# 6.A: Gauss's Law (Answers)

- Page ID
- 10250

## Check Your Understanding

**6.1. **Place it so that its unit normal is perpendicular to \(\displaystyle \vec{E}\).

**6.2. **\(\displaystyle mab^2/2\)

**6.3 **a. \(\displaystyle 3.4×10^5N⋅m^2/C;\)

b. \(\displaystyle −3.4×10^5N⋅m^2/C;\)

c. \(\displaystyle 3.4×10^5N⋅m^2/C;\)

d. 0

**6.4. **In this case, there is only \(\displaystyle \vec{E}_{out}\). So,yes.

**6.5. **\(\displaystyle \vec{E}=\frac{λ_0}{2πε_0}\frac{1}{d}\hat{r}\); This agrees with the calculation of Example 5.5 where we found the electric field by integrating over the charged wire. Notice how much simpler the calculation of this electric field is with Gauss’s law.

**6.6. **If there are other charged objects around, then the charges on the surface of the sphere will not necessarily be spherically symmetrical; there will be more in certain direction than in other directions.

## Conceptual Questions

**1. **a. If the planar surface is perpendicular to the electric field vector, the maximum flux would be obtained. b. If the planar surface were parallel to the electric field vector, the minimum flux would be obtained.

**3. **true

**5. **Since the electric field vector has a \(\displaystyle \frac{1}{r^2}\) dependence, the fluxes are the same since \(\displaystyle A=4πr^2\).

**7. **a. no;

b. zero

**9. **Both fields vary as \(\displaystyle \frac{1}{r^2}\). Because the gravitational constant is so much smaller than \(\displaystyle \frac{1}{4πε_0}\), the gravitational field is orders of magnitude weaker than the electric field.

**11. **No, it is produced by all charges both inside and outside the Gaussian surface.

**13. **No, since the situation does not have symmetry, making Gauss’s law challenging to simplify.

**15. **Any shape of the Gaussian surface can be used. The only restriction is that the Gaussian integral must be calculable; therefore, a box or a cylinder are the most convenient geometrical shapes for the Gaussian surface.

**17. **yes

**19. **Since the electric field is zero inside a conductor, a charge of \(\displaystyle −2.0μC\) is induced on the inside surface of the cavity. This will put a charge of \(\displaystyle +2.0μC\) on the outside surface leaving a net charge of \(\displaystyle −3.0μC\) on the surface.

## Problems

**21. **\(\displaystyle Φ=\vec{E}⋅\vec{A}→EAcosθ=2.2×10^4N⋅m^2/C\) electric field in direction of unit normal; \(\displaystyle Φ=\vec{E}⋅\vec{A}→EAcosθ=−2.2×10^4N⋅m^2/C\) electric field opposite to unit normal

**23. **\(\displaystyle \frac{3×10^{−5}N⋅m^2/C}{(0.05m)^2}=E⇒σ=2.12×10^{−13}C/m^2\)

**25. **a. \(\displaystyle Φ=0.17N⋅m^2/C\);

b. \(\displaystyle Φ=0\);

c. \(\displaystyle Φ=EAcos0°=1.0×10^3N/C(2.0×10^{−4}m)^2cos0°=0.20N⋅m^2/C\)

**27. **\(\displaystyle Φ=3.8×10^4N⋅m^2/C\)

**29. **\(\displaystyle \vec{E}(z)=\frac{1}{4πε_0}\frac{2λ}{z}\hat{k},∫\vec{E}⋅\hat{n}dA=\frac{λ}{ε_0}l\)

**31. **a. \(\displaystyle Φ=3.39×10^3N⋅m^2/C\);

b. \(\displaystyle Φ=0\);

c. \(\displaystyle Φ=−2.25×10^5N⋅m^2/C\);

d. \(\displaystyle Φ=90.4N⋅m^2/C\)

**33.** \(\displaystyle Φ=1.13×10^6N⋅m^2/C\)

**35. **Make a cube with **q** at the center, using the cube of side a. This would take four cubes of side a to make one side of the large cube. The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be \(\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}\).

**37. **\(\displaystyle q=3.54×10^{−7}C\)

**39. **zero, also because flux in equals flux out

**41. **\(\displaystyle r>R,E=\frac{Q}{4πε_0r^2};r<R,E=\frac{qr}{4πε_0R^3}\)

**43. **\(\displaystyle EA=\frac{λl}{ε_0}⇒E=4.50×10^7N/C\)

**45. **a. 0;

b. 0;

c. \(\displaystyle \vec{E}=6.74×10^6N/C(−\hat{r})\)

**47. **a. 0;

b. \(\displaystyle E=2.70×10^6N/C\)

**49. **a. Yes, the length of the rod is much greater than the distance to the point in question.

b. No, The length of the rod is of the same order of magnitude as the distance to the point in question.

c. Yes, the length of the rod is much greater than the distance to the point in question.

d. No. The length of the rod is of the same order of magnitude as the distance to the point in question.

**51. **a. \(\displaystyle \vec{E}=\frac{Rσ_0}{ε_0}\frac{1}{r}\hat{r}⇒σ_0=5.31×10^{−11}C/m^2, λ=3.33×10^{−12}C/m\);

b. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}=\frac{3.33×10^{−12}C/m(0.05m)}{ε+0}=0.019N⋅m^2/C\)

**53. **\(\displaystyle E2πrl=\frac{ρπr^2l}{ε_0}⇒E=\frac{ρr}{2ε_0}(r≤R); E2πrl=\frac{ρπR^2l}{ε_0}⇒E=\frac{ρR^2}{2ε_0r}(r≥R)\)

**55. **\(\displaystyle Φ=\frac{q_{enc}}{ε_0}⇒q_{enc}=−1.0×10^{−9}C\)

**57. **\(\displaystyle q_{enc}=\frac{4}{5}παr^5, E4πr^2=\frac{4παr^5}{5ε_0}⇒E=\frac{αr^3}{5ε_0}(r≤R), q_{enc}=\frac{4}{5}παR^5,E4πr^2=\frac{4παR^5}{5ε_0}⇒E=\frac{αR^5}{5ε_0r^2}(r≥R)\)

**59. **integrate by parts: \(\displaystyle q_{enc}=4πρ_0[−e^{−αr}(\frac{(r)^2}{α}+\frac{2r}{α^2}+\frac{2}{α^3})+\frac{2}{α^3}]⇒E=\frac{ρ_0}{r^2ε_0}[−e^{−αr}(\frac{(r)^2}{α}+\frac{2r}{α^2}+\frac{2}{α^3})+\frac{2}{α^3}]\)

**61.**

**63. **a. Outside: \(\displaystyle E2πrl=\frac{λl}{ε_0}⇒E=\frac{3.0C/m}{2πε_0r}\); Inside \(\displaystyle E_{in}=0\);

b.

**65. **a. \(\displaystyle E2πrl=\frac{λl}{ε_0}⇒E=\frac{λ}{2πε_0r}r≥R\) **E** inside equals 0;

b.

**67. **\(\displaystyle E=5.65×10^4N/C\)

**69. **\(\displaystyle λ=\frac{λl}{ε_0}⇒E=\frac{aσ}{ε_0r}r≥a, E=0\) inside since \(\displaystyle q\) **enclosed**=0

**71. **a. \(\displaystyle E=0\);

b. \(\displaystyle E2πrL=\frac{Q}{ε_0}⇒E=\frac{Q}{2πε_0rL}\); c. \(\displaystyle E=0\) since **r** would be either inside the second shell or if outside then **q** enclosed equals 0.

## Additional Problems

**73. **\(\displaystyle ∫\vec{E}⋅\hat{n}dA=a^4\)

**75. **a. \(\displaystyle ∫\vec{E}⋅\hat{n}dA=E_0r^2π\); b. zero, since the flux through the upper half cancels the flux through the lower half of the sphere

**77. **\(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); There are two contributions to the surface integral: one at the side of the rectangle at \(\displaystyle x=0\) and the other at the side at \(\displaystyle x=2.0m\); \(\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C\)

where the minus sign indicates that at \(\displaystyle x=0\), the electric field is along positive** x** and the unit normal is along negative **x**. At \(\displaystyle x=2\), the unit normal and the electric field vector are in the same direction: \(\displaystyle q_{enc}=ε_0Φ=−8.85×10^{−10}C\)

**79. **didn’t keep consistent directions for the area vectors, or the electric fields

**81.** a. \(\displaystyle σ=3.0×10^{−3}C/m^2, +3×10^{−3}C/m^2\) on one and \(\displaystyle −3×10^{−3}C/m^2\) on the other;

b. \(\displaystyle E=3.39×10^8N/CE=3.39×10^8N/C\)

**83. **Construct a Gaussian cylinder along the **z**-axis with cross-sectional area **A.**

\(\displaystyle |z|≥\frac{a}{2}q_{enc}=ρAa,Φ=\frac{ρAa}{ε_0}⇒E=\frac{ρa}{2ε_0}\),

\(\displaystyle |z|≤\frac{a}{2}q_{enc}=ρA2z,E(2A)=\frac{ρA2z}{ε_0}⇒E=\frac{ρz}{ε_0}\)

**85. **a. \(\displaystyle r>b_2\) \(\displaystyle E4πr^2=\frac{\frac{4}{3}π[ρ_1(b^3_1−a^3_1)+ρ_2(b^3_2−a^3_2)}{ε_0}⇒E=\frac{ρ_1(b^3_1−a^3_1)+ρ_2(b^3_2−a^3_2)}{3ε_0r^2}\);

b. \(\displaystyle a_2<r<b_2\) \(\displaystyle E4πr^2=\frac{\frac{4}{3}π[ρ_1(b^3_1−a^3_1)+ρ_2(r^3−a^3_2)]}{ε_0}⇒E=\frac{ρ_1(b^3_1−a^3_1)+ρ_2(r^3−a^3_2)}{3ε_0r^2}\);

c. \(\displaystyle b_1<r<a_2\) \(\displaystyle E4πr^2=\frac{\frac{4}{3}πρ_1(b^3_1−a^3_1)}{ε_0}⇒E=\frac{ρ_1(b^3_1−a^3_1)}{3ε_0r^2}\);

d. \(\displaystyle a_1<r<b_1\) \(\displaystyle E4πr^2=\frac{\frac{4}{3}πρ_1(r^3_−a^3_1)}{ε_0}⇒E=\frac{ρ_1(r^3−a^3_1)}{3ε_0r^2}\);

e. 0

**87. **Electric field due to plate without hole: \(\displaystyle E=\frac{σ}{2ε_0}\).

Electric field of just hole filled with \(\displaystyle −σE=\frac{−σ}{2ε_0}(1−\frac{z}{\sqrt{R^2+z^2}}).

Thus, \(\displaystyle E_{net}=\frac{σ}{2ε_0}\frac{h}{\sqrt{R^2+h^2}}\)

**89. **a. \(\displaystyle E=0\); b. \(\displaystyle E=\frac{q_1}{4πε_0r^2}\); c. \(\displaystyle E=\frac{q_1+q_2}{4πε_0r^2}\); d. 0 \(\displaystyle q_1−q_1, q_1+q_2\)

## Challenge Problems

**91. **Given the referenced link, using a distance to Vega of \(\displaystyle 237×10^{15} m\)^{4} and a diameter of 2.4 m for the primary mirror,^{5} we find that at a wavelength of 555.6 nm, Vega is emitting \(\displaystyle 2.44×10^{24}J/s\) at that wavelength. Note that the flux through the mirror is essentially constant.

**93. **The symmetry of the system forces \(\displaystyle \vec{E}\) to be perpendicular to the sheet and constant over any plane parallel to the sheet. To calculate the electric field, we choose the cylindrical Gaussian surface shown. The cross-section area and the height of the cylinder are **A** and **2x**, respectively, and the cylinder is positioned so that it is bisected by the plane sheet. Since **E **is perpendicular to each end and parallel to the side of the cylinder, we have **EA** as the flux through each end and there is no flux through the side. The charge enclosed by the cylinder is \(\displaystyle σA\), so from Gauss’s law, \(\displaystyle 2EA=\frac{σA}{ε_0}\), and the electric field of an infinite sheet of charge is

\(\displaystyle E=\frac{σ}{2ε_0}\), in agreement with the calculation of in the text.

**95. **There is **Q/2** on each side of the plate since the net charge is \(\displaystyle Q: σ=\frac{Q}{2A}\),

\(\displaystyle ∮_S\vec{E}⋅\hat{n}dA=\frac{2σΔA}{ε_0}⇒E_P=\frac{σ}{ε_0}=\frac{Q}{ε_02A}\)

## Contributors and Attributions

Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).