Imibuzo Eyisibonelo Exoxa Ngomthetho KaCoulomb
Umthetho kaCoulomb uyisimiso esiyisisekelo ku-physics esichaza amandla aphakathi kwamacala amabili kagesi. Ufundwe ngaphakathi kwesigaba esingaphansi kwe-electrostatics, lo mthetho uchaza ukuthi amacala kagesi adonselana noma axoshana kanjani. Lo mthetho waqala ukuphakanyiswa yisazi sesayensi yefiziksi saseFrance uCharles-Augustin de Coulomb ngekhulu le-18. Lesi sihloko sizoxoxa ngoMthetho kaCoulomb ngokuhlola izibonelo nezixazululo zazo, ngaleyo ndlela sisize abafundi baqonde ukusetshenziswa okusebenzayo kwalesi simiso esiyisisekelo.
Isisekelo Semfundiso: Uyini Umthetho KaCoulomb?
UMthetho kaCoulomb uthi ubukhulu bamandla kagesi \( F \) phakathi kwamacala amabili \( q_1 \) kanye \( q_2 \) bulingana ngqo nomkhiqizo wobukhulu bamacala amabili futhi bulingana ngokuphambene nesikwele sebanga \( r \) eliphakathi kwawo. Ngokwezibalo, ifomula yoMthetho kaCoulomb ingachazwa kanje:
\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
Di mana:
– \( F \) ubukhulu bamandla kagesi aqinile
– \( k_e \) kuyinto engaguquki kaCoulomb (\( 8.9875 \times 10^9 \, N \cdot m^2 \cdot C^{-2} \))
– \( q_1 \) kanye \( q_2 \) ubukhulu beshaja kagesi
– \( r \) ibanga eliphakathi kwamacala amabili
Isibonelo Umbuzo 1
Umbuzo:
Kunezinkokhelo ezimbili zamaphuzu: \( q_1 = 2 \times 10^{-6} \, C \) kanye \( q_2 = -3 \times 10^{-6} \, C \). Ziqhelelene ngamamitha angu-0,05. Bala ubukhulu bamandla kagesi asebenza phakathi kwezinkokhelo ezimbili.
Ingxoxo:
Isinyathelo sokuqala ukubhala kabusha ifomula yoMthetho kaCoulomb:
\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
Manje sifaka amanani aziwayo esikhundleni sefomula:
\[ k_e = 8.9875 \izikhathi 10^9 \, N \cdot m^2 \cdot C^{-2} \]
\[q_1 = 2 \izikhathi eziyi-10^{-6} \, C \]
\[ q_2 = -3 \izikhathi eziyi-10^{-6} \, C \]
\[r = 0,05 \, m \]
Sixhuma la manani kufomula:
\[ F = 8.9875 \izikhathi 10^9 \, \frac{|2 \izikhathi 10^{-6} \cdot -3 \izikhathi 10^{-6}|}{(0,05)^2} \]
\[ F = 8.9875 \izikhathi 10^9 \, \frac{6 \izikhathi 10^{-12}}{0,0025} \]
\[ F = 8.9875 \izikhathi 10^9 \, \frac{6 \izikhathi 10^{-12}}{2,5 \izikhathi 10^{-3}} \]
\[ F = 8.9875 \izikhathi 10^9 \, \izikhathi 2,4 \izikhathi 10^{-9} \]
\[ F = 21.57 \, N \]
Ngenxa yokuthi ishaja \(q_2 \) ayilungile, amandla kagesi asebenzayo angamandla akhangayo, ngoba amacala amahle nangalungile ayakhangana.
Isibonelo Umbuzo 2
Umbuzo:
Amashaja amabili kagesi \( q_1 = 5 \times 10^{-9} \, C \) kanye \( q_2 = 10 \times 10^{-9} \, C \) ahlukaniswe ngebanga lamamitha angu-0,1. Bala ubukhulu bamandla kagesi asebenza ku-\( q_1 \).
Ingxoxo:
Sebenzisa ifomula yoMthetho kaCoulomb futhi:
\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
Faka amanani aziwayo esikhundleni sawo:
\[ k_e = 8.9875 \izikhathi 10^9 \, N \cdot m^2 \cdot C^{-2} \]
\[q_1 = 5 \izikhathi eziyi-10^{-9} \, C \]
\[q_2 = 10 \izikhathi eziyi-10^{-9} \, C \]
\[r = 0,1 \, m \]
\[ F = 8.9875 \izikhathi 10^9 \, \frac{(5 \izikhathi 10^{-9}) (10 \izikhathi 10^{-9})}{(0,1)^2} \]
\[ F = 8.9875 \izikhathi 10^9 \, \frac{50 \izikhathi 10^{-18}}{0,01} \]
\[ F = 8.9875 \izikhathi 10^9 \, \izikhathi 5 \izikhathi 10^{-15} \]
\[ F = 44.9375 \, N \]
Amandla asebenzayo angamandla axoshayo, ngoba womabili amacala \( q_1 \) kanye \( q_2 \) anophawu olufanayo, okungukuthi amahle.
Isibonelo Umbuzo 3
Umbuzo:
Amashaja amaphuzu amathathu asemgqeni oqondile. Ishaja yokuqala ingu-\( q_1 = 2 \mu C \), ishaja yesibili ingu-\( q_2 = -1 \mu C \), kanti ishaja yesithathu ingu-\( q_3 = 3 \mu C \). Ibanga eliphakathi kuka-\( q_1 \) no-\( q_2 \) lingu-0,1 m, kanti ibanga eliphakathi kuka-\( q_2 \) no-\( q_3 \) lingu-0,2 m. Bala ubukhulu kanye nesiqondiso samandla kagesi aphelele asebenza ku-\( q_2 \).
Ingxoxo:
Okokuqala, sibala amandla kagesi aphakathi kuka-\( q_1 \) no-\( q_2 \):
\[ F_{12} = k_e \frac{|q_1 \cdot q_2|}{r^2} \]
\[ k_e = 8.9875 \izikhathi 10^9 \, N \cdot m^2 \cdot C^{-2} \]
\[q_1 = 2 \izikhathi eziyi-10^{-6} \, C \]
\[ q_2 = -1 \izikhathi eziyi-10^{-6} \, C \]
\[r = 0,1 \, m \]
\[ F_{12} = 8.9875 \izikhathi 10^9 \, \frac{(2 \izikhathi 10^{-6}) (-1 \izikhathi 10^{-6})}{(0,1)^2} \]
\[ F_{12} = 8.9875 \izikhathi 10^9 \, \frac{2 \izikhathi 10^{-12}}{0,01} \]
\[ F_{12} = 8.9875 \izikhathi 10^9 \, \izikhathi 2 \izikhathi 10^{-10} \]
\[ F_{12} = 1.7975 \, N \]
Njengoba u-\( q_1 \) elungile futhi u-\( q_2 \) engalungile, amandla \( F_{12} \) angamandla akhangayo aqondiswe ku-\( q_1 \).
Okwesibili, sibala amandla kagesi aphakathi kuka-(q_2 \) no-(q_3 \):
\[ F_{23} = k_e \frac{|q_2 \cdot q_3|}{r^2} \]
\[ q_2 = -1 \izikhathi eziyi-10^{-6} \, C \]
\[q_3 = 3 \izikhathi eziyi-10^{-6} \, C \]
\[r = 0,2 \, m \]
\[ F_{23} = 8.9875 \izikhathi 10^9 \, \frac{(-1 \izikhathi 10^{-6}) (3 \izikhathi 10^{-6})}{(0,2)^2} \]
\[ F_{23} = 8.9875 \izikhathi 10^9 \, \frac{3 \izikhathi 10^{-12}}{0,04} \]
\[ F_{23} = 8.9875 \izikhathi 10^9 \, \izikhathi 7.5 \izikhathi 10^{-11} \]
\[ F_{23} = 0.67406 \, N \]
Njengoba u-\( q_2 \) engemuhle futhi u-\( q_3 \) emuhle, amandla \( F_{23} \) angamandla akhangayo aqondiswe ku-\( q_3 \).
Ekugcineni, sihlanganisa womabili la mandla ukuze sithole umphumela wokugcina:
Amandla aphelele ku-\( q_2 \):
\[ F_{\text{total}} = F_{12} – F_{23} \]
\[ F_{\text{total}} = 1.7975 \, N – 0.67406\, N \]
\[ F_{\text{total}} = 1.12344 \, N \]
Isiqondiso samandla aphelele sibheke ku-\( q_1 \) ngoba amandla \( F_{12} \) makhulu kuno-\( F_{23} \).
Isiphetho
UMthetho kaCoulomb unikeza ukuqonda okubalulekile mayelana nokusebenzisana phakathi kwamacala kagesi, kokubili akhangayo nanyanyekayo. Izibonelo ezixoxwe ngazo zibonisa ukusetshenziswa kwalo mthetho ekubaleni amandla kagesi ngokubheka ubukhulu bamacala kanye nebanga eliphakathi kwawo. Ukuqonda kahle uMthetho kaCoulomb kungasisiza siqonde izenzakalo ezahlukene zikagesi kanye namagnetic ezenzeka eduze kwethu.
