Imibuzo Eyisibonelo Exoxa Ngamandla Kagesi Aphumelelayo
Ugesi uhlobo lwamandla olubalulekile empilweni yethu yansuku zonke. Omunye wemibono eyisisekelo ku-physics kagesi yi-Coulomb force, echaza ukusebenzisana phakathi kwamacala kagesi. Ukuqonda amandla kagesi aphumayo kubalulekile ezindleleni eziningi zokusebenza, kusukela kubuchwepheshe be-elekthronikhi kuya ekuqondeni izenzakalo endaweni yonke. Okulandelayo ingxoxo ejulile yezinkinga zezibonelo kanye nendlela yokunquma amandla kagesi aphumayo.
Isingeniso kuMqondo Wamandla kaCoulomb
Ngaphambi kokuba singene enkingeni yesibonelo, ake sibukeze umthetho kaCoulomb. Umthetho kaCoulomb uthi amandla kagesi aphakathi kwamacala amabili ayalingana nomkhiqizo wobukhulu bawo futhi alingana ngokuphambene nesikwele sebanga eliphakathi kwawo. Ngokwezibalo, amandla kaCoulomb \( F \) angachazwa kanje:
\[ F = k_e \frac{|q_1 q_2|}{r^2} \]
Di mana:
– \( F \) ubukhulu bamandla kagesi aqinile.
– \( k_e \) kuyinto engaguquki kaCoulomb, \( k_e \cishe 8,99 \izikhathi 10^9 \, \text{Nm}^2/\text{C}^2 \).
– \( q_1 \) kanye \( q_2 \) yizilinganiso zamanani.
– \( r \) ibanga eliphakathi kwamacala.
Imibuzo Eyisibonelo Nengxoxo
Umbuzo 1: Ukushaja Okubili Emgqeni Oqondile
Amashaja amabili amaphoyinti \( q_1 = 4 \mu C \) kanye \( q_2 = -3 \mu C \) atholakala ebangeni elingama-3 cm ukusuka komunye nomunye. Bala ubukhulu kanye nesiqondiso samandla e-Coulomb phakathi kwala mashaja amabili.
Ingxoxo:
Isinyathelo sokuqala ukuguqula ishaja nebanga kube amayunithi e-SI:
– \( q_1 = 4 \izikhathi ezingu-10^{-6} C \)
– \( q_2 = -3 \izikhathi ezingu-10^{-6} C \)
– \( r = 3 \izikhathi ezingu-10^{-2} m \)
Ukusebenzisa i-Coulomb's law equation:
\[ F = k_e \frac{|q_1 q_2|}{r^2} \]
Faka amanani aziwayo esikhundleni sawo:
\[ F = (8,99 \izikhathi 10^9) \frac{(4 \izikhathi 10^{-6})(3 \izikhathi 10^{-6})}{(3 \izikhathi 10^{-2})^2} \]
Bala ubukhulu bamandla:
\[ F = (8,99 \izikhathi 10^9) \frac{12 \izikhathi 10^{-12}}{9 \izikhathi 10^{-4}} \]
\[ F = (8,99 \izikhathi 10^9) \izikhathi (1,33 \izikhathi 10^{-8}) \]
\[ F = 1,197 \izikhathi ezingu-10^2 \]
\[ F = 119,7 \, N \]
Isiqondiso samandla: Njengoba \( q_1 \) ilungile futhi \( q_2 \) ingalungile, amandla ayakhanga. Ngakho-ke, amandla aku-\( q_1 \) aqondiswe ku-\( q_2 \) kanye nokuphikisana nalokho.
Umbuzo 2: Amacala Amathathu Kunxantathu Olinganayo
Amashaja amathathu, ngalinye \( q = 2 \mu C \), abekwa emaphethelweni kanxantathu olinganayo onezinhlangothi ezinde ezingama-5 cm. Bala amandla aphumayo asebenza kushaja ngayinye.
Ingxoxo:
Guqula ukushaja kanye nebanga kube amayunithi e-SI:
– \( q = 2 \izikhathi 10^{-6} C \)
– \( r = 5 \izikhathi ezingu-10^{-2} m \)
Amandla phakathi kwamacala amabili:
\[ F = k_e \frac{q^2}{r^2} \]
Faka amanani aziwayo esikhundleni sawo:
\[ F = (8,99 \izikhathi 10^9) \frac{(2 \izikhathi 10^{-6})^2}{(5 \izikhathi 10^{-2})^2} \]
\[ F = (8,99 \izikhathi 10^9) \frac{4 \izikhathi 10^{-12}}{25 \izikhathi 10^{-4}} \]
\[ F = (8,99 \izikhathi 10^9) \izikhathi (1,6 \izikhathi 10^{-10}) \]
\[ F = 1,4384 N \]
La mandla asebenza kubhangqa ngalinye lamashaji kunxantathu. Ukuze uthole amandla aphumayo asebenza kushaja ngayinye, kubalulekile ukuhlaziya amavektha amandla amabutho amabili asebenza kushaja ngayinye eziqondisweni eziqondile neziqondile zonxantathu.
Ake sithi ukushaja ku-\( A \) kuthola amandla avela ku-\( B \) kanye no-\( C \):
– Amandla avela ku-\( B \) kuya ku-\( A \) angu-\( F_{AB} = 1,4384 \, N \).
– Amandla avela ku-\( C \) kuya ku-\( A \) angu-\( F_{AC} = 1,4384 \, N \).
Njengoba la mandla amabili akha i-engeli engu-60° komunye nomunye (ngenxa yonxantathu olinganayo), singasebenzisa ukuhlaziywa kwengxenye yevektha ukuthola umphumela.
Izingxenye zamandla ku-x-axis kanye ne-y-axis:
\[ F_{Ax} = F_{AB} \cos 30^\circ + F_{AC} \cos 30^\circ \]
\[ F_{Ay} = F_{AB} \sin 30^\circ - F_{AC} \sin 30^\circ \]
Kodwa-ke, njengoba womabili amandla elingana futhi elingana ncamashi kuzo zombili izinhlangothi, zonke izingxenye zika-y zizokhanselwa futhi ngempumelelo, ishaja \( A \) ithintwa kuphela amandla avundlile asebenza phakathi.
\[
F_{Ax} = 2F_{AB} \cos 30^\circ \\
= 2(1,4384 \ N \cdot \ 0,866) \\
= 2 \cdot 1.2467 \\
= 2.4934 uN
]
Ngakho-ke amandla kagesi aphumela kwenye yamacala angu-\( F = 2.4934 N.
Isiphetho
Ukunquma amandla kagesi aphumayo ohlelweni lwama-point charges kudinga ukuqonda okujulile komthetho kaCoulomb kanye nekhono lokuhlakaza i-force vector ibe yizingxenye zayo. Ezimweni eziyinkimbinkimbi kakhulu ezinama-charges amathathu noma ngaphezulu, ukuhlaziywa kwe-vector kuvame ukuba yithuluzi elibalulekile lokuthola imiphumela enembile. Ngezinkinga zokuzijwayeza ezinjengalezi, singaqonda futhi sisebenzise imiqondo yefiziksi ngempumelelo enkulu.
