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

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

+ True - False

This partners with the similar one that is false.

### 7

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

- True + False

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

### 8

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

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

! Public Domain CC0 user:Guy vandegrift
? In this description of the flux element, ${\displaystyle d{\vec {S}}={\hat {n}}dA_{j}}$ (j=1,2,3) where ${\displaystyle {\hat {n}}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at ${\displaystyle S_{1}}$ and ${\displaystyle S_{3}}$ but enter at ${\displaystyle S_{2}}$. In this figure, ${\displaystyle {\vec {E}}_{1}\cdot dA_{1}+{\vec {E}}_{3}\cdot dA_{3}=0}$
- 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, ${\displaystyle d{\vec {S}}={\hat {n}}dA_{j}}$ (j=1,2,3) where ${\displaystyle {\hat {n}}}$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at ${\displaystyle S_{1}}$ and ${\displaystyle S_{3}}$ but enter at ${\displaystyle S_{2}}$. In this figure, ${\displaystyle {\vec {E}}_{1}\cdot dA_{1}+{\vec {E}}_{2}\cdot dA_{3}=0}$ + True - False$ True because they are of opposite sign.