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5.A: Mashtaka ya umeme na Mashamba (Jibu)

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    Angalia Uelewa Wako

    5.1. Nguvu ingekuwa uhakika nje.

    5.2. Nguvu ya wavu ingeelekeza 58° chini ya - x -axis.

    5.3. \(\displaystyle \vec{E} = \frac{1}{4πε_0}\frac{q}{r^2}\hat{r}\)

    5.4. Hatuwezi tena kuchukua faida ya ulinganifu. Badala yake, tutahitaji kuhesabu kila sehemu mbili za uwanja wa umeme na muhimu yao wenyewe.

    5.5. Malipo ya uhakika yatakuwa\(\displaystyle Q=σab\) ambapo a na b ni pande za mstatili lakini vinginevyo vinafanana.

    5.6. Shamba la umeme litakuwa sifuri kati, na kuwa na ukubwa\(\displaystyle \frac{σ}{ε_0}\) kila mahali pengine.

    Maswali ya dhana

    1. Kuna idadi kubwa sawa ya mashtaka mazuri na hasi sasa, na kufanya kitu umeme neutral.

    3. a. ndiyo;

    b. ndiyo

    5. Chukua kitu kwa malipo inayojulikana, ama chanya au hasi, na uiletee karibu na fimbo. Ikiwa kitu kinachojulikana cha kushtakiwa ni chanya na kinachukuliwa kutoka fimbo, fimbo inashtakiwa chanya. Ikiwa kitu kilichopakiwa vyema kinavutiwa na fimbo, fimbo inashtakiwa vibaya.

    7. Hapana, vumbi huvutiwa na wote kwa sababu molekuli za chembe za vumbi huwa polarized katika mwelekeo wa hariri.

    9. Ndiyo, malipo ya polarization yanatokana na conductor ili malipo mazuri iko karibu na fimbo ya kushtakiwa, na kusababisha nguvu ya kuvutia.

    11. Kulipia kwa uendeshaji ni malipo kwa kuwasiliana ambapo malipo huhamishiwa kwenye kitu. Kulipia kwa induction kwanza kunahusisha kuzalisha malipo ya ubaguzi katika kitu na kisha kuunganisha waya chini ili kuruhusu baadhi ya malipo kuondoka kitu, na kuacha kitu kushtakiwa.

    13. Hii ni hivyo kwamba malipo yoyote ya ziada yamehamishiwa chini, kuweka vikwazo vya petroli vya neutral. Ikiwa kuna malipo ya ziada kwenye chombo cha petroli, cheche inaweza kuwaka.

    15. Dryer inashutumu nguo. Ikiwa ni uchafu, uwepo wa molekuli za maji huzuia malipo.

    17. Kuna aina mbili tu za malipo, zinazovutia na zinazopuuza. Ikiwa unaleta kitu kilichopakiwa karibu na quartz, moja tu ya madhara haya mawili yatatokea, kuthibitisha hakuna aina ya tatu ya malipo.

    19. a. hapana, tangu malipo ubaguzi ni ikiwa. b. ndiyo, tangu malipo ubaguzi bila kuzalisha tu nguvu ya kuvutia.

    21. Nguvu inayoshikilia kiini pamoja lazima iwe kubwa zaidi kuliko nguvu ya kupuuza umeme kwenye protoni.

    23. Aidha ishara ya malipo ya mtihani inaweza kutumika, lakini mkataba ni kutumia chanya mtihani malipo.

    25. Mashtaka ni ya ishara sawa.

    27. Kwa infinity, tunatarajia shamba kwenda sifuri, lakini kwa sababu karatasi haipatikani kwa kiwango, hii sio kesi. Kila mahali ulipo, unaona ndege isiyo na kipimo katika pande zote.

    29. usio kushtakiwa sahani ingekuwa\(\displaystyle E=\frac{σ}{2ε_0}\) kila mahali. Shamba lingeelekeza kuelekea sahani ikiwa limeshtakiwa vibaya na kuelekeza mbali na sahani ikiwa ingekuwa imeshtakiwa vyema. Sehemu ya umeme ya sahani sambamba ingekuwa sifuri kati yao ikiwa walikuwa na malipo sawa, na E ingekuwa\(\displaystyle E=\frac{σ}{ε_0}\) kila mahali pengine. Ikiwa mashtaka yalikuwa kinyume, hali hiyo inabadilishwa, sifuri nje ya sahani na\(\displaystyle E=\frac{σ}{ε_0}\) kati yao.

    31. ndiyo; hapana

    33. Katika uso wa Dunia, shamba la mvuto daima linaelekezwa kuelekea katikati ya Dunia. Shamba la umeme lingeweza kusonga chembe iliyoshtakiwa katika mwelekeo tofauti kuliko kuelekea katikati ya Dunia. Hii itaonyesha shamba umeme ni sasa.

    35. 10

    Matatizo

    37. a\(\displaystyle 2.00×10^{−9}C(\frac{1}{1.602×10^{−19}}e/C)=1.248×10^{10}electrons2\);.

    b.\(\displaystyle 0.500×10^{−6}C(\frac{1}{1.602×10^{−19}}e/C)=3.121×10^{12}electrons\)

    39. \(\displaystyle \frac{3.750×10^{21}e}{6.242×10^{18}e/C}=-600.8C\)

    41. a\(2.0×10^{−9}C(6.242×10^{18}e/C)=1.248×10^{10}e\);.

    b.\(\displaystyle 9.109×10^{−31}kg(1.248×10^{10}e)=1.137×10^{−20}kg, \frac{1.137×10^{−20}kg}{2.5×10^{−3}kg}=4.548×10^{−18}\) au\(\displaystyle 4.545×10^{−16}%\)

    43. \(\displaystyle 5.00×10^{−9}C(6.242×10^{18}e/C)=3.121×10^{10}e; 3.121×10^{10}e+1.0000×10^{12}e=1.0312×10^{12}e\).

    45. umati wa atomiki wa nyakati za atomi za shaba\(\displaystyle 1u=1.055×10^{−25}kg\); idadi ya atomi za shaba =\(\displaystyle 4.739×10^{23}atoms\); idadi ya elektroni ni sawa na mara 29 idadi ya atomi au\(\displaystyle 1.374×10^{25}electrons\);\(\displaystyle \frac{2.00×10^{−6}C(6.242×10^{18}e/C)}{1.374×10^{25}e}=9.083×10^{−13}\) au\(\displaystyle 9.083×10^{−11}%\).

    47. \(\displaystyle 244.00u(1.66×10^{−27}kg/u)=4.050×10^{−25}kg\);\(\displaystyle \frac{4.00kg}{4.050×10^{−25}kg}=9.877×10^{24}atoms\)\(\displaystyle 9.877×10^{24}(94)=9.284×10^{26}protons\)\(\displaystyle 9.284×10^{26}protons; 9.284×10^{26}(1.602×10^{−19}C/p)=1.487×10^8C\)

    49. a. malipo 1 ni\(\displaystyle 3μC\); malipo 2 ni\(\displaystyle 12μC\),\(\displaystyle F_{31}=2.16×10^{−4}N\) upande wa kushoto,

    \(\displaystyle F_{32}=8.63×10^{−4}N\)kwa haki,

    \(\displaystyle F_{net}=6.47×10^{−4}N\)kwa haki;

    b.\(\displaystyle F_{31}=2.16×10^{−4}N\) kwa haki,

    \(\displaystyle F_{32}=9.59×10^{−5}N\)kwa haki,

    \(\displaystyle F_{net}=3.12×10^{−4}N\)kwa haki,

    Mashtaka matatu yanaonyeshwa. Malipo 1 ni malipo ya 3 micro Coulomb chini kushoto. Charge 2 ni 12 micro Coulomb malipo chini kulia, 1 mita na haki ya malipo 1. Malipo 3 ni minus 2 nano Coulomb malipo 0.5 mita juu ya malipo 2. Mashtaka yanafafanua pembetatu sahihi, na malipo 2 kwenye pembe ya kulia. Pembe kwenye vertex na malipo moja ni theta. Majeshi ya malipo ya tatu yanaonyeshwa. F 3 1 anasema chini na kushoto, kuelekea malipo 1. Nguvu F 3 2 pointi wima chini.

    c.\(\displaystyle \vec{F}_{31x}=−2.76×10^{−5}N\hat{i},\)

    \(\displaystyle \vec{F}_{31y}=−1.38×10^{−5}N\hat{j}\),

    \(\displaystyle \vec{F}_{32y}=−8.63×10^{−4}N\hat{j}\),

    \(\displaystyle \vec{F}_{net}=−2.76×10^{−5}N\hat{i}−8.77×10^{−4}N\hat{j}\)

    51. \(\displaystyle F=230.7N\)

    53. \(\displaystyle F=53.94N\)

    55. Mvutano ni\(\displaystyle T=0.049N\). Sehemu ya usawa ya mvutano ni\(\displaystyle 0.0043N\)

    \(\displaystyle d=0.088m,q=6.1×10^{−8}C\).

    Mashtaka yanaweza kuwa chanya au hasi, lakini wote wawili wanapaswa kuwa ishara sawa.

    57. Hebu malipo kwenye moja ya nyanja kuwa rQ, ambapo r ni sehemu kati ya 0 na 1. Katika namba ya sheria ya Coulomb, neno linalohusisha mashtaka ni\(\displaystyle rQ(1−r)Q\). Hii ni sawa na\(\displaystyle (r−r^2)Q^2\). Kupata upeo wa muda huu inatoa\(\displaystyle 1−2r=0⇒r=\frac{1}{2}\)

    59. Kufafanua haki ya kuwa mwelekeo chanya na hivyo kushoto ni mwelekeo hasi, basi\(\displaystyle F=−0.05N\)

    61. Chembe huunda pembetatu ya pande 13, 13, na cm 24. x -vipengele kufuta, ambapo kuna mchango kwa y-sehemu kutoka mashtaka yote 24 cm mbali. Mhimili wa y unaopitia malipo ya tatu hupunguza mstari wa 24-cm, na kuunda pembetatu mbili za kulia za pande 5, 12, na 13 cm. \(\displaystyle F_y=2.56N\)katika hasi y -mwelekeo tangu nguvu ni ya kuvutia. Nguvu ya wavu kutoka kwa mashtaka yote ni\(\displaystyle \vec{F}_{net}=−5.12N\hat{j}\)

    63. Ulalo ni\(\displaystyle \sqrt{2}a\) na vipengele vya nguvu kutokana na malipo ya diagonal ina sababu\(\displaystyle cosθ=\frac{1}{\sqrt{2}}\);\(\displaystyle \vec{F}_{net}=[k\frac{q^2}{a^2}+k\frac{q^2}{2a^2}\frac{1}{\sqrt{2}}]\hat{i}−[k\frac{q^2}{a^2}+k\frac{q^2}{2a^2}\frac{1}{\sqrt{2}}]\hat{j}\)

    65. \(\displaystyle a. E=2.0×10^{−2}\frac{N}{C}\);

    \(\displaystyle b. F=2.0×10^{−19}N\)

    67. a\(\displaystyle E=2.88×10^{11}N/C\);.

    b\(\displaystyle E=1.44×10^{11}N/C\);

    c.\(\displaystyle F=4.61×10^{−8}N\) juu ya chembe ya alpha

    \(\displaystyle F=4.61×10^{−8}N\)juu ya elektroni

    69. \(\displaystyle E=(−2.0\hat{i}+3.0\hat{j})N\)

    71. \(\displaystyle F=3.204×10^{−14}N\),

    \(\displaystyle a=3.517×10^{16}m/s^2\)

    73. \(\displaystyle q=2.78×10^{−9}C\)

    75. a\(\displaystyle E=1.15×10^{12}N/C\);.

    b.\(\displaystyle F=1.47×10^{−6}N\)

    77. Kama\(\displaystyle q_2\) ni haki ya\(\displaystyle q_1\), umeme shamba vector kutoka mashtaka yote uhakika na haki.

    a\(\displaystyle E=2.70×10^6N/C\);.

    b.\(\displaystyle F=54.0N\)

    79. Kuna jiometri ya pembetatu ya kulia ya 45°. Vipengele vya x vya shamba la umeme wakati wa\(\displaystyle y=3m\) kufuta. Ya y -vipengele kutoa\(\displaystyle E(y=3m)=2.83×10^3N/C\).

    Katika asili tuna malipo hasi ya magnitide\(\displaystyle q=−2.83×10^{−6}C\)

    81. \(\displaystyle \vec{E}(z)=3.6×10^4N\hat{k}\)

    83. \(\displaystyle dE=\frac{1}{4πε_0}\frac{λdx}{(x+a)^2},E=\frac{λ}{4πε_0}[\frac{1}{l+a}−\frac{1}{a}]\)

    85. \(\displaystyle σ=0.02C/m^2\)\(\displaystyle E=2.26×10^9N/C\)

    87. Katika\(\displaystyle P_1: \vec{E}(y)=\frac{1}{4πε_0}\frac{λL}{y\sqrt{y^2+\frac{L^2}{4}}}\hat{j}⇒\frac{1}{4πε_0}\frac{q}{\frac{a}{2}\sqrt{(\frac{a}{2})^2+\frac{L^2}{4}}}\hat{j}=\frac{1}{πε_0}\frac{q}{a\sqrt{a^2+L^2}}\hat{j}\)

    Katika\(\displaystyle P_2\): Weka asili mwishoni mwa L.

    \(\displaystyle dE=\frac{1}{4πε_0}\frac{λdx}{(x+a)^2},\vec{E} =−\frac{q}{4πε_0l}[\frac{1}{l+a}−\frac{1}{a}]\hat{i}\)

    89. a\(\displaystyle \vec{E}(\vec{r})=\frac{1}{4πε_0}\frac{2λ_x}{a}\hat{i}+\frac{1}{4πε_0}\frac{2λ_y}{b}\hat{j}\);.

    b.\(\displaystyle \frac{1}{4πε_0}\frac{2(λ_x+λ_y)}{c}\hat{k}\)

    91. a.\(\displaystyle \vec{F}=3.2×10^{−17}N\hat{i}\),

    \(\displaystyle \vec{a}=1.92×10^{10}m/s^2\hat{i}\);

    b.\(\displaystyle \vec{F} =−3.2×10^{−17}N\hat{i}\),

    \(\displaystyle \vec{a} =−3.51×10^{13}m/s^2\hat{i}\)

    93. \(\displaystyle m=6.5×10^{−11}kg\),

    \(\displaystyle E=1.6×10^7N/C\)

    95. \(\displaystyle E=1.70×10^6N/C\),

    \(\displaystyle F=1.53×10^{−3}NTcosθ=mgTsinθ=qE\),

    \(\displaystyle tanθ=0.62⇒θ=32.0°\),

    Hii ni huru ya urefu wa kamba.

    97. arc mviringo\(\displaystyle dE_x(−\hat{i})=\frac{1}{4πε_0}\frac{λds}{r^2}cosθ(−\hat{i}\),

    \(\displaystyle \vec{E}_x=\frac{λ}{4πε_0r}(−\hat{i})\),

    \(\displaystyle dEy(−\hat{i}ˆ)=\frac{1}{4πε_0}\frac{λds}{r^2}sinθ(−\hat{j})\),

    \(\displaystyle \vec{E}_y=\frac{λ}{4πε_0r}(−\hat{j})\);

    y -axis:\(\displaystyle \vec{E}_x=\frac{λ}{4πε_0r}(−\hat{i})\);

    x -mhimili:\(\displaystyle \vec{E}_y=\frac{λ}{4πε_0r}(−\hat{j})\),

    \(\displaystyle \vec{E}=\frac{λ}{2πε_0r}(−\hat{i})+\frac{λ}{2πε_0r}(−\hat{j})\)

    99. a\(\displaystyle W=\frac{1}{2}m(v^2−v^2_0), \frac{Qq}{4πε_0}(\frac{1}{r}−\frac{1}{r_0})=\frac{1}{2}m(v^2−v^2_0)⇒r_0−r=\frac{4πε_0}{Qq}\frac{1}{2}rr_0m(v^2−v^2_0)\);.

    b.\(\displaystyle r_0−r\) ni hasi; kwa hiyo,\(\displaystyle v_0>v, r→∞\), na\(\displaystyle v→0:\frac{Qq}{4πε_0}(−\frac{1}{r_0})=−\frac{1}{2}mv^2_0⇒v_0=\sqrt{\frac{Qq}{2πε_0mr_0}}\)

    101.

    Kielelezo a inaonyesha chanya 20 micro Coulomb malipo upande wa kushoto, hasi 20 micro Coulomb malipo upande wa kulia, na mistari shamba kutokana na mashtaka. Mstari wa shamba hutoka kwa malipo mazuri na hujiunga na kuingia kwenye malipo mabaya. Mstari wa shamba la nje hupanua zaidi ya eneo la kuchora na hivyo tunawaona wakipiga kwa haki, kuelekea malipo mabaya, lakini tu kuona sehemu ya mstari. Uzito wa mistari inayotoka kwa chanya ni sawa na wiani unaoingia hasi. Kielelezo b inaonyesha chanya 20 micro Coulomb malipo upande wa kushoto, chanya 20 micro Coulomb malipo upande wa kulia, na mistari shamba kutokana na mashtaka. Mistari ya shamba hutoka kwenye mashtaka mazuri na hutofautiana, ikitengana mbali na malipo ya mbali. Uzito wa mistari ni sawa karibu na kila mashtaka. Kielelezo c inaonyesha chanya 20 micro Coulomb malipo upande wa kushoto, hasi 30 micro Coulomb malipo upande wa kulia, na mistari shamba kutokana na mashtaka. Mstari wa shamba hutoka kwa malipo mazuri. Zaidi mistari kwenda katika hasi 20 micro Coulomb malipo ya kuja nje ya chanya 20 micro Coulomb malipo. Mstari wote unaotoka kwenye malipo mazuri hukoma kwa hasi, wakati mistari ya nje inakwenda kwenye hasi kuanza kwa infinity.

    103.

    Mashtaka manne yanaonyeshwa kwenye pembe za mraba. Kwenye kushoto juu ni chanya 10 nano Coulombs. Kwenye haki ya juu ni hasi 10 nano Coulombs. Chini kushoto ni hasi 10 nano Coulombs. Chini ya kulia ni chanya 10 nano Coulombs. Mstari wa shamba pia umeonyeshwa. Wao kuja nje ya mashtaka chanya na Curve kuelekea na mwisho katika mashtaka hasi. Uzito wa chini kabisa ni karibu katikati ya mraba.

    105. \(\displaystyle E_x=0, E_y=\frac{1}{4πε_0}[\frac{2q}{(x^2+a^2})\frac{a}{\sqrt{(x^2+a^2)}}⇒x≫a⇒\frac{1}{2πε_0}\frac{qa}{x^3}\)

    \(\displaystyle E_y=\frac{q}{4πε_0}[\frac{2ya+2ya}{(y−a)^2(y+a)^2}]⇒y≫a⇒\frac{1}{πε_0}\frac{qa}{y^3}\)

    107. Wakati wa dipole wa wavu wa molekuli ni jumla ya vector ya wakati wa dipole ya mtu binafsi kati ya mbili O-H. Kutenganishwa O-H ni 0.9578 angstroms:

    \(\displaystyle \vec{p} =1.889×10^{−29}Cm\hat{i}\)

    Matatizo ya ziada

    109. \(\displaystyle \vec{F}_{net}=[−8.99×10^9\frac{3.0×10^{−6}(5.0×10^{−6})}{(3.0m)^2}−8.99×10^9\frac{9.0×10^{−6}(5.0×10^{−6})}{(3.0m)^2}]\hat{i}, −8.99×10^9\frac{6.0×10^{−6}(5.0×10^{−6})}{(3.0m)^2}\hat{j}=−0.06N\hat{i}−0.03N\hat{j}\)

    111. Malipo Swali na q huunda pembetatu sahihi ya pande 1 m na\(\displaystyle 3+\sqrt{3}m.\) Mashtaka 2Q na q huunda pembetatu sahihi ya pande 1 m na\(\displaystyle \sqrt{3}m\).

    \(\displaystyle F_x=0.049N,\)

    \(\displaystyle F_y=0.09N\),

    \(\displaystyle \vec{F}_{net}=0.049N\hat{i}+0.09N\hat{j}\)

    113. W=0.054J

    115. a\(\displaystyle \vec{E}=\frac{1}{4πε_0}(\frac{q}{(2a)^2}−\frac{q}{a^2})\hat{i}\);.

    b\(\displaystyle \vec{E}=\frac{\sqrt{3}}{4πε_0}\frac{q}{a^2}(−\hat{j})\);

    c.\(\displaystyle \vec{E}=\frac{2}{πε_0}\frac{q}{a^2}\frac{1}{\sqrt{2}}(−\hat{j})\)

    117. \(\displaystyle \vec{E}=6.4×10^6(\hat{i})+1.5×10^7(\hat{j})N/C\)

    119. \(\displaystyle F=qE_0(1+x/a)\)\(\displaystyle W=\frac{1}{2}m(v^2−v^2_0)\),

    \(\displaystyle \frac{1}{2}mv^2=qE_0(\frac{15a}{2})J\)

    121. Umeme uwanja wa waya katika x:\(\displaystyle \vec{E}(x)=\frac{1}{4πε_0}\frac{2λ_y}{x}\hat{i}\),

    \(\displaystyle dF=\frac{λ_yλ_x}{2πε_0}(lnb−lna)\)

    123.

    Fimbo ya urefu L inavyoonyeshwa, iliyokaa na x-axis na mwisho wa kushoto katika asili. Hatua P inavyoonyeshwa kwenye mhimili wa z, umbali a juu ya mwisho wa kushoto wa fimbo. Sehemu ndogo ya fimbo imeandikwa kama d x na ni umbali x upande wa kulia wa mwisho wa kushoto wa fimbo. Mstari kutoka dx hadi kumweka P hufanya angle ya theta na mhimili x. Vector d E, inayotolewa na mkia wake kwenye hatua P, inaonyesha mbali na sehemu d x.

    \(\displaystyle dEx=\frac{1}{4πε_0}\frac{λdx}{(x^2+a^2)}\frac{x}{\sqrt{x^2+a^2}}\),

    \(\displaystyle \vec{E}_x=\frac{λ}{4πε_0}[\frac{1}{\sqrt{L^2+a^2}}−\frac{1}{a}]\hat{i}\),

    \(\displaystyle dE_z=\frac{1}{4πε_0}\frac{λdx}{(x^2+a^2)}\frac{a}{\sqrt{x^2+a^2}}\),

    \(\displaystyle \vec{E}_z=\frac{λ}{4πε_0a}\frac{L}{\sqrt{L^2+a^2}}\hat{k}\),

    Kubadilisha z kwa, tuna:

    \(\displaystyle \vec{E}(z)=\frac{λ}{4πε_0}[\frac{1}{\sqrt{L^2+z^2}}−\frac{1}{z}]\hat{i}+\frac{λ}{4πε_0z}\frac{L}{\sqrt{L^2+z^2}}\hat{k}\)

    125. Kuna nguvu ya wavu tu katika y -mwelekeo. Hebu\(\displaystyle θ\) kuwa angle vector kutoka dx kwa q hufanya na x -axis. Vipengele kando ya x -axis kufuta kutokana na ulinganifu, na kuacha y -sehemu ya nguvu.

    \(\displaystyle dF_y=\frac{1}{4πε_0}\frac{aqλdx}{(x^2+a^2)^{3/2}}\),

    \(\displaystyle Fy=\frac{1}{2πε_0}\frac{qλ}{a}[\frac{l/2}{((l/2)^2+a^2)^{1/2}}]\)

    Wachangiaji na Majina

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