Talk:QB/d cp2.gaussC

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

QB/e_cp2.6

1

• If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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

• If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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

• If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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

• If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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

• In this description of the flux element, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, $dA_{1}=dA_{3}$

  

- 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

• In this description of the flux element, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, ${\vec {E}}_{1}\cdot dA_{1}={\vec {E}}_{3}\cdot dA_{3}$

  

+ True - False

              This partners with the similar one that is false.

7

• In this description of the flux element, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, ${\vec {E}}_{1}\cdot dA_{1}+{\vec {E}}_{3}\cdot dA_{3}=0$

  

- True + False

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

8

• In this description of the flux element, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, ${\vec {E}}_{1}\cdot dA_{1}+{\vec {E}}_{2}\cdot dA_{3}=0$

  

+ True - False

              True because they are of opposite sign.

Raw Text

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 $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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 $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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 $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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 $(\varepsilon _{0}EA^{*}=\rho V^{*})$, ${\vec {E}}$ 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, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, $dA_{1}=dA_{3}$

- 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, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, ${\vec {E}}_{1}\cdot dA_{1}={\vec {E}}_{3}\cdot dA_{3}$

+ True

- False

$ This partners with the similar one that is false.

! Public Domain CC0 user:Guy vandegrift

? In this description of the flux element, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, ${\vec {E}}_{1}\cdot dA_{1}+{\vec {E}}_{3}\cdot dA_{3}=0$

- 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, $d{\vec {S}}={\hat {n}}dA_{j}$ (j=1,2,3) where ${\hat {n}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $S_{1}$ and $S_{3}$ but enter at $S_{2}$. In this figure, ${\vec {E}}_{1}\cdot dA_{1}+{\vec {E}}_{2}\cdot dA_{3}=0$

+ True

- False

$ True because they are of opposite sign.