments where the thicker filaments take more power to fuel.
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22:47:15 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 --> Greater because the electrons are already moving faster but they are also colliding more often with other particles and losing kinetic energy so they need more energy to get through the longer distance.
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22:52:28 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 --> Assuming the charge of a proton is 1 coulomb. using e= f/q , 3.75 * 10^-14 N/1= 3.75 * 10^-14 n/c
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22:56:30 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.
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RESPONSE --> The assumption for the charge of a proton was wrong and messed up the answer. The direction of the field would of course be south since the force is equal and opposite and the charge is in the north.
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23:02:33 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 --> If both charges are equal and opposite and the field force is 745 n/c then E=Q*k / R^2 745*.08^2 / 9x10^9 = 5.29x10^-10 C This is the magnitude for both charges but one charge is negative.
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23:05:25 ** 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 --> The process is correct but the idea that the field is double at the mid point was overlooked. Because both field are affecting the midpoint they both have resultant forces on the middle so the middle field value is half since both fields are of the same magnitude.
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23:07:07 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> E= 2*[k*|Q| / [distance between both charges/2]^2
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23:07:19 ** 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 -->
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23:09:33 Query Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
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RESPONSE --> If you are looking for the electric field. then use E=k*|q| / r^2 E=33.0 * 10^-6 C * 9*19^9 / .2^2 = 7425000 N/C
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23:09:39 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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23:16:28 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 --> Because the object is a cude and that all of the inward flux is also accompanied by outward flux, there is no net flux. The flux through each face of the cube cannot be calculated because the area or dimensions of the cube are not given even though E is.
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23:23:06 **** 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 --> The question here was very difficult to undertand and was not in the book as numbered. If the length of the box was given then I am wrong and the area was included. I understand the dot product is used and not the cross product. The flux is then simply the components of the flux in the i,j,k times the area. The total flux is the sum of all of the flux of all of the sides. Once the total flux is obtained it is equal to 4 pi k Q where Q is the charge inclosed by the surface asked for.
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23:24:43 query univ 22.37 (23.27 10th edition) Spherical conducting shell inner radius a outer b, concentric with larger conducting shell inner radius c outer d. Total charges +2q, +4q. Give your solution.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. What is the question? Where is it in the book?
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23:27:11 ** The electric field inside either shell must be zero, so the charge enclosed by any sphere concentric with the shells and lying within either shell must be zero, and the field is zero for a < r < b and for c < r < d. Thus the total charge on the inner surface of the innermost shell is zero, since this shell encloses no charge. The entire charge 2q of the innermost shell in concentrated on its outer surface. For any r such that b < r < c the charge enclosed by the corresponding sphere is the 2 q of the innermost shell, so that the electric field is 4 pi k * 2q / r^2 = 8 pi k q / r^2. Considering a sphere which encloses the inner but not the outer surface of the second shell we see that this sphere must contain the charge 2q of the innermost shell. Since this sphere is within the conducting material the electric field on this sphere is zero and the net flux thru this sphere is zero. Thus the total charge enclosed by this sphere is zero. Since the charge enclosed by the sphere includes the 2q of the innermost shell, the sphere must also enclose a charge -2 q, which by symmetry must be evenly distributed on the inner surface of the second shell. Any sphere which encloses both shells must enclose the total charge of both shells, which is 6 q. Since we have 2q on the innermost shell and -2q on the inner surface of the second shell the charge on the outer surface of this shell must be 6 q. For any r such that d < r the charge enclosed by the corresponding sphere is the 6 q of the two shells, so that the electric field is 4 pi k * 6q / r^2 = 24 pi k q / r^2. **
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RESPONSE --> Any sphere with a net charge of zero with an external nonuniform shape also has the zero net charge due to the law flux= E*Acos{angle].
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23:28:53 query univ 23.46 (23.34 10th edition). Long conducting tube inner radius a, outer b. Lin chg density `alpha. Line of charge, same density along axis.
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RESPONSE --> What is the question?
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23:35:15 **The Gaussian surfaces appropriate to this configuration are cylinders of length L which are coaxial with the line charge. The symmetries of the situation dictate that the electric field is everywhere radial and hence that the field passes through the curved surface of each cylinder at right angle to that surface. The surface area of the curved portion of any such surface is 2 pi r L, where r is the radius of the cylinder. For r < a the charge enclosed by the Gaussian surface is L * alpha so that the flux is charge enclosed = 4 pi k L * alpha and the electric field is electric field = flux / area = 4 pi k L * alpha / (2 pi r L ) = 2 k alpha / r. For a < r < b, a Gaussian surface of radius r lies within the conductor so the field is zero (recall that if the field wasn't zero, the free charges inside the conductor would move and we wouldn't be in a steady state). So the net charge enclosed by this surface is zero. Since the line charge enclosed by the surface is L * alpha, the inner surface of the conductor must therefore contain the equal and opposite charge -L * alpha, so that the inner surface carries charge density -alpha. For b < r the Gaussian surface encloses both the line charge and the charge of the cylindrical shell, each of which has charge density alpha, so the charge enclosed is 2 L * alpha and the electric field is radial with magnitude 4 pi k * 2 L * alpha / (2 pi r L ) = 4 k alpha / r. Since the enclosed charge that of the line charge (L * alpha) as well as the inner surface of the shell (L * (-alpha) ), which the entire system carries charge L * alpha, we have line charge + charge on inner cylinder + charge on outer cylinder = alpha * L, we have alpha * L - alpha * L + charge on outer cylinder = alpha * L, so charge on outer cylinder = 2 alpha * L, so the outer surface of the shell has charge density 2 alpha. **
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RESPONSE --> Since th object described resembles a cylinder it is necessary to use the equation E=K* lambda /r where lambda is the change per unit length and r is the radius of the conducting cylinder. When the charge is within the cylinder E is always = 0. Beyond this I have no further understand of this explanation. I need help.
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