Query 26

course Phy 232

026. Query 27Note 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 electric field is equal to the voltage/length. The bigger the cross-sectional area the greater the volume. I think I got that right.

confidence rating:

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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.

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Self-critique (if necessary):

I didn’t really describe it, but I think I got it right.

Your statements were valid, and they were relevant to the question. They didn't provide a complete explanation, but it wasn't a bad attempt.

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Self-critique Rating:

3

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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 charge carriers will travel the length in a time which would be determined by the calculated average drift velocity.

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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 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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Self-critique (if necessary):

Once again my answer lacks explanation, but I got the beginning part right.

An improvement based on your wording:

All the charge carriers present in the wire will travel the length in a time which would be determined by the calculated average drift velocity.

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Self-critique Rating:

ok

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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: The bigger the cross sectional area the greater the available charge. This means greater current for a given voltage. Greater current for voltage means less electrical resistance.

confidence rating:

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Given Solution: 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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Self-critique (if necessary):

ok

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Self-critique Rating:

ok

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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: The bigger the length the less the electrical field will be for resistance. This means less current flow and greater electrical resistance I believe. I may have that backwards I’m not sure.

Confidence rating: 2

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Given Solution: Greater length implies lesser electrical field for a given resistance, which implies less current flow. This implies greater electrical resistance.

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Self-critique (if necessary):

I thought that was right.

You were.

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Self-critique Rating:

ok

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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 midpoint would be 2 *k*Q/r^2.

confidence rating:

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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

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Self-critique Rating:

ok

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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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Your Solution: The area of each face would be 0.09 m^2. So it would be (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0 for S1 flux. Then for S2 it would be 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.

S3: (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0.

S4:(-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.

S5: (-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.

S6: (-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 would get flux .027 N m^2 / C on S2 and flux 0 on S4.

On S5 and S6 we would get 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.

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. **

Confidence rating: 3

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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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Self-critique (if necessary):

ok

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Self-critique Rating:

ok

&#Your work looks good. See my notes. Let me know if you have any questions. &#