course Phy 202
1:15am 7/25
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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Your solution:
The difference in cross-sectional area between 2 different wires can be multiplied by the current of one of the wires, to find the current of another. This only applies if the lengths of the two wires are identical.
confidence rating:
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2
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.
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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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Your solution:
The difference in the lengths of 2 wires is multiplied by the drift velocity of electrons, to find the current in a wire.
confidence rating:
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2
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 L of the conductor and the drift velocity is v, the N charges will pass the end of the length in time interval `dt = L / 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, is N / (L / v) = N / L * v.
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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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Your solution:
A greater cross-sectional area in a wire will have less electrical resistance because it will be less restrictive to the electrons that flow throw through, being that there is more space.
confidence rating:
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2
Greater cross-sectional area implies greater available charge, which implies greater current for a given voltage.
Greater current for a given voltage implies less 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?
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Your Solution:
I am not sure about this answer, but I am going to give it an educated guess. I think that resistance is like a blockage. It’s something that makes it harder for whatever substance we’re working with to get by. I don’t think that resistance is something that is influenced by the length because although one is longer than the other, the electricity should go through at the same rate.
confidence rating:
Greater length implies lesser electrical field for a given resistance, which implies less current flow. This implies greater electrical resistance.
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1
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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?
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Your Solution:
Magnitude= force/ charge
Mag= (3.75*10^-14 N)/(1.6*10^-19 Coulombs)
Mag= 2.36*10^5 N/C
confidence rating:
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3
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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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Self-critique (if necessary):
ok
Self-critique Rating:
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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?
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Your Solution:
Field contribution= K*q/r^2
Field contribution= .08 meters
Q= E*r^2/k
Q= (373 N/C)*(.08 m)^2/(9*10^9 N m^2/C^2)
Q= (373 N/C)*(.0064 m^2)/(9*10^9 N m^2/C^2)
Q= 2.6*10^-10 C
confidence rating:
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3
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Given Solution:
* 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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Self-critique (if necessary):
ok
Self-critique Rating:
3
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Question: `qIf the charges are represented by Q and -Q, what is the electric field at the midpoint?
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Your Solution:
The electric field would be,
2kQ/r^2
R= .08 m
It is multiplied by 2 because there are two charges of magnitude Q
confidence rating:
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1
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Given Solution:
** 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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Self-critique (if necessary):
ok
Self-critique Rating:
3
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Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
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Your Solution:
E= (kqQ/r^2)/ Q
E= (9*10^9 N m^2/C^2)* (33.0*10^-6 C)/ (.200 m)^2
E= 7.43*10^6 N/C
confidence rating:
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2
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Given Solution:
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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Self-critique (if necessary):
ok
Self-critique Rating:
3
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