1. a and d. Symmetry requires the electric field to be unchanged if front and back are reversed, if left and right are reversed, or if the field is rotated about the wire’s axis. Fields a and d both have the proper symmetry. Other factors would now need to be considered to determine the correct field.

2. e. The net flux is into the box.

3. c. There’s no flux through the four sides. The flux is positive $1{\text{N m}}^2\text{/C}$ through both the top and bottom because $\vec{E}$ and $\vec{A}$ both point outward.

4. $\text{Φ}{}_{\text{B}}={\text{Φ}}_{\text{E}}>{\text{Φ}}_{\text{A}}={\text{Φ}}_{\text{C}}={\text{Φ}}_{\text{D}}$. The flux through a closed surface depends only on the amount of enclosed charge, not the size or shape of the surface.

5. d. A cube doesn’t have enough symmetry to use Gauss’s law. The electric field of a charged cube is not constant over the face of a cubic Gaussian surface, so we can’t evaluate the surface integral for the flux.

1.

3.

5. Negative

7. Inward, $\left|E\right|>10\text{N/C}$

9. $-2.0{\text{ N m}}^2\text{/C}$

11.

1. $2.4\text{N}{\text{m}}^2\text{/C}$

2. 0

13. $1.8{\text{ N m}}^2\text{/C}$

15. $3.5\times {10}^{-4}{\text{ N m}}^2\text{/C}$

19. $+2q\text{,}+q\text{,}-3q$

21. $0.11{\text{ kN m}}^2\text{/C}$

23. $-8.85\text{nC}$

25.

1. $2.5\times {10}^4\text{ N/C,}$ downward

2. 0 N/C

3. $2.5\times {10}^4\text{N/C,}$ upward

27. 0.90 kN/C, 0, 0

29. ${\text{Φ}}_1=-3.2{\text{ kN m}}^2\text{/C},{\text{ Φ}}_2={\text{Φ}}_3={\text{Φ}}_5=0.0{\text{ N m}}^2\text{/C},{\text{ Φ}}_4=3.2{\text{ kN m}}^2\text{/C}$

31. $0.19{\text{ kN m}}^2\text{/C}$

33.

1. $-100\text{nC}$

2. $\pm 50\text{nC}$

35.

1. ${\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{Q}{r^2}}\widehat{r}\text{,}\text{0}\text{.0 N/C,}{\displaystyle \frac{1}{4\pi {ϵ}_0}}\left({\displaystyle \frac{3Q}{r^2}}\right)\widehat{r}$

2. $-Q\text{, }+3Q$

37.

1. $-1\times {10}^{-8}\text{C}$

2. $+1\times {10}^{-8}\text{ C}$

3. $4.8\times {10}^{-8}\text{C}$

39.

41. $700{\text{nC/m}}^2$

43. $0\text{N/C,}{\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{Q}{r^2}}\widehat{r}$

45.

• a. $\vec{E}(\left|z\right|\le z_0)={\displaystyle \frac{\rho z}{{ϵ}_0}}\widehat{k}$

• b. $\vec{E}(z\ge z_0)={\displaystyle \frac{\rho z_0}{{ϵ}_0}}\widehat{k}$

• c.

  

  Full Alternative Text

47.

1. $3Q\text{/(}2{ϵ}_{0}A\text{),}0\text{,}Q\text{/(}2{ϵ}_{0}A\text{),}0\text{,}3Q\text{/(}2{ϵ}_{0}A)$

2. $+{\displaystyle \frac{3}{2}}{\displaystyle \frac{Q}{A}}$, $-{\displaystyle \frac{1}{2}}{\displaystyle \frac{Q}{A}}$, $+Q\text{/(}2A)$, $+3Q\text{/(}2A)$

49.

1. $\left({\displaystyle \frac{\text{λ}}{2\pi {ϵ}_0}}{\displaystyle \frac{1}{r}}\right)\widehat{r}$

2. ${\displaystyle \frac{1}{2\pi {ϵ}_0}}{\displaystyle \frac{\text{λ}r}{R^2}}\widehat{r}$

51. $6.2\times {10}^{-11}{\text{ C}}^2{\text{/N m}}^2$

53. ${\displaystyle \frac{\rho }{6{ϵ}_0}}r$

55.

• b. 0, because this is a neutral atom

• c. $4.6\times {10}^{13}\text{ N/C}$

57.

1. ${\displaystyle \frac{\text{λ}{L}^{2}dy}{4\pi {ϵ}_0[y^2+{(L/2)}^2]}}$

2. $\text{λ}L/\left(4{ϵ}_0\right)Q_{\text{in}}/{ϵ}_0$

59.

1. $C={\displaystyle \frac{Q}{4\pi R}}$

2. ${\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{Q}{Rr}}\widehat{r}$

3. Yes

61.

1. ${\displaystyle \frac{Q}{4\pi {ϵ}_0{R}^2}}$

2. ${\displaystyle \frac{3Q{r}^3}{2\pi R^6}}$