course Phy 232
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
026. Query 27
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).
Introductory Problem Set 2
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Question: `qBased 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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The longer a wire, the less of a potential gradient occurs; however, drift velocity and current are proportionally larger. The current in a wire of a larger diameter will be (d2/d1)^2 more than the current in the wire with the smaller diameter.
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Question: `qHow 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?
By knowing the number of current-carrying electrons in a wire of a length, L, and if we know that they drift a certain distance in one second, dL. Then we can calculate that dL/L number of electrons will pass a point in a second. Each electron carries 1.6*10^-19 Coulombs so:
Current = 1.6*10^-19 Coulombs/electron * (dL/L electrons/sec) * #electrons = amps
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Question: `qWill a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why?
If a wire has a greater cross-sectional area it will have a lesser electrical resistance because there is more surface area for the electrons to pass through at a given time and during a given length.
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Question: `qWill a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why?
It will have a greater electrical resistance if its length is greater because it requires the electrons to pass through a longer length, thus reducing the current which is inversely proportional to resistance.
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Question: `qQuery 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?
I’m in university physics
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Question: `q`qThe 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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Question: `qQuery 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?
I’m in university physics
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Question: `q`q** 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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Question: `qIf the charges are represented by Q and -Q, what is the electric field at the midpoint?
I’m in university physics
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Question: `q`q** 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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Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
I’m in university physics
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Question: `q`qA 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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Question: `qquery 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?
Area of each side = 0.3 m^2 = 0.09 m^2
E = -5xi + 3zk
We are using the dot product to find the flux at each face:
S1: -5xi + 3zk • -j * 0.09 = 0
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Question: `q`q**** 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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Question: `qquery 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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Question: `q`q** 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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Question: `qquery 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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Question: `q`q**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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The formatting of the last several queries was confused, as I discovered just a little earlier this evening. I think I've repaired the formatting. I apologize for the confusion.
Be sure to let me know if you have questions.