# 6.A: Sheria ya Gauss (Majibu)

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

## Angalia Uelewa Wako

6.1. Weka ili kitengo chake cha kawaida kinapingana$$\displaystyle \vec{E}$$.

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. Katika kesi hii, kuna tu$$\displaystyle \vec{E}_{out}$$. Kwa hiyo, ndiyo.

6.5. $$\displaystyle \vec{E}=\frac{λ_0}{2πε_0}\frac{1}{d}\hat{r}$$; Hii inakubaliana na hesabu ya Mfano 5.5 ambapo tulipata uwanja wa umeme kwa kuunganisha juu ya waya iliyoshtakiwa. Angalia jinsi rahisi hesabu ya uwanja huu wa umeme ni pamoja na sheria ya Gauss.

6.6. Ikiwa kuna vitu vingine vya kushtakiwa karibu, basi mashtaka juu ya uso wa nyanja haitakuwa lazima kuwa spherically symmetrical; kutakuwa na zaidi katika mwelekeo fulani kuliko kwa njia nyingine.

## Maswali ya dhana

1. a Ikiwa uso wa mipango ni perpendicular kwa vector ya shamba la umeme, kiwango cha juu kitapatikana. b Kama uso wa planar ulikuwa sawa na vector ya shamba la umeme, kiwango cha chini cha flux kitapatikana.

3. kweli

5. Kwa kuwa vector shamba la umeme lina$$\displaystyle \frac{1}{r^2}$$ utegemezi, fluxes ni sawa tangu$$\displaystyle A=4πr^2$$.

7. a. hapana;

b. sifuri

9. Wote mashamba kutofautiana kama$$\displaystyle \frac{1}{r^2}$$. Kwa sababu mara kwa mara ya mvuto ni ndogo sana kuliko$$\displaystyle \frac{1}{4πε_0}$$, uwanja wa mvuto ni amri ya ukubwa dhaifu kuliko uwanja wa umeme.

11. Hapana, huzalishwa na mashtaka yote ndani na nje ya uso wa Gaussia.

13. Hapana, kwa kuwa hali haina ulinganifu, na kufanya sheria ya Gauss changamoto kurahisisha.

15. Sura yoyote ya uso wa Gaussia inaweza kutumika. Vikwazo pekee ni kwamba sehemu ya Gaussia inapaswa kuhesabiwa; kwa hiyo, sanduku au silinda ni maumbo ya kijiometri rahisi zaidi kwa uso wa Gaussia.

17. ndiyo

19. Kwa kuwa uwanja wa umeme ni sifuri ndani ya kondakta, malipo ya$$\displaystyle −2.0μC$$ hutolewa kwenye uso wa ndani wa cavity. Hii itaweka malipo ya$$\displaystyle +2.0μC$$ juu ya uso wa nje na kuacha malipo ya wavu$$\displaystyle −3.0μC$$ juu ya uso.

## Matatizo

21. $$\displaystyle Φ=\vec{E}⋅\vec{A}→EAcosθ=2.2×10^4N⋅m^2/C$$uwanja wa umeme katika mwelekeo wa kitengo cha kawaida; uwanja wa$$\displaystyle Φ=\vec{E}⋅\vec{A}→EAcosθ=−2.2×10^4N⋅m^2/C$$ umeme kinyume na kitengo cha kawaida

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. Kufanya mchemraba na q katika kituo, kwa kutumia mchemraba wa upande a Hii itachukua cubes nne za upande a kufanya upande mmoja wa mchemraba kubwa Upande wa kivuli wa mchemraba mdogo ungekuwa 1/24 ya jumla ya eneo la mchemraba mkubwa; kwa hiyo, mtiririko kupitia eneo la kivuli ungekuwa$$\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}$$.

37. $$\displaystyle q=3.54×10^{−7}C$$

39. sifuri, pia kwa sababu flux katika sawa flux nje

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. Ndiyo, urefu wa fimbo ni kubwa zaidi kuliko umbali wa uhakika katika swali.

b Hapana, urefu wa fimbo ni wa utaratibu sawa wa ukubwa kama umbali wa uhakika katika swali.

c Ndiyo, urefu wa fimbo ni kubwa zaidi kuliko umbali wa uhakika katika swali.

d. Urefu wa fimbo ni wa utaratibu sawa wa ukubwa kama umbali wa uhakika katika swali.

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. kuunganisha na sehemu:$$\displaystyle q_{enc}=4πρ_0[−e^{−αr}(\frac{(r)^2}{α}+\frac{2r}{α^2}+\frac{2}{α^3})+\frac{2}{α^3}]⇒E=\frac{ρ_0}{r^2ε_0}[−e^{−αr}(\frac{(r)^2}{α}+\frac{2r}{α^2}+\frac{2}{α^3})+\frac{2}{α^3}]$$

61.

63. a. nje:$$\displaystyle E2πrl=\frac{λl}{ε_0}⇒E=\frac{3.0C/m}{2πε_0r}$$; Ndani$$\displaystyle E_{in}=0$$;

b.

65. ndani ya$$\displaystyle E2πrl=\frac{λl}{ε_0}⇒E=\frac{λ}{2πε_0r}r≥R$$ E ni sawa na 0;

b.

67. $$\displaystyle E=5.65×10^4N/C$$

69. $$\displaystyle λ=\frac{λl}{ε_0}⇒E=\frac{aσ}{ε_0r}r≥a, E=0$$ndani tangu$$\displaystyle q$$ iliyoambatanishwa =0

71. a$$\displaystyle E=0$$;.

b.$$\displaystyle E2πrL=\frac{Q}{ε_0}⇒E=\frac{Q}{2πε_0rL}$$; c.$$\displaystyle E=0$$ tangu r itakuwa ama ndani ya shell pili au kama nje kisha q iliyoambatanishwa sawa 0.

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

75. a.$$\displaystyle ∫\vec{E}⋅\hat{n}dA=E_0r^2π$$; b. sifuri, tangu kuongezeka kwa nusu ya juu kufuta mtiririko kupitia nusu ya chini ya nyanja

77. $$\displaystyle Φ=\frac{q_{enc}}{ε_0}$$; Kuna michango miwili kwa uso muhimu: moja upande wa mstatili$$\displaystyle x=0$$ na nyingine upande wa$$\displaystyle x=2.0m$$;$$\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C$$

ambapo ishara ndogo inaonyesha kuwa saa$$\displaystyle x=0$$, uwanja wa umeme ni pamoja na x nzuri na kitengo cha kawaida ni pamoja na x hasi. Katika$$\displaystyle x=2$$, kitengo cha kawaida na vector uwanja wa umeme ni katika mwelekeo huo:$$\displaystyle q_{enc}=ε_0Φ=−8.85×10^{−10}C$$

79. hakuwa na kuweka maelekezo thabiti kwa wadudu eneo hilo, au mashamba ya umeme

81. a.$$\displaystyle σ=3.0×10^{−3}C/m^2, +3×10^{−3}C/m^2$$ kwa moja na kwa$$\displaystyle −3×10^{−3}C/m^2$$ upande mwingine;

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

83. Kujenga silinda ya Gaussia kando ya z -axis na eneo la msalaba 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. Umeme shamba kutokana na sahani bila shimo:$$\displaystyle E=\frac{σ}{2ε_0}$$.

Umeme uwanja wa shimo tu kujazwa na\ (\ displaystyle -σe=\ frac {-σ} {2ε_0} (1-\ frac {z} {\ sqrt {r ^ 2+z ^ 2}}).

Hivyo,$$\displaystyle E_{net}=\frac{σ}{2ε_0}\frac{h}{\sqrt{R^2+h^2}}$$

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$$\displaystyle q_1−q_1, q_1+q_2$$

## Changamoto Matatizo

91. Kutokana na kiungo kinachotazamwa, kwa kutumia umbali wa Vega ya$$\displaystyle 237×10^{15} m$$ 4 na kipenyo cha 2.4 m kwa kioo cha msingi, 5 tunaona kwamba kwa wavelength ya 555.6 nm, Vega inatoa$$\displaystyle 2.44×10^{24}J/s$$ kwa wavelength hiyo. Kumbuka kuwa mtiririko kupitia kioo kimsingi ni mara kwa mara.

93. Ulinganifu wa nguvu$$\displaystyle \vec{E}$$ za mfumo kuwa perpendicular kwa karatasi na mara kwa mara juu ya ndege yoyote sambamba na karatasi. Ili kuhesabu shamba la umeme, tunachagua uso wa Gaussia wa cylindrical umeonyeshwa. Sehemu ya msalaba na urefu wa silinda ni A na 2x, kwa mtiririko huo, na silinda imewekwa ili iingizwe na karatasi ya ndege. Kwa kuwa E ni perpendicular kwa kila mwisho na sambamba na upande wa silinda, tuna EA kama flux kupitia kila mwisho na hakuna flux kupitia upande. Malipo yaliyofungwa na silinda ni$$\displaystyle σA$$, hivyo kutokana na sheria ya Gauss$$\displaystyle 2EA=\frac{σA}{ε_0}$$,, na uwanja wa umeme wa karatasi isiyo na kipimo cha malipo ni

$$\displaystyle E=\frac{σ}{2ε_0}$$, kwa makubaliano na hesabu ya maandishi.

95. Kuna Q/2 kila upande wa sahani tangu malipo wavu ni$$\displaystyle Q: σ=\frac{Q}{2A}$$,

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

## Wachangiaji na Majina

Template:ContribOpenStaxUni