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
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?
Your solution:
Confidence rating:
Given Solution: The electric field in the wire is equal to the voltage divided by the length of the wire. So a longer wire has a lesser electric field, which results in less acceleration of the free charges (in this case the electrons in the conduction band), and therefore a lower average charge velocity and less current.
The greater the cross-sectional area the greater the volume of wire in any given length, so the greater the number of charge carriers (in this case electrons), and the more charges to respond to the electric field. This results in a greater current, in proportion to the cross-sectional area.
Self-critique (if necessary):
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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?
Your solution:
Given Solution: The charge carriers in a unit length will travel that length in a time determined by the average drift velocity. The higher the drift velocity the more quickly they will travel the unit length. This will result in a flow of current which is proportional to the drift velocity.
Specifically if there are N charges in length interval `dL of the conductor and the drift velocity is v, all of the N charges will pass the end of the length interval in time interval `dt = `dL / v. The current can be defined as
current = # of charges passing a point / time required to pass the point
Thus the current, in charges / unit of time passing the end of the length interval, is
current = N / `dt = N / (`dL / v) = (N / `dL) * v.
N / `dL is the number of charges per unit length, and v is the drift velocity, so we can also say that
current = number of charges per unit length * drift velocity
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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?
Your solution:
Confidence rating:
Given Solution: Greater cross-sectional area implies greater number of available charge carriers.
For a given voltage and a given length of wire the electric field (equal to `dV / `dL) will be the same.
Since it is the electric field that accelerates the charge carriers each charge will experience the same acceleration, independent of the cross-sectional area. The average drift velocity of the charge carriers will therefore be the same, regardless of the cross-sectional area.
The result will be greater current for a given voltage.
Greater current for a given voltage implies lesser electrical resistance.
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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?
Your Solution:
Confidence rating:
Given Solution: The electric field is E = `dV / `dL, so greater length implies lesser electrical field for a given voltage, which implies less current flow. This implies greater electrical 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?
Your Solution:
Confidence rating:
Given Solution:
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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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?
Your Solution:
Confidence rating:
Given Solution:
* We make the following conceptual observations:
At a point halfway between two opposite charges, a positive test charge will be repelled from one and attracted to the other. The repelling force on the test charge will be in the direction from the positive charge toward the negative charge, and the attracting force will also be toward the negative charge, so the two forces will reinforce one another.
Thus the electric field at the halfway will be directed from the positive charge toward the negative, and will be double the field produced by either of the charges.
The halfway point is 8 cm from each of the two charges.
If the magnitude of the charge is q then the field
contribution of each 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 the field of one of the charges is E = 373 N/C.
Another expression for this field is E = k q / r^2.
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?
Your Solution:
Confidence rating:
Given Solution:
** This calls for a symbolic expression in terms of the symbol Q.
The field from either charge is k Q / r^2, directed toward the negative charge.
The field of both charges together is therefore
E_total = 2 k Q / r^2,
where r=.08 meters. **
STUDENT COMMENT:
That is a tough one. I will have to read up on this one. I
guess you just added the 2
because they are two charges?
INSTRUCTOR RESPONSE:
There are two charges and you are asked for the field at
their midpoint.
We find the field due to each of the two charges, then we add the two fields.
Had the charges been of the same sign, rather than equal and opposite, the two
fields would have been equal, but opposite, and would therefore have added up to
zero.
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Question:
`qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C
charge.
Your Solution:
Confidence rating:
Given Solution:
A positive test charge Q 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.
The field is the force per unit test charge, in this case (k q Q / r^2) / Q = k q / r^2.
Substituting our given values 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 22.34 / 22.32 11th edition 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?
Your Solution:
Confidence rating:
Given Solution:
**** 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 22.47/22.45 11th edition 22.39 (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.
Your Solution:
Confidence rating:
Given Solution:
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.
**
STUDENT COMMENT
with that said it now becomes a little wierd that the net
charge would be 6q with a negative 2q on the outer surface of the
inner shell and a -2q on the inner of the outer shell. i would think that these
cancel each other, They really wouldnt cancel
each other though because they dont have the same radius.
I'll just trust that the field is 6q on the outside and that the charge is also
k*6q/r^2
INSTRUCTOR RESPONSE
The -2q of charge on the inside of the outer shell stays
there, rather than migrating to the outside of the shell, because of the
electric field created by the +2q charge on the inner shell.
If the inner shell was removed, or the charge on it somehow neutralized, then
the -2q on the inside of the outer shell would migrate to the outside of the
shell, leaving 0 charge on the inside and charge +4q on the outside of that
shell.
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Question:
`qquery univ 22.40 / 22.30 / 22.38 11th edition 23.46 (23.34 10th edition). Long conducting tube
inner radius a, outer b. Lin chg density
`alpha (or possibly `lambda, depending on which edition of the text you are
using). Line of charge, same density along axis.
Your Solution:
Confidence rating:
Given Solution:
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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Question:
`qquery univ phy 13.64/23.62 11th edition 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow
cylinder radius 1.8 cm.
What potential difference between the wire in the cylinder
will produce an electric field of 20,000 volts/m at 1.2 cm from the wire?
Your Solution:
Confidence rating:
Given Solution: ** The voltage V_ab is obtained by integrating the
electric field from the radius of the central wire to the outer radius.
From this we determine that E = Vab / ln(b/a) * 1/r, where a
is the inner radius and b the outer radius.
If E = 20,000 V/m at r = 1.2 cm then
Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) *
.012 m = 1157 V. **
STUDENT QUESTION:
Can you tell me what you integrated to get: E = Vab / ln(b/a)
* 1/r ?
INSTRUCTOR RESPONSE:
Sure. The following assumes you know how to use Gaussian surfaces for axially symmetric charge distributions. If necessary see your text to fill in the details, but given the basic knowledge the explanation that follows is complete. I'll also be glad to clarify anything you wish to ask about:
If the charge per unit length on the inner cylinder is lambda, then a coaxial cylinder of length L will contain charge Q = lamda * L.
So the flux through the cylinder will be 4 pi k Q = 4 pi k lambda * L.
Using symmetry arguments and assuming edge effects to be negligible, the electric field penetrates the curved surface of the cylinder at right angles.
The area of the curved surface of such a coaxial cylinder of radius r is 2 pi r * L, so the electric field is
field E = flux / area = 4 pi k lambda * L / (2 pi r L) = 2 k lambda / r.
Integrating this field from inner radius a to outer radius b, we get the
potential difference Vab:
Our antiderivative function is 2 k lambda ln | r |, so the change in the antiderivative is
Vab = 2 k lambda ( ln(b) - ln(a) ) = 2 k lambda ln(b /
a).
Thus Vab = 2 k lambda ln(b/a).
This gives us 2 k lambda = Vab / (ln(b/a)), which will be
used below.
Since E = 2 k lambda / r, we substitute to get
E = Vab / (ln(b/a)) * 1 / r, the expression about which you asked, and which
we might wish to simplify into the form
E = Vab / (r ln(b/a) ).
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