Tambayoyi Misali Game da Dokar Coulomb
Dokar Coulomb ƙa'ida ce ta asali a fannin kimiyyar lissafi wadda ke bayyana ƙarfin da ke tsakanin cajin lantarki guda biyu. An yi nazari a cikin ɓangaren electrostatics, wannan doka ta bayyana yadda cajin lantarki ke jawo hankali ko tunkuɗe juna. Masanin kimiyyar lissafi ɗan Faransa Charles-Augustin de Coulomb ne ya fara gabatar da wannan doka a ƙarni na 18. Wannan labarin zai tattauna Dokar Coulomb ta hanyar bincika misalai da mafita, don haka zai taimaka wa masu karatu su fahimci aikace-aikacen wannan ƙa'ida ta asali.
Tushen Ka'ida: Menene Dokar Coulomb?
Dokar Coulomb ta bayyana cewa girman ƙarfin lantarki mai ƙarfin lantarki \( F \) tsakanin cajin maki biyu \( q_1 \) da \( q_2 \) yana daidai gwargwado kai tsaye da samfurin girman cajin biyu kuma yana daidai gwargwado da murabba'in nisan \( r \) da ke tsakaninsu. A lissafi, ana iya bayyana dabarar Dokar Coulomb kamar haka:
\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
Ina:
– \( F \) shine girman ƙarfin lantarki
– \( k_e \) shine madaidaicin Coulomb (\( 8.9875 \sau 10^9 \, N \cdot m^2 \cdot C^{-2} \))
– \( q_1 \) da \( q_2 \) su ne girman cajin wutar lantarki
– \( r \) shine nisan da ke tsakanin caji biyu
Misali Tambaya ta 1
Tambaya:
Akwai cajin maki guda biyu: \( q_1 = 2 \sau 10^{-6} \, C \) da \( q_2 = -3 \sau 10^{-6} \, C \). Suna da nisan mita 0,05. Lissafa girman ƙarfin lantarki da ke aiki tsakanin cajin biyu.
Tattaunawa:
Mataki na farko shine sake rubuta dabarar Dokar Coulomb:
\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
Yanzu mun maye gurbin dabi'un da aka sani a cikin dabarar:
\[ k_e = 8.9875 \sau 10^9 \, N \cdot m^2 \cdot C^{-2} \]
\[ q_1 = sau 2 10^{-6} \, C \]
\[ q_2 = -3 \sau 10^{-6} \, C \]
\[ r = 0,05 \, m \]
Mun haɗa waɗannan dabi'u a cikin dabarar:
\[ F = 8.9875 \sau 10^9 \, \frac{|2 \sau 10^{-6} \cdot -3 \sau 10^{-6}|}{(0,05)^2} \]
\[ F = 8.9875 \sau 10^9 \, \frac{6 \sau 10^{-12}}{0,0025} \]
\[ F = 8.9875 \sau 10^9 \, \frac{6 \sau 10^{-12}}{2,5 \sau 10^{-3}} \]
\[ F = 8.9875 \sau 10^9 \, \sau 2,4 \sau 10^{-9} \]
\[ F = 21.57 \, N \]
Domin kuwa cajin \( q_2 \) yana da rashin inganci, ƙarfin lantarki da ke aiki yana da ƙarfi mai jan hankali, domin kuwa caji mai kyau da mara kyau suna jawo hankalin juna.
Misali Tambaya ta 2
Tambaya:
An raba cajin lantarki guda biyu \( q_1 = 5 \sau 10^{-9} \, C \) da \( q_2 = 10 \sau 10^{-9} \, C \) da nisan mita 0,1. Lissafa girman ƙarfin lantarki da ke aiki akan \( q_1 \).
Tattaunawa:
Yi amfani da dabarar Dokar Coulomb kuma:
\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
Maye gurbin dabi'un da aka sani:
\[ k_e = 8.9875 \sau 10^9 \, N \cdot m^2 \cdot C^{-2} \]
\[ q_1 = sau 5 10^{-9} \, C \]
\[ q_2 = sau 10 10^{-9} \, C \]
\[ r = 0,1 \, m \]
\[ F = 8.9875 \sau 10^9 \, \frac{(sau 5 10^{-9}) (sau 10 10^{-9})}{(0,1)^2} \]
\[ F = 8.9875 \sau 10^9 \, \frac{50 \sau 10^{-18}}{0,01} \]
\[ F = 8.9875 \sau 10^9 \, \sau 5 \sau 10^{-15} \]
\[ F = 44.9375 \, N \]
Ƙarfin da ke aiki ƙarfi ne mai ƙyama, domin duka caji \( q_1 \) da \( q_2 \) suna da alama iri ɗaya, wato tabbatacce.
Misali Tambaya ta 3
Tambaya:
Cajin maki uku suna cikin layi madaidaiciya. Cajin farko shine \( q_1 = 2 \mu C \), cajin na biyu shine \( q_2 = -1 \mu C \), kuma cajin na uku shine \( q_3 = 3 \mu C \). Nisa tsakanin \( q_1 \) da \( q_2 \) shine 0,1 m, yayin da nisan da ke tsakanin \( q_2 \) da \( q_3 \) shine 0,2 m. Lissafa girma da alkiblar jimlar ƙarfin lantarki da ke aiki akan \( q_2 \).
Tattaunawa:
Da farko, muna ƙididdige ƙarfin lantarki tsakanin \( q_1 \) da \( q_2 \):
\[ F_{12} = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
\[ k_e = 8.9875 \sau 10^9 \, N \cdot m^2 \cdot C^{-2} \]
\[ q_1 = sau 2 10^{-6} \, C \]
\[ q_2 = -1 \sau 10^{-6} \, C \]
\[ r = 0,1 \, m \]
\[ F_{12} = 8.9875 \sau 10^9 \, \frac{(sau 2 \sau 10^{-6}) (-1 \sau 10^{-6})}{(0,1)^2} \]
\[ F_{12} = 8.9875 \sau 10^9 \, \frac{2 \sau 10^{-12}}{0,01} \]
\[ F_{12} = 8.9875 \sau 10^9 \, \sau 2 \sau 10^{-10} \]
\[ F_{12} = 1.7975 \, N \]
Tunda \( q_1 \) yana da kyau kuma \( q_2 \) yana da korau, ƙarfin \( F_{12} \) shine ƙarfin jan hankali wanda aka mayar da hankali zuwa ga \( q_1 \).
Na biyu, muna ƙididdige ƙarfin lantarki tsakanin \( q_2 \) da \( q_3 \):
\[ F_{23} = k_e \frac{|q_2 \cdot q_3|}{r^2} \]
\[ q_2 = -1 \sau 10^{-6} \, C \]
\[ q_3 = sau 3 10^{-6} \, C \]
\[ r = 0,2 \, m \]
\[ F_{23} = 8.9875 \sau 10^9 \, \frac{(-1 \sau 10^{-6}) (sau 3 10^{-6})}{(0,2)^2} \]
\[ F_{23} = 8.9875 \sau 10^9 \, \frac{3 \sau 10^{-12}}{0,04} \]
\[ F_{23} = 8.9875 \sau 10^9 \, \sau 7.5 \sau 10^{-11} \]
\[ F_{23} = 0.67406 \, N \]
Tunda \( q_2 \) yana da korau kuma \( q_3 \) yana da kyau, ƙarfin \( F_{23} \) shine ƙarfin jan hankali wanda aka mayar da hankali zuwa ga \( q_3 \).
A ƙarshe, mun haɗa dukkan ƙarfin don samun sakamakon ƙarshe:
Jimlar ƙarfi akan \( q_2 \):
\[ F_{\text{total}} = F_{12} – F_{23} \]
\[ F_{\text{jimlar}} = 1.7975 \, N – 0.67406\, N \]
\[ F_{\text{jimlar}} = 1.12344 \, N \]
Alkiblar jimlar ƙarfin tana zuwa ne ga \( q_1 \) saboda ƙarfin \( F_{12} \) ya fi \( F_{23} \).
Kammalawa
Dokar Coulomb ta ba da muhimman bayanai game da hulɗar da ke tsakanin cajin lantarki, mai kyau da kuma abin ƙyama. Misalan da aka tattauna sun nuna yadda wannan doka ke amfani da ita wajen ƙididdige ƙarfin lantarki ta hanyar la'akari da girman cajin da kuma nisan da ke tsakaninsu. Cikakken fahimtar Dokar Coulomb zai iya taimaka mana mu fahimci nau'ikan abubuwan lantarki da maganadisu daban-daban da ke faruwa a kusa da mu.
