20 Tusaalooyin su'aalo koronto oo aan joogto ahayn
Xoogga Korontada
1. Barta A waxay ku taal garoonka korontadaXoogga goobta korantada ee barta A = 0,5 NC-1Haddii shay koronto leh oo ah 0,25 C la dhigo barta A, markaas shaqada ayaa laga qaban doonaa shayga. gaya Coulomb sida weyn…
A. 0,125 N
B. 0,25 N
C. 0,35 N
D. 0,40 N
E. 0,70 N
Dood
Waa la ogyahay in:
Xoogga goobta korontada ee barta A = 0,5 NC-1
Dareeraha korontada ee barta A = 0,25 C
La weydiiyay: Xoogga Coulomb ee ku shaqeeya walxaha korontada lagu dallaco
Jawaab:
Qaacidada sheegaysa xiriirka ka dhexeeya xoogga korontada (F), goobta korontada (E) iyo dallacaadda korontada (q) waa:
F = q E
F = (0,25 C)(0,5 NC)-1)
F = 0,125N
Jawaabta saxda ah waa A.
2. Laba dallacaad oo ah 5 C iyo 4 C waxay u dhexeeyaan 3 m. Haddii k = 9 × 109 Nm2 C-2 , markaas baaxadda ciidanka Coulomb ee ay la kulmeen labada dacwadood waa...
A. 2 × 109 N
B. 60 × 109 N
C. 2 × 1010 N
D. 6 × 1010 N
E. 20 × 1010 N
Dood
Waa la ogyahay in:
Kharash 1 (q1) = 5 C
Kharash 2 (q2) = 4 C
Masaafada u dhaxaysa culaysyada 1 iyo 2 (r) = 3 mitir.
Joogtada ah ee Coulomb (k) = 9 × 109 Nm2 C-2
La weydiiyay: Baaxadda xoogga Coulomb (F)
Jawaab:
Jawaabta saxda ah waa C.
3. Kharashka korontada + q1 = 10 μC; +q2 = 20 μC; iyo q3 sida sawirka hoose ku qoran. Si ciidanka Coulomb ay u fuliyaan dacwadda q2 = eber; ka dibna kharashka q3 waa…
A. +2,5 μC
B. –2,5 μC
C. +25 μC
D. –25 μC
E. +4 μC
Dood
Waa la ogyahay in:
Kharash 1 (q1) = 10 μC = 10 x 10-6 C
Kharash 2 (q2) = 20 μC = 20 x 10-6 C
La weydiiyay: Waa maxay kharashka q?3 si ciidanka Coulomb ee ku dhaqmaya dacwadda q2 la mid ah eber (F)2 = 0).
Jawaab:
Waxaa jira laba awoodood oo ku shaqeeya + q2.
Xoogga koowaad waa xoogga iska caabiya ee u dhexeeya khidmadaha + q1 iyo dallac + q2 gaar ahaan F12 taas oo dhanka midig u jirta.
Si ay awoodda korantada ee ka dhalata ay u dhaqanto q2 waxay la mid tahay eber markaa q3 waa in si xun loo eegaaMarkaa xoogga labaad waa soo jiidashada u dhaxaysa khidmadaha +q2 iyo -q3 gaar ahaan F23 taas oo bidixda u jirta. Labadan awoodood waxay ku dhaqmaan q2, waxay leeyihiin cabbir isku mid ah laakiin jiho liddi ku ah.
Xoogga ka dhashay + q2 la mid ah eber.
Jawaabta saxda ah waa B.
4. Dhibcaha A iyo B waxay leeyihiin koronto koronto oo ah −10 μC iyo +40 μC, siday u kala horreeyaan. Marka hore labada dallac waxaa la dhigaa meel 0,5 mitir u jirta si xoog Coulomb F Newton ah u soo baxo. Haddii masaafada u dhaxaysa A iyo B loo beddelo 1,5 mitir, markaa xoogga Coulomb ee soo baxa waa...
A. 1/9 F
B. 1/3 F
C. 3/2 F
D. 3 F
E. 9 F
Dood
Isbarbardhig doodda su'aasha lambarka 9.
Masaafada u dhaxaysa A iyo B waxaa loo beddelaa 1,5 mitir ama 3 jeer masaafada asalka ah.
Xoogga ayaa si liddi ku ah u dhigma labajibbaaranaha masaafada:
Xoogga Coulomb ee soo baxa waa 1/9 F.
Jawaabta saxda ah waa A.
5. Nidaam leh 3 lacag oo bilaash ah oo cabbir isku mid ah ayaa la dhigayaa si uu u dheellitiro sida sawirka ka muuqata.3 1/3 x ayaa loo wareejiyay meel u dhow Q2, ka dibna saamiga baaxadda xoogga Coulomb F2 : F1 noqo….
A. 1: 3
B. 2: 3
C. 3: 4
D. 9: 1
E. 9: 4
Dood
Waa la garanayaa :
Masaafada u dhaxaysa q1 iyo q2 =x
Masaafada u dhaxaysa q2 iyo q3 = 2/3 x
La weydiiyay : F2 : F1 = ….?
Jawab :
Qaacidada sharciga ee Coulomb:
Sharaxaad: k = joogto ah, q1 = dallac 1, q2 = dallac 2, r = masaafada u dhaxaysa dallac 1 iyo dallac 2
Isbarbardhigga baaxadda xoogga Coulomb
q1, q2 iyo q3 cabbirkoodu waa isku mid sidaa darteed waxaa laga saarayaa isla'egta. k iyo x2 sidoo kale waa isku cabbir waxayna ku yaalliin dhinaca bidix iyo midig sidaa darteed waa laga saarayaa isla'egta.
Jawaabta saxda ah waa E.
6. Fiiri sawirka hoose. Saddexda dallac ee korontada q1, q, iyo q2 waa xariiq toosan. Haddii q = 5,0 μC iyo d = 30 cm, markaa baaxadda iyo jihada xoogga korontada ee ku shaqeeya dallacaadda q waa… (k = 9 x 109 N m2 C-2)
A. 7,5 N dhanka q1
B. 7,5 N dhanka q2
C. 15 N dhanka q1
D. 22,5 N dhanka q1
E. 22,5 N dhanka q2
Dood
Waa la ogyahay in:
Kharash 1 (q1) = 30 μC = 30 x 10-6 C
Kharash 2 (q2) = 60 μC = 60 x 10-6 C
Dalac 3 (q) = 5 μC = 5 x 10-6 C
Masaafada u dhaxaysa q1 iyo q = d
Masaafada u dhaxaysa q2 iyo q = 2d
d = 30 cm = 0,3 mitir
d2 = (0,3)2 = 0,09
Joogtada ah ee Coulomb (k) = 9 x 109 N m2 C-2
La weydiiyay: Cabbirka iyo jihada xoogga korontada ee ku shaqeeya dallacaadda korantada
Jawaab:
Waxaa jira laba awoodood oo ku shaqeeya q, kuwaas oo kala ah F1 jihada waa dhanka midig (q iyo q)1 si togan ayaa loogu soo oogay sidaas darteed F1 ka fogow q iyo q1) iyo F2 jihada waxay u socotaa bidixda (q iyo q)2 si togan ayaa loogu soo oogay sidaas darteed F2 ka fogow q iyo q2) Marka hore xisaabi F1 iyo F2.
Xoogga natiijada:
ΣF = 15 – 7,5 = 7,5
Xoogga ka dhashay waa 7,5 Newtons. Jihada uu u socdo waa la mid F.1 taas oo ah dhanka midig ee u jeeda q2.
Jawaabta saxda ah waa B.
Goobta Korontada
7. Barta leh dallacaadda q waxay ku taal barta P ee goob koronto oo ay soo saarto dallacaadda (+) si ay ula kulanto xoog dhan 0,05 N jihada dallacaadda. Haddii xoogga goobta ee barta P uu yahay 2 x 10 -2 NC -1, markaa cabbirka iyo nooca kharashka keena goobta waa...
A. 5,0 C, togan
B. 5,0 C, taban
C. 3,0 C, togan
D. 2,5 C, taban
E. 2,5 C, togan
Dood
Waa la ogyahay in:
Xoogga korontada (F) = 0,05 N
Xoogga goobta korontada (E) = 2 x 10 -2 NC -1 = 0,02 NC -1
La weydiiyay: Cabbirka iyo nooca kharashka abuura goobta
Jawaab:
Kharashka korontada waxaa lagu xisaabiyaa iyadoo la adeegsanayo qaacido sheegaysa xiriirka ka dhexeeya xoogga korontada (F), goobta korontada (E) iyo kharashka korontada (q):
F = q E
q = F / E = 0,05 N / 0,02 NC -1 = 2,5 Coulombs
Kharash q wuxuu la kulmaa xoog koronto oo u jeeda dhanka dallacaadda (+) taasoo abuurta goob koronto, sidaas darteed dallacaadda q waxay leedahay calaamad taban.
Jawaabta saxda ah waa D.
8. Masaafada u dhaxaysa laba dallac A iyo B waa 4 m. Barta C waxay u dhaxaysaa labada dallac waana 1 m A. Haddii Q ay tahayA = –300 μC, QB = 600 μC. 1/4 π ε0 = 9 × 109 N m2 C-2 , markaa xoogga goobta korantada ee barta C sababtoo ah saameynta labada dallacaad waa...
A. 9 × 105 NC -1
B. 18 × 105 NC -1
C. 33 × 105 NC -1
D. 45 × 105 NC -1
E. 54 × 105 NC -1
Dood
Waa la ogyahay in:
Masaafada u dhaxaysa khidmadaha A iyo B (r)AB) = 4 mitir
Masaafada u dhaxaysa barta C iyo dallacaadda A (r)AC) = 1 mitir
Masaafada u dhaxaysa barta C iyo dallacaadda B (r)BC) = 3 mitir
Dalac A (q)A) = –300 μC = -300 x 10-6 C = -3 x 10-4 Coulomb
Dalac B (q)B) = 600 μC = 600 x 10-6 C = 6 x 10-4 Coulomb
Joogto ah (k) = 9 × 109 N m2 C-2
La weydiiyay: xoogga goobta korantada ee barta C
Jawaab:
Goobta korantada ee ay soo saarto dallacaadda A barta C:
Dalac A waa taban sidaa darteed jihada goobta korontada waxay u socotaa dalac A iyo meel ka fog dalac B (bidixda).
Goobta korantada ee ay soo saarto dallacaadda B ee barta C:
Dalac B waa togan sidaa darteed jihada goobta korontada waa ka fog tahay dalac B iyo jihada dalac A (bidixda).
Goobta korantada ee natiijada ka dhalatay barta A:
EA iyo EB isla jihadaas ayaa la isku daraa.
E = EA +EB
E = (27 x 105) + (6 x 105)
E = 33 x 105 N / C
Jihada goobta korontada waxay u socotaa dhanka dallacaadda A iyo dhanka dallacaadda B (dhinaca bidix).
Jawaabta saxda ah waa C.
9. Qashin 1 milligram ah oo boodh ah ayaa hawada ku dul sabeyn kara sababtoo ah joogitaanka goob koronto oo meesha ku haysa. Haddii dallacaadda walaxda ay tahay 0,5 μC iyo dardargelinta cufisjiidadka dhulku uu yahay 10 m/s.2 , go'aami baaxadda xoogga goobta korantada ee qaban kara boodhka.
A. 5 N/C
B. 10 N/C
C. 20 N/C
D. 25 N/C
E. 40 N/C
Dood
Waa la ogyahay in:
Cufka boodhka (m) = 1 milligram = 1 x 10-6 kg
Dareeraha boodhka (q) = 0,5 μC = 0,5 x 10-6 C
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s2
La weydiiyay: Goob koronto oo xooggan oo haysa boodhka
Jawaab:
Qaacidada miisaanka:
w = mg
Sharaxaad: w = miisaanka boodhka, m = cufnaanta boodhka, g = dardargelinta cufisjiidadka awgeed
Xoogga cufisjiidadka ee ku shaqeeya boodhka ama miisaanka boodhka waxaa lagu xisaabiyaa iyadoo la adeegsanayo qaacidada miisaanka:
w = mg = (1 x 10-6 kg) (10 m/s2) = 10 x 10-6 kg m/s2 = 10x10-6 Newton
Qaacidada xoogga goobta korontada:
E = F/q
Sharaxaad: E = xoogga goobta korantada, F = xoogga korontada, q = dallacaadda korontada
Boorku wuxuu ku dul sabeeyaa hawada, sidaa darteed xoogga ka dhasha ee ku shaqeeya boodhka waa inuu ahaadaa eber. Cufisjiidadka boodhka waxaa loo jiheeyaa hoos, sidaa darteed xoogga korontada waa in kor loo jeediyaa, baaxadda cufisjiidadka boodhkana waa inay la mid noqotaa baaxadda xoogga korontada, si xoogga ka dhasha boodhka uu noqdo eber. Sidaa darteed, F ee qaacidada xoogga goobta korontada waxaa lagu beddeli karaa w qaacidada miisaanka.
E = F/q = w/q
E = (10 x 10-6 N) / (0,5 x 10-6 C)
E = 10 N / 0,5 C
E = 20 N/C
Jawaabta saxda ah waa C.
10. Laba khidmadood oo q ah midkiiba1 = 32 μC iyo q2 = -214 μC waxaa lagu kala saaraa masaafad x ah oo midba midka kale ka fog yahay sida ku cad sawirka kore. Haddii ay tahay barta p oo ah 10 cm u jirta q.2 Xoogga goobta korantada ee ka dhalatay waa eber. Markaa baaxadda x waa….
A. 20 cm
B. 30 cm
C. 40 cm
D. 50 cm
E. 60 cm
Dood
Waa la ogyahay in:
Kharash 1 (Q)1) = 32 μC
Kharash 2 (Q)2) = -214 μC
Masaafada barta p laga bilaabo q1 = x + 10 cm
Masaafada barta p laga bilaabo q2 = 10cm
La weydiiyay: x
Jawaab:
E1 ma goobta korontada ee ay soo saarto dallacaadda Q?1Jihada garoonka korontadu way ka fog tahay Q1 sababtoo ah Q1 si togan ayaa loo dallacay. E2 ma goobta korontada ee ay soo saarto dallacaadda Q?2Jihada garoonka korontadu waxay u socotaa Q2 sababtoo ah Q2 si xun ayaa loogu eedeeyay.
Barta p oo ah 10 cm u jirta Q2, xoogga goobta korantada ee ka dhalatay waa eber.
Isticmaal qaacidada ABC:
11. Barta leh dallacaadda q waxay ku taal barta P ee goob koronto oo uu abuuray dallacaadda (+), sidaas darteed waxay la kulantaa xoog dhan 0,05 N. Haddii baaxadda dallacaaddu ay tahay +5 × l0-6 Coulomb, markaa baaxadda goobta korantada ee barta P waa…
A. 2,5 × 103 NC-1
B. 3.0 × 103 NC-1
C. 4,5 × l03 NC-1
D. 8,0 × 103 NC-1
E. 10kii4 NC-1
Dood
Waa la ogyahay in:
Xoogga korontada (F) = 0,05 Newton
Dakhli koronto (Q) = +5 × l0-6 Coulomb = 0,000005
La weydiiyay: baaxadda goobta korantada ee barta P
Jawaab:
Qaacidada sheegaysa xiriirka ka dhexeeya goobta korontada, xoogga korontada iyo dallacaadda korontada:
E = F / Q
E = 0,05 Newton / 0,000005 Coulomb
E = 5 Newton / 0,0005 Coulomb
E = 10.000 Newton/Coulomb
E = 104 N / C
E = 104 NC-1
Jawaabta saxda ah waa E.
12. Saddex dacwadood ayaa loo habeeyey sida ku cad jaantuska hoose. Xoogga Coulomb ee lagu arkay khidmadda B waa …. (k = 9 x 109 Nm2C-2, 1 μC = 10-6 C)
A. 09 x 101 Lacag N ilaa C ah
B. 09 x 101 N si loogu dallaco A
C. 18 x 101 Lacag N ilaa C ah
D. 18 x 101 N si loogu dallaco A
E. 36 x 101 Lacag N ilaa C ah
Dood
Waa la garanayaa :
qA = 10 µC = 10 x 10-6 C = 10-5 Coulomb
qB = 10 µC = 10 x 10-6 = 10-5 Coulomb
qC = 20 µC = 20 x 10-6 = 2x10-5 Coulomb
rAB = 0,1 mitir = 10-1 meter
rBC = 0,1 mitir = 10-1 meter
k = 9 x 109 Nm2C-2
La weydiiyay Ciidanka Coulomb oo ay khibrad u leeyihiin B
Jawab :
Waxaa jira laba awoodood oo Coulomb ah ama awood koronto oo ku shaqeeya dallacaadda B, kuwaas oo kala ah xoogga Coulomb ee u dhexeeya dallacaadaha A iyo B (F)AB) iyo sidoo kale xoogga Coulomb ee u dhexeeya dallacaadaha B iyo C (FBC) Xoogga Coulomb ee uu soo maray darajada B waa natiijada FAB iyo FBC.
Xoogga Coulomb ee u dhexeeya dallacaadaha A iyo B:
Kharash A wuxuu leeyahay calaamad togan halka kharash B uu leeyahay calaamad togan sidaa darteed FAB dhanka kharashka C.
Xoogga Coulomb ee u dhexeeya dallacaadaha B iyo C:
Dalacaadda B waa togan, Dalacaadda C waa togan, sidaa darteed F waa toganBC dhanka kharashka A.
Ciidanka Coulomb oo uu khibrad u leeyahay darajada B:
FB =FBC - FAB = 180 – 90 = 90 N
Baaxadda xoogga Coulomb ee uu soo maray kharashka B (F)B) waa 90 Newtons. Jihada FB la mid ah jihada FBC gaar ahaan dhanka A.
Jawaabta saxda ah waa B.
13. Cabbirka iyo jihada xoogga Coulomb ee ku jira mas'uuliyadda B waa... (k = 9 x 109 Nm2C-2, 1 μC = 10-6 C)
A. 2,5 k Q2 r-2 dhanka bidix
B. 2,5 k Q2 r-2 dhanka midig
C. 2k Q2 r-2 dhanka bidix
D. 2k Q2 r-2 dhanka midig
E. 1 k Q2 r-2 dhanka bidix
Dood
Waa la garanayaa :
Dalac A (q)A) = +Q
Dalac B (q)B) = -2Q
Dalacaadda C (q)C) = -Q
Masaafada u dhaxaysa khidmadaha A iyo B (r)AB) = r
Masaafada u dhaxaysa khidmadaha B iyo C (r)BC) = 2r
k = 9 x 109 Nm2C-2
La weydiiyay : baaxadda iyo jihada ciidanka Coulomb ee ku jira kharashka B
Jawab :
Xoogga Coulomb ee u dhexeeya dallacaadda A iyo dallacaadda B:
Dalacaadda A waa togan, dalacaadda B-na waa taban, marka jihada waa FAB dhanka kharashka A
Xoogga Coulomb ee u dhexeeya dallacaadda B iyo dallacaadda C:
Dalacaadda B waa taban, Dalacaadda C-na waa taban, sidaa darteed jihada F waa tabanBC dhanka kharashka A
Xoogga ka dhashay ee ku shaqeynaya dacwadda B:
F = FAB +FBC = 2 k Q2/r2 + 0,5 k Q2/r2 = 2,5 k Q2/r2 = 2,5 k Q2 r-2
Jihada ciidanka Coulomb waxay u socotaa dhanka A ama dhanka bidix.
Jawaabta saxda ah waa A.
Goobta Korontada
14. Fiiri sawirka laba dalac oo hoose! Xaggee buu ku yaal barta P si xoogga goobta korantada ee barta P uu ula mid noqdo eber? (k = 9 x 109 Nm2C-2, 1 μC = 10-6 C)
A. midig bartamaha Q1 iyo Q2
B. 6 cm dhanka midig ee Q2
C. 6 cm dhanka bidix ee Q1
D. 2 cm dhanka midig ee Q2
E. 2 cm dhanka bidix ee Q1
Dood
Si aad u xisaabiso xoogga goobta korantada ee barta P, u qaado in ay jirto dallac tijaabo oo togan oo ku yaal barta P. Q1 togan iyo Q2 taban, sidaa darteed dhibicda P waa inay ku taal dhinaca midig ee Q2 ama dhanka bidix ee Q1. Haddii dhibicda P ay bidixda ka tahay Q1; goobta korantada ee ay soo saarto dhibicda Q1 barta P jihada waxay u socotaa bidix (ka fog Q)1) iyo goobta korontada ee Q-da laga soo saaray2 barta P jihada waxay u socotaa dhanka midig (dhinaca Q)1Maadaama jihada garoonka korontadu ay ka soo horjeeddo, labadooduba way isdhaafsadaan si xoogga goobta korontadu ee barta P uu noqdo eber.
Waa la garanayaa :
Q1 = +9 μC = +9 x 10-6 C
Q2 = -4 μC = -4 x 10-6 C
k = 9 x 109 Nm2C-2
Masaafada u dhaxaysa dallacaadda 1 iyo dallacaadda 2 = 3 cm
Masaafada u dhaxaysa Q1 iyo dhibic P (r)1P) = a
Masaafada u dhaxaysa Q2 iyo dhibic P (r)2P) = 3 + a
La weydiiyay : Xaggee bay ku taal barta P si xoogga goobta korantada ee barta P uu ula mid noqdo eber?
Jawab :
Barta P waxay bidixda ka xigtaa Q1.
Goobta korantada ee ay soo saarto Q1 barta P :
Kharashka baaritaanka togan iyo Q1 togan si jihada goobta korontada ay u socoto dhanka bidix.
Goobta korantada ee ay soo saarto Q2 barta P :
Kharashka baaritaanka togan iyo Q2 taban si jihada garoonka korontada ay u noqoto midig.
Goobta korantada ee natiijada ka dhalatay barta A :
E1 iyo E2 jihada ka soo horjeeda.
E1 - E2 = 0
E1 =E2
Isticmaal qaacidada ABC si aad u go'aamiso qiimaha a.
a = -1,25, b = -13,5, c = -20,25
Ma noqon karto mid taban.
Masaafada u dhaxaysa Q2 iyo dhibic P (r)2P) = 3 + a = 3 – 1,8 = 1,2 cm.
Barta P waxay ku taal masaafo dhan 1,2 cm dhanka midig ee Q.2.
15. Fiiri sawirka soo socda! Lacag q3 meel fog oo 5 cm u jirta q2, ka dibna xoogga goobta korantada ee dallacaadda q3 waa… (1 µC = 10-6 C)
A. 4,6 x 107 NC-1
B. 3,6 x 107 NC-1
C. 1,6 x 107 NC-1
D. 1,4 x 107 NC-1
E. 1,3 x 107 NC-1
Dood
Kharashka q3 meel fog oo 5 cm u jirta q2, taasoo la micno ah in aan loo jeedin dhanka bidix ee q2 laakiin dhinaca midig q2Haddii dhinaca bidix uu ku yaal q2 markaas goobta korontada ee ka dhalata waa eber. Tani waa sababta oo ah masaafada u dhaxaysa dallacaadaha q3 oo leh kharash q1 iyo q2 waa 5 cm, baaxadda kharashkana waa q1 la mid ah qiimaha q2.
Sababtoo ah kharashka q3 togan ka dibna jihada garoonka korontada kharashka ku baxaya q3 dhanka qiimaha taban q2 (E2) iyo ka fog qiimaha togan q1 (E1) Goobta korantada ee ka dhalata waa wadarta awoodaha goobta korantada E1 iyo E2.
Waa la garanayaa :
Kharashka q1 = 5 µC = 5 x 10-6 Coulomb
Kharashka q2 = 5 µC = -5 x 10-6 Coulomb
Masaafada u dhaxaysa khidmadaha q1 iyo kharashka q3 (r1) = 15 cm = 0,15 m = 15 x 10-2 meter
Masaafada u dhaxaysa khidmadaha q2 iyo kharashka q3 (r2) = 5 cm = 0,05 m = 5 x 10-2 meter
k = 9 x 109 N m2 C-2
La weydiiyay : Xoogga garoonka korontada kharashka ku baxaya q3
Jawab :
Xoogga goobta korontada 1
E1 = kq1 /r12
E1 = (9 x 109)(5 x 10-6) / (15 x 10-2)2
E1 = (45 x 103) / (225 x 10-4)
E1 = 0,2x107 N / C
Xoogga goobta korontada 2
E2 = kq2 /r22
E2 = (9 x 109)(5 x 10-6) / (5 x 10-2)2
E2 = (45 x 103) / (25 x 10-4)
E2 = 1,8x107 N / C
Xoogga garoonka korantada ee ka dhashay
Xoogga goobta korantada ee ka dhalatay qiimaha q3 waa:
E = E2 - E1 = (1,8 x 107)– (0,2 x 107) = 1,6 x 107 N / C
Jihada goobta korontada waxay u socotaa dhanka bidix ama dhanka E.2.
Jawaabta saxda ah waa C.
16. Laba dallacaad koronto ayaa loo kala saaray sida ku cad sawirka. Xoogga goobta ee barta P waa… (k = 9 x 109 N m2 C-2)
A. 9,0 x 109 NC-1
B. 4,5 x 109 NC-1
C. 3,6 x 109 NC-1
D. 5,4 x 109 NC-1
E. 4,5 x 109 NC-1
Dood
Waa la garanayaa :
Kharashka qA = +2,5 C
Kharashka qB = -2 C
Masaafada u dhaxaysa khidmadaha qA iyo dhibic P (r)A) = 5 m
Masaafada u dhaxaysa khidmadaha qB iyo dhibic P (r)B) = 2 m
k = 9 x 109 N m2 C-2
La weydiiyay Xoogga goobta korontada ee barta P
Jawab :
Xoogga goobta korontada A
EA = kqA /rA2
EA = (9 x 109)(2,5) / (5)2
EA = (22,5 x 109) / 25
EA = 0,9x109 N / C
Xoogga goobta korontada B
EB = kqB /rB2
EB = (9 x 109)(2) / (2)2
EB = (18 x 109) / 4
EB = 4,5x109 N / C
Xoogga garoonka korantada ee ka dhashay
Xoogga goobta korantada ee ka dhalatay barta P waa:
E = EB - EA = (4,5 – 0,9) x 109 = 3,6x109 N / C
Jihada goobta korontada waxay u socotaa dhanka bidix ama dhanka E.B.
Jawaabta saxda ah waa C.
17. Laba dallac oo koronto ah midkiiba wuxuu leeyahay dallac Q.1 = -40 µC iyo Q2 = +5 µC waxay ku taal booska sida ku cad sawirka (k = 9 x 109 Nm2.C-2 iyo 1 µC = 10-6 C), xoogga goobta korantada ee barta P waa…
A. 2,25 x 106 NC-1
B. 2,45 x 106 NC-1
C. 5,25 x 106 NC-1
D. 6,75 x 106 NC-1
E. 9,00 x 106 NC-1
Dood
Waa la garanayaa :
Kharashka q1 = -40 µC = -40 x 10-6 C
Kharashka q2 = +5 µC = +5 x 10-6 C
Masaafada u dhaxaysa khidmadaha q1 iyo dhibic P (r)1) = 40 cm = 0,4 m = 4 x 10-1 m
Masaafada u dhaxaysa khidmadaha q2 iyo dhibic P (r)2) = 10 cm = 0,1 = 1 x 10-1 m
k = 9 x 109 N m2 C-2
La weydiiyay Xoogga goobta korontada ee barta P
Jawab :
Xoogga goobta korontada 1
E1 = kq1 /r12
E1 = (9 x 109)(40 x 10-6) / (4 x 10-1)2
E1 = (360 x 103) / (16 x 10-2)
E1 = 22,5x105 N / C
Xoogga goobta korontada 2
E2 = kq2 /r22
E2 = (9 x 109)(5 x 10-6) / (1 x 10-1)2
E2 = (45 x 103) / 1 x 10-2
E2 = 45x105 N / C
Xoogga garoonka korantada ee ka dhashay
Xoogga goobta korantada ee ka dhalatay barta P waa:
E = E2 - E1 = (45 – 22,5) x 105 = 22,5x105 N / C
E = 2,25 x 106 N / C
Jihada goobta korontadu waxay u jirtaa dhanka midig ama dhanka E.2.
Jawaabta saxda ah waa A.
18. Laba dallac koronto ayaa si gaar ah loo dhigay sida ku cad jaantuska. Dallaca A waa 8 µC, xoogga soo jiidashada leh ee labada dallacna waa 45 N. Haddii dallaca A loo wareejiyo dhanka midig 1 cm iyo k = 9.109 Nm2.C-2, markaa awoodda soo jiidashada leh ee ka shaqeysa labada dacwadood waa...
A. 45 N
B. 60 N
C. 80 N
D. 90 N
E. 120 N
Dood
Waa la garanayaa :
Korontada ku dallacda A (q)A) = 8 µC = 8 x 10-6 Coulomb
Xoogga korantada ee u dhexeeya labada dallacaad (F) = 45 Newton
Masaafada u dhaxaysa labada dallacaad (r)AB) = 4 cm = 0,04 mitir = 4 x 10-2 meter
Joogto ah (k) = 9 x 109 Nm2.C-2
La weydiiyay : Xoogga korontada inta u dhaxaysa labada dacwadood haddii dallacaadda A loo wareejiyo dhanka midig 1 cm ama 0,01 mitir
Jawab :
Marka hore xisaabi kharashka korontada ee B, ka dibna xisaabi xoogga korontada ee u dhexeeya labada dallac ee korontada, haddii dallacyada korontada ee A loo wareejiyo dhanka midig 1 cm.
Koronto laga helo B :
Qaacidda sharciga ee Coulomb :
F = k (q)A)(qB) / r2
F r2 = k (q)A)(qB)
qB = F r2 / k (q)A)
Koronto laga helo B :
qB = (45)(4 x 10-2)2 / (9 x 109)(8 x 10-6)
qB = (45)(16 x 10-4) / 72 x 103
qB = (720 x 10-4) / (72 x 103)
qB = 10x10-7 Coulomb
Xoogga korontada ee u dhexeeya dallacaadaha korantada A iyo B :
Haddii dallacaadda A loo wareejiyo dhanka midig 1 cm, masaafada u dhaxaysa labada dallac waxay noqonaysaa 3 cm = 0,03 mitir = 3 x 10-2 meter
F = k (q)A)(qB) / r2
F = (9 x 109)(8 x 10-6)(10 x 10-7) / (3 x 10-2)2
F = (9 x 109)(80 x 10-13) / (9 x 10-4)
F = (1 x 109)(80 x 10-13) / (1 x 10-4)
F = (80 x 10-4) / (1 x 10-4)
F = 80 Newton
Jawaabta saxda ah waa C.
19. Laba dallac koronto oo P iyo Q ah oo 10 cm u jira waxay la kulmaan xoog soo jiidasho leh oo ah 8 N. Haddii dallac Q loo wareejiyo 5 cm dhanka dallacsiinta P (1 µC = 10-6 C iyo k = 9 x 109 Nm2.C-2), markaa xoogga korontada ee dhaca waa...
A. 8 N
B. 16 N
C. 32 N
D. 40 N
E. 56 N
Dood
Waa la garanayaa :
Masaafada u dhaxaysa khidmadaha P iyo Q (r)PQ) = 10 cm = 0,1 m = 1 x 10-1 m
Xoogga korontada ee u dhexeeya dallacaadaha P iyo Q (F) = 8 N
Kharash koronto Q (q)Q) = 40 µC = 40 x 10-6 C
Joogto ah (k) = 9 x 109 Nm2.C-2
La weydiiyay Awoodda korantada ee u dhaxaysa dallacaadaha P iyo Q haddii dallacaadda Q loo wareejiyo 5 cm dhanka dallacaadda P
Jawab :
Marka hore xisaabi lacagta korontada ee P, ka dibna xisaabi xoogga korontada ee u dhexeeya labada dallac ee korontada, haddii lacagta korontada ee Q loo wareejiyo 5 cm dhanka dallacda P.
Dalacaadda korontada P :
qP = F r2 / k (q)Q)
qP = (8)(1 x 10-1)2 / (9 x 109)(40 x 10-6)
qP = (8)(1 x 10-2) / 360 x 103
qP = (8 x 10-2) / (36 x 104)
qP = (1 x 10-2) / (4,5 x 104)
qP = (1/4,5) x 10-6 Coulomb
Awoodda korantada ee u dhaxaysa dallacaadaha korantada P iyo Q :
Haddii dallacaadda Q loo wareejiyo bidixda 5 cm, masaafada u dhaxaysa labada dallac waxay noqonaysaa 5 cm = 0,05 mitir = 5 x 10-2 meter
F = k (q)P)(qQ) / r2
F = (9 x 109)( (1/4,5) x 10-6)(40 x 10-6) / (5 x 10-2)2
F = (2 x 103)(40 x 10-6) / (25 x 10-4)
F = (80 x 10-3) / (25 x 10-4)
F = 3,2 x 101
F = 32 Newton
Jawaabta saxda ah waa C.
20. Fiiri sawirka soo socda ee dallacaadda korontada. Xoogga korontada ee uu la kulmay dallacaadda q waaB waa 8 N (1 µC = 10-6 C) iyo (k = 9.109 Nm2.C-2) Haddii kharashka qB loo wareejiyay 4 cm laga bilaabo A, ka dibna xoogga korontada ee la arkay waa qB hadda waa…
A. 2 N
B. 4 N
C. 6 N
D. 8 N
E. 10 N
Dood
Waa la garanayaa :
Masaafada u dhaxaysa khidmadaha A iyo B (r)AB) = 2 cm = 0,02 m = 2 x 10-2 m
Xoogga korontada ee u dhexeeya dallacaadaha A iyo B (F) = 8 N
Dalacaadda korontada A (q)A) = 2 µC = 2 x 10-6 C
Joogto ah (k) = 9 x 109 Nm2.C-2
La weydiiyay : Awoodda korantada ee u dhaxaysa dallacaadaha A iyo B haddii masaafada u dhaxaysa labada dallacaadood ay tahay 4 cm
Jawab :
Marka hore xisaabi lacagta korantada B, ka dibna xisaabi xoogga korontada ee u dhexeeya labada dallac ee korontada haddii masaafada u dhaxaysa labada dallac ee korontada ay tahay 4 cm = 0,04 mitir = 4 x 10-2 mitir
Lacag koronto B :
qB = F r2 / k (q)A)
qB = (8)(2 x 10-2)2 / (9 x 109)(2 x 10-6)
qB = (8)(4 x 10-4)/ (18 x 103)
qB = (32 x 10-4) / (18 x 103)
qB = (32/18) x 10-7
qB = (16/9) x 10-7 Coulomb
Xoogga korontada ee u dhexeeya dallacaadaha A iyo B :
F = k (q)A)(qB) / r2
F = (9 x 109)(2 x 10-6)( (16/9) x 10-7) / (4 x 10-2)2
F = (18 x 103)( (16/9) x 10-7) / (16 x 10-4)
F = (2 x 103)(16 x 10-7) / (16 x 10-4)
F = (2 x 103)(1 x 10-7) / (1 x 10-4)
F = (2 x 10-4) / (1 x 10-4)
F = 2 Newton
Jawaabta saxda ah waa A.
Isha su'aasha:
Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda
