course Phy 232
July 20, at 6:20 p.m.
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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Self-critique (if necessary):
Current in a material is proportional to cross-sectional area and inversely proportional to length. So as the length of a wire increases, the amount of current running through it decreases. As the cross sectional area increases, the amount of electrons that can run throughout the material increases, so the current increases as well.
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Question: `q`q
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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?
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Self-critique (if necessary):
The current in a wire is a measurement of the amount of charge per second that runs through the wire. If we divide the velocity of the charge carriers by the number of charge carriers per length, it will give us the speed of the electrons per amount of carriers per unit length. This will give us the current in charges per second.
So, R = V*L/(# of electrons) #################
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Question: `q`q
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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?
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Self-critique (if necessary):
A wire with greater cross-sectional area will present less electrical resistance than one which a smaller area. This is true because the electrons move more freely with a larger wire, so there is less conflicting force. A smaller wire will cause the electrons to be squeezed tighter together, requiring more force to move them.
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Question: `q`q
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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?
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Self-critique (if necessary):
If the rod is longer the electrical resistance is less because the electrical current has to travel a longer the distance. So electrical resistance is inversely proportional to length.
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Question: `q`q
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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?
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Self-critique (if necessary):
The Electric Field vector only points in the x and z directions, so the sides, S1 and S3 will have an electric flux of 0.
The electric flux of a side of the cube is the magnitude of the field pointing in that direction multiplied by the area of the side.
For side S2: E2 = (0.3 m)(0.3 m)(3 N/(C*m)) = 0.27 N*m/C
For side S4: E4 = -E2 = -0.27 N*m/C
For side S5: E5 = (0.3 m)(0.3 m)(-5 N/(C*m)) = -0.45 N*m/C
For side S6: E6 = -E5 = 0.45 N*m/C
Total Flux: Im not sure about this, Im guessing that the total flux would be the magnitude of the electric field vector time the total surface area of the cube:
Magn(E) = sqrt(25+9) = 5.83 N*m/C
Surface Area of cube: 0.09 m^2*6 = 0.45 m^2
Total Flux = 0.45 m^2*5.83 N*m/C = 2.62 N*m/C #########################
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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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Self-critique (if necessary):
Im a little confused on the total flux question. Is it just Ex*area – Ey*area?
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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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Self-critique (if necessary):
We need to find the electric field in three different regions.
- Inside the solid shell
- Between the solid shell and the outer shell
- Outside the outer shell
The electric field inside a solid sphere is:
E = (2q)*r/(R^3)*k
Outside a solid sphere it is:
E = k(2q)/r^2
Inside Outer shell it would be
E = k(2q)/r^2 + (4q)(r)k/(2R – R) #####################
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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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Self-critique (if necessary):
I am a little confused here it seems like this problem is different than the one which is assigned in the book. In the book the problem says we are to take the formula for potential difference of a solid rod and differentiate it with respect to x to find the electric field of the rod.
I got: dV/dx = E_x = 2x*Q*k/(2*a*sqrt(a^2+x^2))(1( sqrt(a^2+x^2) +a) – 1/ (sqrt(a^2+x^2)-a)
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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 for this document was changed as of 7/19/09, the day before you submitted it. I suspect that you copied the document before the change. I'm sorry I didn't get the changes made sooner, but only became aware of the formatting problem on the 19th.
Let me know if there's anything you don't understand in the above. If you want to check it out with the corrected formatting, the document should now be in order.