Su'aalo Tusaale ah oo Ka Hadlaya Xoogga Korontada
Qeexidda xoogga korontadu waa xoogga ka dhasha isdhexgalka u dhexeeya dallacaadaha korontada. Xooggan waxaa lagu bartaa fiisigiska, gaar ahaan laanta electrodynamics. Saamaynta xoogga korontadu waxaa lagu arki karaa dhacdooyin badan oo maalinle ah, laga bilaabo isku xoqidda walxaha qaarkood ilaa hawlgalka aaladaha elektaroonigga ah. Maqaalkani wuxuu ka hadli doonaa dhowr tusaale oo ku saabsan dhibaatooyinka xoogga korontadu iyo xalalkooda, iyo sidoo kale fikradaha ugu muhiimsan ee la xiriira xoogga korontadu.
Hordhac ku saabsan Xoogga Korontada
Xoogga korontada ee u dhexeeya laba dallacaad koronto waxaa bixiya Sharciga Coulomb, kaas oo lagu qeexay isla'egta:
$$ F = k_e \frac{|q_1 q_2|}{r^2} $$
Halkee:
– \( F \) waa xoogga korontada ee u dhexeeya laba dallacaad.
– \( q_1 \) iyo \( q_2 \) waa baaxadda khidmadaha.
– \( r \) waa masaafada u dhaxaysa labada dallacaad.
– \( k_e \) waa joogtada Coulomb (qiyaastii \(8.99 \jeer 10^9 \, N \cdot m^2 / C^2\)).
Sharciga Coulomb wuxuu la mid yahay sharciga Newton ee cufisjiidadka, laakiin cufisjiidadka had iyo jeer waa mid soo jiidasho leh halka xoogga korontadu uu noqon karo mid soo jiidasho leh ama laga yaqyaqsado iyadoo ku xiran calaamadda ku dul-saarista.
Su'aalo iyo Doodo Tusaale ah
Su'aal 1: Laba Lacagood oo Isku Mid ah
Laba dhibic oo dallac ah \( q_1 = 3 \, \mu C \) iyo \( q_2 = -4 \, \mu C \) waxay ku jiraan masaafo dhan 0.5 mitir midba midka kale. Xisaabi xoogga korontada ee u dhexeeya labada dallac.
Dood:
1. Aqoonsiga Culayska iyo Masaafada:
– \( q_1 = 3 \, \mu C = 3 \jeer 10^{-6} \, C \)
– \( q_2 = -4 \, \mu C = -4 \jeer 10^{-6} \, C \)
– \( r = 0.5 \, m \)
2. Beddelka Sharciga Coulomb:
\[
F = k_e \frac{|q_1 q_2|}{r^2}
\]
\[
F = 8.99 \ jeer 10^9 \, \frac{(3 \ jeer 10^{-6})(4 \ jeer 10^{-6})}{(0.5)^2}
\]
3. Xisaabi Natiijooyinka:
\[
F = 8.99 \jeer 10^9 \, \frac{12 \jeer 10^{-12}}{0.25}
\]
\[
F = 8.99 \jeer 10^9 \jeer 48 \jeer 10^{-12}
\]
\[
F = 431.52 \jeer 10^{-3} \, N
\]
\[
F \ qiyaastii 0.432 \, N
\]
Markaa, xoogga korontada ee u dhexeeya labada dallacaad waa qiyaastii 0.432 Newtons.
Su'aal 2aad: Lacag ku dallac badan
Saddex dhibcood oo isku xiga, \( q_1 = 2 \, \mu C \), \( q_2 = 3 \, \mu C \), iyo \( q_3 = -1 \, \mu C \) waxaa la dhigayaa xariiq toosan iyadoo masaafada u dhaxaysa \( q_1 \) iyo \( q_2 \) ay tahay 0.6 mitir iyo inta u dhaxaysa \( q_2 \) iyo \( q_3 \) ay tahay 0.4 mitir. Xisaabi xoogga ka dhashay \( q_2 \).
Dood:
1. Aqoonsiga Culayska iyo Masaafada:
– \( q_1 = 2 \, \mu C = 2 \jeer 10^{-6} \, C \)
– \( q_2 = 3 \, \mu C = 3 \jeer 10^{-6} \, C \)
– \( q_3 = -1 \, \mu C = -1 \jeer 10^{-6} \, C \)
– Masaafada u dhaxaysa \( q_1 \) iyo \( q_2 \) waa 0.6 m
– Masaafada u dhaxaysa \( q_2 \) iyo \( q_3 \) waa 0.4 m
2. Ku qasbo \( q_1 \) ka dhanka ah \( q_2 \):
\[
F_{12} = k_e \frac{|q_1 q_2|}{r_{12}^2}
\]
\[
F_{12} = 8.99 \jeer 10^9 \, \frac{(2 \jeer 10^{-6})(3 \jeer 10^{-6})}{(0.6)^2}
\]
\[
F_{12} = 8.99 \jeer 10^9 \, \frac{6 \jeer 10^{-12}}{0.36}
\]
\[
F_{12} = 8.99 \jeer 10^9 \jeer 16.67 \jeer 10^{-12}
\]
\[
F_{12} \qiyaastii 0.15 \, N
\]
Jihada xoogga \( F_{12} \) way ka fog tahay \( q_1 \) (midigta).
3. Ku qasbo \( q_3 \) ka dhanka ah \( q_2 \):
\[
F_{32} = k_e \frac{|q_3 q_2|}{r_{32}^2}
\]
\[
F_{32} = 8.99 \jeer 10^9 \, \frac{(1 \jeer 10^{-6})(3 \jeer 10^{-6})}{(0.4)^2}
\]
\[
F_{32} = 8.99 \jeer 10^9 \, \frac{3 \jeer 10^{-12}}{0.16}
\]
\[
F_{32} = 8.99 \jeer 10^9 \jeer 18.75 \jeer 10^{-12}
\]
\[
F_{32} \qiyaastii 0.17 \, N
\]
Jihada xoogga \( F_{32} \) waxay u jiidanaysaa dhanka \( q_3 \) (bidixda).
4. Xoogga Natiijada:
Maadaama \(F_{12} \) uu midig yahay iyo \(F_{32} \) uu bidix yahay, xoogga ka dhasha \(F_{res} = F_{32} – F_{12} \):
\[
F_{res} = 0.17 \,N – 0.15 \,N
\]
\[
F_{res} = 0.02 \, N (bidixda)
\]
Markaa, xoogga ka dhashay \( q_2 \) waa 0.02 Newton dhanka bidix.
Su'aal 3: Nidaamka Rarka ee Saddex Xagalka ah
Saddexda dhibcood ee dallacaadda \( q_1 = 5 \, \mu C \), \( q_2 = -3 \, \mu C \), iyo \( q_3 = 4 \, \mu C \) waxaa la dhigayaa saddex gees oo saddexagal isle'eg ah oo dhererkiisu yahay 0.2 mitir. Xisaabi wadarta guud ee xoogga korantada ee ku shaqeeya dallacaadda \( q_1 \).
Dood:
1. Ku qasbo \( q_2 \) ka dhanka ah \( q_1 \):
\[
F_{12} = k_e \frac{|q_1 q_2|}{r^2}
\]
\[
F_{12} = 8.99 \jeer 10^9 \, \frac{(5 \jeer 10^{-6})(3 \jeer 10^{-6})}{(0.2)^2}
\]
\[
F_{12} = 8.99 \jeer 10^9 \, \frac{15 \jeer 10^{-12}}{0.04}
\]
\[
F_{12} = 8.99 \jeer 10^9 \jeer 375 \jeer 10^{-12}
\]
\[
F_{12} \qiyaastii 3.37 \, N
\]
Jihada xoogga \( F_{12} \) waa mid soo jiidasho leh oo ku wajahan \( q_2 \).
2. Ku qasbo \( q_3 \) ka dhanka ah \( q_1 \):
\[
F_{13} = k_e \frac{|q_1 q_3|}{r^2}
\]
\[
F_{13} = 8.99 \jeer 10^9 \, \frac{(5 \jeer 10^{-6})(4 \jeer 10^{-6})}{(0.2)^2}
\]
\[
F_{13} = 8.99 \jeer 10^9 \, \frac{20 \jeer 10^{-12}}{0.04}
\]
\[
F_{13} = 8.99 \jeer 10^9 \jeer 500 \jeer 10^{-12}
\]
\[
F_{13} \qiyaastii 4.50 \, N
\]
Jihada xoogga \( F_{13} \) waa mid laga naxo \( q_3 \).
3. Xoogga Natiijada:
Waxaan u baahanahay inaan isticmaalno habka vector-ka si aan u helno natiijada \( F_{12} \) iyo \( F_{13} \), sababtoo ah xagasha u dhaxaysa xoogagga waa 60 darajo (saddex-xagal isle'eg).
Annagoo adeegsanayna qaybaha x iyo y, waxaan go'aamin karnaa xoogga guud. Si kastaba ha ahaatee, si fudud, waxaan isticmaali doonnaa hababka garaafyada ama xisaabinta tirooyinka si aan u ogaano in xoogga guud uu ka weyn yahay 4.50 N oo uu ku yaal meel u dhaxaysa xariiqda isku xirta \( q_2 \) iyo \( q_3 \) ilaa \( q_1 \).
Gabagabo
Iyada oo loo marayo dhibaatada tusaalaha ah iyo doodda kor ku xusan, waxaan fahmi karnaa sida loo xisaabiyo xoogga korontada iyadoo la adeegsanayo Sharciga Coulomb. Adigoo dhibaatada u kala qaybinaya tallaabooyin fudud oo adeegsanaya tixgelinta vector-ka, waxaan go'aamin karnaa xoogga guud ee ku jira kharash kasta oo ku jira nidaamka. Xoogga korontada, inkastoo uu fudud yahay qaab-dhismeedkiisa xisaabta, wuxuu yeelan karaa saameyn aad u muhiim ah oo adag oo ku saabsan codsiyada dhabta ah.
