Talk:QB/d cp2.gaussC

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First to allow and display discussion of each question, and second, to store the quiz in raw-script for.

QB/e_cp2.6[edit source]

1[edit source]

  • If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , was calculated inside the Gaussian surface

- True + False

              True false questions are easy to write, but a collection of them forces the student to remember the answer (undestanding it is the student's responsibility)

2[edit source]

  • If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , was calculated outside the Gaussian surface

- True + False

              Part of a triplet of similar TF questions, designed to force those who memorize to work harder than those who understand.

3[edit source]

  • If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , was calculated on the Gaussian surface

+ True - False

              Think of the Gaussian surface as a "probe" that you place where you want to know the field

4[edit source]

  • If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , had

- constant direction and magnitude over the entire Gaussian surface + constant magnitude over a portion of the Gaussian surface - constant direction over a portion of the Gaussian surface - constant in direction over the entire Gaussian surface

              students need to be coached on the meaning of "necessary" and "sufficient".

5[edit source]

  • In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
    GAUSS2.svg

- True + False

              Sort of a trick question, perhaps, but this is an important topic. Also, we pair this on up with one where it is true.

6[edit source]

  • In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
    GAUSS2.svg

+ True - False

              This partners with the similar one that is false.

7[edit source]

  • In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
    GAUSS2.svg

- True + False

              False because they are equal in absolute magnitude but of the same sign.

8[edit source]

  • In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
    GAUSS2.svg

+ True - False

              True because they are of opposite sign.

Raw Text[edit source]

t QB/e_cp2.6
! Public Domain CC0 user:Guy vandegrift
? If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , was calculated inside the Gaussian surface
- True
+ False
$ True false questions are easy to write, but a collection of them forces the student to remember the answer (undestanding it is the student's responsibility)
! Public Domain CC0 user:Guy vandegrift
? If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , was calculated outside the Gaussian surface
- True
+ False
$ Part of a triplet of similar TF questions, designed to force those who memorize to work harder than those who understand.
! Public Domain CC0 user:Guy vandegrift
? If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , was calculated on the Gaussian surface
+ True
- False
$ Think of the Gaussian surface as a "probe" that you place where you want to know the field
! Public Domain CC0 user:Guy vandegrift
? If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field , had
- constant direction and magnitude over the entire Gaussian surface
+ constant magnitude over a portion of the Gaussian surface
- constant direction over a portion of the Gaussian surface
- constant in direction over the entire Gaussian surface
$ students need to be coached on the meaning of "necessary" and "sufficient".


! Public Domain CC0 user:Guy vandegrift
? In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
GAUSS2.svg
- True
+ False
$ Sort of a trick question, perhaps, but this is an important topic. Also, we pair this on up with one where it is true.


! Public Domain CC0 user:Guy vandegrift
? In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
GAUSS2.svg
+ True
- False
$ This partners with the similar one that is false.


! Public Domain CC0 user:Guy vandegrift
? In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
GAUSS2.svg
- True
+ False
$ False because they are equal in absolute magnitude but of the same sign.
! Public Domain CC0 user:Guy vandegrift
? In this description of the flux element, (j=1,2,3) where is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at and but enter at . In this figure,
GAUSS2.svg
+ True
- False
$ True because they are of opposite sign.