Misalai 20 na tambayoyin wutar lantarki mai tsauri
Ƙarfin Wutar Lantarki
1. Maki A yana cikin filin lantarki . Ƙarfin filin lantarki a wurin A shine 0,5 NC -1 . Idan aka sanya abu mai caji na 0,25 C a wurin A, to ƙarfin Coulomb na...
A. 0,125 N
B. 0,25 N
C. 0,35 N
D. 0,40 N
E. 0,70 N
Tattaunawa
An san cewa:
Ƙarfin filin lantarki a wuri A = 0,5 NC -1
Cajin wutar lantarki a wurin A = 0,25 C
Tambaya: Ƙarfin Coulomb da ke aiki a kan wani abu da aka yi masa caji ta hanyar lantarki
Amsa:
Tsarin da ke nuna alaƙar da ke tsakanin ƙarfin lantarki (F), filin lantarki (E) da cajin lantarki (q) shine:
F = q E
F = (0,25 C)(0,5 NC -1 )
F = 0,125N
Amsar da ta dace ita ce A.
2. Caji biyu na 5 C da 4 C suna da nisan mita 3 tsakanin juna. Idan k = 9 × 10 9 Nm 2 C –2 , to girman ƙarfin Coulomb da aka samu a caji biyu shine…
A. 2 × 10 9 N
B. 60 × 10 9 N
C. 2 × 10 10 N
D. 6 × 10 10 N
E. 20 × 10 10 N
Tattaunawa
An san cewa:
Caji 1 (q 1 ) = 5 C
Caji 2 (q 2 ) = 4 C
Nisa tsakanin lodi 1 da 2 (r) = mita 3.
Madaidaitan Coulomb (k) = 9 × 10 9 Nm 2 C –2
Tambaya: Girman ƙarfin Coulomb (F)
Amsa:
Amsar da ta dace ita ce C.
3. An raba cajin wutar lantarki +q 1 = 10 μC; +q 2 = 20 μC; da q 3 kamar yadda aka nuna a cikin hoton da ke ƙasa. Don haka ƙarfin Coulomb da ke aiki akan cajin q 2 = sifili; to cajin q 3 shine…
A. +2,5 μC
B. –2,5 μC
C. +25 μC
D. –25 μC
E. +4 μC
Tattaunawa
An san cewa:
Caji 1 (q 1 ) = 10 μC = 10 x 10 -6 C
Caji 2 (q 2 ) = 20 μC = 20 x 10 -6 C
Tambaya: Menene cajin q 3 don haka ƙarfin Coulomb da ke aiki akan cajin q 2 yayi daidai da sifili (F 2 = 0).
Amsa:
Akwai ƙarfi biyu da ke aiki akan +q 2.
Ƙarfin farko shine ƙarfin da ke tsakanin caji +q 1 da caji +q 2 , wato F 12 , wanda aka nuna zuwa dama.
Domin ƙarfin lantarki da ke aiki akan q 2 ya zama sifili, dole ne a caji q 3 ta hanyar da ba ta dace ba . Saboda haka, ƙarfin na biyu shine ƙarfin jan hankali tsakanin caji + q 2 da -q 3 , wato F 23, wanda aka nuna a hagu. Waɗannan ƙarfin biyu da ke aiki akan q 2 suna da girma iri ɗaya amma alkibla akasin haka.
Ƙarfin da aka samu akan +q 2 daidai yake da sifili.
Amsar da ta dace ita ce B.
4. Maki A da B suna da wutar lantarki na −10 μC da +40 μC bi da bi. Da farko ana sanya caji biyu a tsakanin mita 0,5 don haka ƙarfin Coulomb F Newton ya taso. Idan an canza nisan da ke tsakanin A da B zuwa mita 1,5, to ƙarfin Coulomb da ke tasowa shine...
A. 1 / 9 F
B. 1 / 3 F
C. 3 / 2 F
D. 3 F
E. 9 F
Tattaunawa
Kwatanta tattaunawar tambaya mai lamba 9.
Ana canza nisan da ke tsakanin A da B zuwa mita 1,5 ko kuma sau 3 na nisan da aka saba.
Ƙarfin yana daidai da murabba'in nisa:
Ƙarfin Coulomb da ke tasowa shine 1/9 F.
Amsar da ta dace ita ce A.
5. An sanya tsarin da ke da caji kyauta guda 3 masu girman daidai gwargwado don a daidaita shi kamar yadda aka nuna a hoton. Idan aka canza Q3 zuwa 1/3 x kusa da Q2 , to rabon girman ƙarfin Coulomb F 2 : F 1 zai zama….
A. 1: 3
B. 2: 3
C. 3: 4
D. 9: 1
E. 9: 4
Tattaunawa
An sani cewa :
Nisa tsakanin q 1 da q 2 = x
Nisa tsakanin q 2 da q 3 = 2/3 x
An tambaya : F 2 : F 1 = …. ?
Amsa :
Tsarin dokar Coulomb:
Bayani: k = akai-akai, q 1 = caji 1, q 2 = caji 2, r = nisan da ke tsakanin caji 1 da caji 2
Kwatanta girman ƙarfin Coulomb
q 1 , q 2 da q 3 daidai suke don haka an cire su daga lissafin. k da x 2 suma girmansu iri ɗaya ne kuma suna gefen hagu da dama don haka an cire su daga lissafin.
Amsar da ta dace ita ce E.
6. Duba hoton da ke ƙasa. Cajin wutar lantarki guda uku q 1 , q, da q 2 suna cikin layi. Idan q = 5,0 μC da d = 30 cm, to girma da alkiblar ƙarfin wutar lantarki da ke aiki akan caji q sune… (k = 9 x 10 9 N m 2 C -2 )
A. 7,5 N zuwa q1
B. 7,5 N zuwa q 2
C. 15 N zuwa q 1
D. 22,5 N zuwa q 1
E. 22,5 N zuwa q 2
Tattaunawa
An san cewa:
Caji 1 (q 1 ) = 30 μC = 30 x 10 -6 C
Caji 2 (q 2 ) = 60 μC = 60 x 10 -6 C
Caji 3 (q) = 5 μC = 5 x 10 -6 C
Nisa tsakanin q 1 da q = d
Nisa tsakanin q 2 da q = 2d
d = 30 cm = mita 0,3
d 2 = (0,3) 2 = 0,09
Madaidaitan Coulomb (k) = 9 x 10 9 N m 2 C -2
Tambaya: Girma da alkiblar ƙarfin lantarki da ke aiki akan cajin lantarki
Amsa:
Akwai ƙarfi biyu da ke aiki akan q, wato F 1 yana gefen dama (q da q 1 suna da caji mai kyau don haka F 1 yana nesa da q da q 1 ) kuma F 2 yana gefen hagu (q da q 2 suna da caji mai kyau don haka F 2 yana nesa da q da q 2 ). Da farko ƙididdige F 1 da F 2.
Ƙarfin da aka samu:
Σ F = 15 – 7,5 = 7,5
Ƙarfin da aka samu shine Newtons 7,5. Alkiblarsa iri ɗaya ce da F 1 , wato zuwa dama zuwa q 2.
Amsar da ta dace ita ce B.
Filin Lantarki
7. Cajin maki q yana wurin P a cikin filin lantarki da aka samar ta hanyar caji (+) don haka yana fuskantar ƙarfin 0,05 N a cikin alkiblar zuwa ga cajin. Idan ƙarfin filin a wurin P shine 2 x 10 -2 NC -1 , to girma da nau'in cajin da ke haifar da filin sune…
A. 5,0 C, tabbatacce
B. 5,0 C, ko kuma korau
C. 3,0 C, tabbatacce
D. 2,5 C, ko kuma korau
E. 2,5 C, tabbatacce
Tattaunawa
An san cewa:
Ƙarfin wutar lantarki (F) = 0,05 N
Ƙarfin filin lantarki (E) = 2 x 10 –2 NC –1 = 0,02 NC –1
An tambaya: Girman da nau'in caji da ke ƙirƙirar filin
Amsa:
Ana ƙididdige cajin lantarki ta amfani da dabarar da ke nuna alaƙar da ke tsakanin ƙarfin lantarki (F), filin lantarki (E) da cajin lantarki (q):
F = q E
q = F / E = 0,05 N / 0,02 NC –1 = 2,5 Coulombs
Cajin q yana fuskantar ƙarfin lantarki a alkiblar zuwa ga cajin (+) wanda ke haifar da filin lantarki, don haka cajin q yana da alamar korau.
Amsar da ta dace ita ce D.
8. Nisa tsakanin caji biyu A da B shine m 4. Ma'ana C tana tsakanin caji biyu, m 1 daga A. Idan Q A = -300 μC, Q B = 600 μC. 1/4 π ε 0 = 9 × 10 9 N m 2 C -2 , to ƙarfin filin lantarki a wurin C saboda tasirin caji biyu shine…
A. 9 × 10 5 NC –1
B. 18 × 10 5 NC –1
C. 33 × 10 5 NC –1
D. 45 × 10 5 NC –1
E. 54 × 10 5 NC –1
Tattaunawa
An san cewa:
Nisa tsakanin lodin A da B (r AB ) = mita 4
Nisa tsakanin maki C da caji A (r AC ) = mita 1
Nisa tsakanin maki C da caji B (r BC ) = mita 3
Cajin A (qA ) = –300 μC = -300 x 10 -6 C = -3 x 10 -4 Coulomb
Cajin B (qB ) = 600 μC = 600 x 10 -6 C = 6 x 10 -4 Coulomb
Daidaitacce (k) = 9 × 10 9 N m 2 C –2
Tambaya: Ƙarfin filin lantarki a wuri C
Amsa:
Filin wutar lantarki da aka samar ta hanyar caji A a wurin C:
Cajin A yana da rashin tabbas don haka alkiblar filin lantarki tana zuwa ga caji A kuma tana nesa da caji B (a hagu).
Filin wutar lantarki da aka samar ta hanyar caji B a wurin C:
Caji B yana da kyau don haka alkiblar filin lantarki tana nesa da caji B kuma tana zuwa ga caji A (a hagu).
Sakamakon filin lantarki a wuri na A:
E A da E B suna kan hanya ɗaya don haka an haɗa su tare.
E = E A + E B
E = (27 x 10 5 ) + (6 x 10 5 )
E = 33 x 10 5 N/C
Alkiblar filin lantarki tana zuwa ga caji A kuma nesa da caji B (zuwa hagu).
Amsar da ta dace ita ce C.
9. Ƙwayar ƙura mai nauyin milligram 1 za ta iya shawagi a sararin samaniya saboda filin lantarki da ke riƙe ƙurar. Idan cajin ƙurar ya kai 0,5 μC kuma saurin gudu saboda nauyi ya kai 10 m/s2 , a tantance girman filin lantarki da zai iya riƙe ƙurar.
A. 5 N/C
B. 10 Ba a San Komai Ba
C. 20 N/C
D. 25 Ba a San Komai Ba
E. 40 N/C
Tattaunawa
An san cewa:
Nauyin ƙura (m) = milligram 1 = 1 x 10 -6 kg
Cajin ƙura (q) = 0,5 μC = 0,5 x 10 -6 C
Saurin gudu saboda nauyi (g) = 10 m/s 2
Tambaya: Ƙarfin filin lantarki wanda ke riƙe ƙurar
Amsa:
Tsarin Nauyi:
w = mg
Bayani: w = nauyin ƙura, m = nauyin ƙura, g = hanzari saboda nauyi
Ana ƙididdige ƙarfin nauyi da ke aiki akan ƙura ko nauyin ƙura ta amfani da dabarar nauyi:
w = mg = (1 x 10 -6 kg)(10 m/s 2 ) = 10 x 10 -6 kg m/s 2 = 10 x 10 -6 Newtons
Tsarin ƙarfin filin lantarki:
E = F/q
Bayani: E = ƙarfin filin lantarki, F = ƙarfin lantarki, q = cajin lantarki
Ƙura tana shawagi a cikin iska, don haka ƙarfin da ke aiki a kan ƙurar dole ne ya zama sifili. Ana karkatar da ƙarfin ƙurar zuwa ƙasa, don haka dole ne a karkatar da ƙarfin wutar lantarki zuwa sama, kuma girman nauyin ƙurar dole ne ya zama daidai da girman ƙarfin wutar lantarki, don haka ƙarfin da ke fitowa a kan ƙurar ba shi da sifili. Don haka, ana iya maye gurbin F a cikin ƙarfin filin lantarki da w a cikin dabarar nauyi.
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
Amsar da ta dace ita ce C.
10. Caji biyu, q 1 = 32 μC da q 2 = -214 μC, an raba su da nisan x daga juna kamar yadda aka nuna a cikin hoton da ke sama. Idan a wurin p, wanda yake 10 cm daga q 2, ƙarfin filin lantarki da aka samu sakamakonsa sifili ne. To girman x shine….
A. 20 cm
B. 30 cm
C. 40 cm
D. 50 cm
E. 60 cm
Tattaunawa
An san cewa:
Caji 1 (Q1 ) = 32 μC
Caji na 2 (Q2 ) = -214 μC
Nisa daga ma'aunin p daga q 1 = x + 10 cm
Nisa daga ma'aunin p daga q 2 = 10 cm
An tambaya: x
Amsa:
E 1 shine filin lantarki da aka samar ta hanyar caji Q 1. Alkiblar filin lantarki tana nesa da Q 1 saboda Q 1 yana da caji mai kyau. E 2 shine filin lantarki da aka samar ta hanyar caji Q 2. Alkiblar filin lantarki tana zuwa ga Q 2 saboda Q 2 yana da caji mara kyau.
A wurin p wanda yake da nisan santimita 10 daga Q2 , ƙarfin filin lantarki da aka samu sakamakonsa sifili ne.
Yi amfani da tsarin ABC:
11. Ma'aunin da ke da caji q yana a wurin P a cikin filin lantarki da aka ƙirƙira ta hanyar caji (+), don haka yana fuskantar ƙarfin 0,05 N. Idan girman cajin shine +5 × l0 –6 Coulomb, to girman filin lantarki a wurin P shine…
A. 2,5 × 10 3 NC –1
B. 3.0 × 10 3 NC –1
C. 4,5 × l0 3 NC –1
D. 8,0 × 10 3 NC –1
E. 10 4 NC –1
Tattaunawa
An san cewa:
Ƙarfin wutar lantarki (F) = 0,05 Newton
Cajin wutar lantarki (Q) = +5 × l0 –6 Coulomb = 0,000005
Tambaya: Menene girman filin lantarki a wurin P?
Amsa:
Tsarin da ke nuna alaƙar da ke tsakanin filin lantarki, ƙarfin lantarki da cajin lantarki:
E = F / Q
E = 0,05 Newtons / 0,000005 Coulombs
E = 5 Newtons / 0,0005 Coulombs
E = 10.000 Newton/Coulomb
E = 10 4 Babu/C
E = 10 4 NC -1
Amsar da ta dace ita ce E.
12. An shirya caji uku kamar yadda aka nuna a cikin hoton da ke ƙasa. Ƙarfin Coulomb da aka samu ta hanyar cajin B shine .... (k = 9 x 10 9 Nm 2 C −2 , 1 μC = 10 −6 C)
A. 09 x 101 Cajin N zuwa C
B. 09 x 101 N don cajin A
C. 18 x 101 Cajin N zuwa C
D. 18 x 101 N don cajin A
E. 36 x 101 Cajin N zuwa C
Tattaunawa
An sani :
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 = 2 x10-5 Coulomb
rAB = mita 0,1 = 10-1 mita
rBC = mita 0,1 = 10-1 mita
ku = 9 x 109 Nm2C-2
An tambaya : Ƙarfin Coulomb wanda aka samu ta hanyar caji B
Jawab :
Akwai ƙarfin Coulomb guda biyu ko ƙarfin lantarki waɗanda ke aiki akan cajin B, wato ƙarfin Coulomb tsakanin caji A da B (F AB ) da ƙarfin Coulomb tsakanin caji B da C (F BC ). Ƙarfin Coulomb da cajin B ke fuskanta shine sakamakon F AB da F BC.
Ƙarfin Coulomb tsakanin caji A da B:
Caji A yana da alama mai kyau, kuma Caji B yana da alama mai kyau, don haka FAB zuwa ga cajin C.
Ƙarfin Coulomb tsakanin caji na B da C:
Caji B yana da kyau kuma Caji C yana da kyau, don haka F yana da kyauBC zuwa ga caji A.
Ƙarfin Coulomb wanda aka samu ta hanyar caji B:
FB = FBC - FAB = 180 – 90 = 90 N
Girman ƙarfin Coulomb da chaji B (F) ke fuskantaB) shine 90 Newtons. Alkiblar FB kamar yadda aka tsara a matsayin FBC wato ga caji A.
Amsar da ta dace ita ce B.
13. Girma da alkiblar ƙarfin Coulomb akan caji B shine... (k = 9 x 10 9 Nm 2 C −2 , 1 μC = 10 −6 C)
A. 2,5 k Q2 r-2 Zuwa hannun hagu
B. 2,5 k Q2 r-2 Zuwa hannun dama
C. 2 k Q2 r-2 Zuwa hannun hagu
D. 2 k Q2 r-2 Zuwa hannun dama
E. 1 k Q2 r-2 Zuwa hannun hagu
Tattaunawa
An sani :
Cajin A (q)A) = +Q
Caji B (q)B) = -2Q
Cajin C (q)C) = -Q
Nisa tsakanin caji A da B (r)AB) = r
Nisa tsakanin caji B da C (r)BC) = 2r
ku = 9 x 109 Nm2C-2
An tambaya girma da alkiblar rundunar Coulomb akan caji B
Jawab :
Ƙarfin Coulomb tsakanin caji A da caji B:
Cajin A yana da kyau kuma Cajin B yana da kyau, don haka alkiblar ita ce FAB zuwa caji A
Ƙarfin Coulomb tsakanin caji B da caji C:
Cajin B yana da ma'ana ta rashin tabbas, kuma Cajin C yana da ma'ana ta rashin tabbas, don haka alkiblar F tana da ma'ana ta rashin tabbas.BC zuwa caji A
Ƙarfin da ya haifar da aiki a kan caji na B:
F = FAB +FBC = 2 k Q2/r2 + 0,5 k Q2/r2 = 2,5 k Q2/r2 = 2,5 k Q2 r-2
Alkiblar ƙarfin Coulomb tana zuwa ga caji A ko hagu.
Amsar da ta dace ita ce A.
Filin Lantarki
14. Kalli hoton cajin maki biyu a ƙasa! Ina aka samo wurin P don ƙarfin filin lantarki a wurin P ya zama daidai da sifili? (k = 9 x 109 Nm2C-2, 1 μC = 10-6 C)
A. daidai a tsakiyar Q1 da Q2
B. 6 cm zuwa dama na Q2
C. 6 cm zuwa hagu na Q1
D. 2 cm zuwa dama na Q2
E. 2 cm zuwa hagu na Q1
Tattaunawa
Don ƙididdige ƙarfin filin lantarki a wurin P, ɗauka cewa akwai cajin gwaji mai kyau a wurin P. Q1 tabbatacce da Q2 korau, don haka dole ne aya ta P ta kasance a gefen dama na Q2 ko kuma a gefen hagu na Q1. Idan batu na P yana gefen hagu na Q1; filin lantarki da aka samar ta hanyar maki Q1 a wurin P alkiblar tana hagu (daga Q)1) da kuma filin lantarki da Q ya samar2 a wurin P alkiblar tana zuwa dama (zuwa Q)1Saboda alkiblar filin lantarki a akasin haka take, su biyun sun soke junansu don ƙarfin filin lantarki a wurin P ya zama sifili.
An sani :
Q1 = +9 μC = +9 x 10-6 C
Q2 = -4 μC = -4 x 10-6 C
ku = 9 x 109 Nm2C-2
Nisa tsakanin caji 1 da caji 2 = 3 cm
Nisa tsakanin Q1 da kuma maki P (r)1P) = a
Nisa tsakanin Q2 da kuma maki P (r)2P) = 3 + a
An tambaya : Ina aka samo wurin P don ƙarfin filin lantarki a wurin P ya zama daidai da sifili?
Jawab :
Maki na P yana gefen hagu na Q1.
Filin wutar lantarki da Q ya samar1 a wurin P :
Cajin gwaji mai kyau da Q1 tabbatacce don haka alkiblar filin lantarki tana hagu.
Filin wutar lantarki da Q ya samar2 a wurin P :
Cajin gwaji mai kyau da Q2 korau don haka alkiblar filin lantarki tana zuwa dama.
Sakamakon filin lantarki a wuri na A :
E1 da kuma E2 akasin alkibla.
E1 - E2 = 0
E1 =E2
Yi amfani da dabarar ABC don tantance ƙimar a.
a = -1,25, b = -13,5, c = -20,25
Ba zai iya zama mummunan ba.
Nisa tsakanin Q2 da kuma maki P (r)2P) = 3 + a = 3 – 1,8 = 1,2 cm.
Ma'aunin P yana da nisan 1,2 cm zuwa dama na Q.2.
15. Kalli hoton da ke ƙasa! Caji q3 an sanya shi a nesa na 5 cm daga q2, sannan ƙarfin filin lantarki a caji q3 shine… (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
Tattaunawa
Cajin q3 an sanya shi a nesa na 5 cm daga q2, ma'ana ba a gefen hagu na q ba2 amma a gefen dama q2Idan a gefen hagu q2 to filin lantarki da ya haifar da hakan sifili ne. Wannan saboda nisan da ke tsakanin caji q3 tare da caji q1 kuma q2 shine 5 cm kuma girman cajin shine q1 daidai da caji q2.
Saboda cajin q3 yana da kyau, alkiblar filin lantarki da ke caji q3 tana zuwa ga cajin negative q2 ( E2 ) kuma tana nesa da cajin positive q1 ( E1 ) . Sakamakon filin lantarki shine jimlar ƙarfin filin lantarki E1 da E2.
An sani :
Cajin q1 = 5 µC = 5 x 10-6 Coulomb
Cajin q2 = 5 µC = -5 x 10-6 Coulomb
Nisa tsakanin caji q1 da cajin q3 (r1) = 15 cm = 0,15 m = 15 x 10-2 mita
Nisa tsakanin caji q2 da cajin q3 (r2) = 5 cm = 0,05 m = 5 x 10-2 mita
ku = 9 x 109 A'a m2 C-2
An tambaya : Ƙarfin filin lantarki a caji q3
Jawab :
Ƙarfin filin lantarki 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 = 0,2 x107 N / C
Ƙarfin filin lantarki 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 = 1,8 x107 N / C
Ƙarfin filin lantarki mai zuwa
Ƙarfin filin lantarki mai sakamako a caji q3 shine:
E = E2 - E1 = (1,8 x 107)– (0,2 x 107= 1,6 x 107 N / C
Alkiblar filin lantarki tana hagu ko kuma a alkiblar E2.
Amsar da ta dace ita ce C.
16. An raba caji biyu na lantarki kamar yadda aka nuna a cikin hoton. Ƙarfin filin a wurin P shine… (k = 9 x 109 A'a 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
Tattaunawa
An sani :
Cajin qA = +2,5 C
Cajin qB = -2 C
Nisa tsakanin caji qA da kuma maki P (r)A= 5m
Nisa tsakanin caji qB da kuma maki P (r)B= 2m
ku = 9 x 109 A'a m2 C-2
An tambaya : Ƙarfin filin lantarki a wurin P
Jawab :
Ƙarfin filin lantarki A
EA = kqA /rA2
EA = (9 x 109)(2,5) / (5)2
EA = (22,5 x 10925
EA = 0,9 x109 N / C
Ƙarfin filin lantarki B
EB = kqB /rB2
EB = (9 x 109)(2) / (2)2
EB = (18 x 1094
EB = 4,5 x109 N / C
Ƙarfin filin lantarki mai zuwa
Ƙarfin filin lantarki da aka samu a wurin P shine:
E = EB - EA = (4,5 – 0,9) x 109 = 3,6 x109 N / C
Alkiblar filin lantarki tana hagu ko kuma a alkiblar EB.
Amsar da ta dace ita ce C.
17. Cajin lantarki guda biyu kowannensu yana da caji ɗaya.1 = -40 µC da Q2 = +5 µC yana nan a wurin kamar yadda aka nuna a cikin hoton (k = 9 x 109 Nm2.C-2 kuma 1 µC = 10-6 C), ƙarfin filin lantarki a wurin P shine…
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
Tattaunawa
An sani :
Cajin q1 = -40 µC = -40 x 10-6 C
Cajin q2 = +5 µC = +5 x 10-6 C
Nisa tsakanin caji q1 da kuma maki P (r)1) = 40 cm = 0,4 m = 4 x 10-1 m
Nisa tsakanin caji q2 da kuma maki P (r)2) = 10 cm = 0,1 = 1 x 10-1 m
ku = 9 x 109 A'a m2 C-2
An tambaya : Ƙarfin filin lantarki a wurin P
Jawab :
Ƙarfin filin lantarki 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 = 22,5 x105 N / C
Ƙarfin filin lantarki 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 = 45 x105 N / C
Ƙarfin filin lantarki mai zuwa
Ƙarfin filin lantarki da aka samu a wurin P shine:
E = E2 - E1 = (45 – 22,5) x 105 = 22,5 x105 N / C
E = 2,25 x 106 N / C
Alkiblar filin lantarki tana zuwa dama ko kuma a alkiblar E2.
Amsar da ta dace ita ce A.
18. An sanya caji biyu na lantarki daban-daban kamar yadda aka nuna a cikin hoton. Cajin da ke A shine 8 µC kuma ƙarfin jan hankali da ke aiki akan caji biyu shine 45 N. Idan aka canza caji A zuwa dama da 1 cm kuma k = 9.109 Nm2.C-2, to, ƙarfin jan hankali da ke aiki akan tuhume-tuhumen guda biyu shine...
A. 45 N
B. 60 N
C. 80 N
D. 90 N
E. 120 N
Tattaunawa
An sani :
Cajin wutar lantarki a A (q)A) = 8 µC = 8 x 10-6 Coulomb
Ƙarfin wutar lantarki tsakanin caji biyu (F) = 45 Newton
Nisa tsakanin caji biyu (r)AB) = 4 cm = mita 0,04 = 4 x 10-2 mita
Daidaitacce (k) = 9 x 109 Nm2.C-2
An tambaya : Ƙarfin wutar lantarki tsakanin caji biyu idan caji A ya koma dama da 1 cm ko mita 0,01
Jawab :
Da farko a ƙididdige cajin lantarki a B, sannan a ƙididdige ƙarfin lantarki tsakanin cajin lantarki guda biyu, idan an canza cajin lantarki a A zuwa dama da 1 cm.
Cajin wutar lantarki a B :
Tsarin dokar Coulomb :
F = k (q)A)(qB) / r2
F r2 = k (q)A)(qB)
qB = F r2 / k (q)A)
Cajin wutar lantarki a 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 = 10 x10-7 Coulomb
Ƙarfin wutar lantarki tsakanin cajin lantarki A da B :
Idan aka canza cajin da ke A zuwa dama da 1 cm, nisan da ke tsakanin cajin biyu zai zama 3 cm = mita 0,03 = 3 x 10-2 mita
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
Amsar da ta dace ita ce C.
19. Cajin lantarki guda biyu P da Q waɗanda suke da nisan santimita 10 suna fuskantar ƙarfin jan hankali na 8 N. Idan aka canza cajin Q zuwa caji P (1 µC = 10)-6 C da k = 9 x 109 Nm2.C-2), to ƙarfin wutar lantarki da ke faruwa shine...
A. 8 N
B. 16 N
C. 32 N
D. 40 N
E. 56 N
Tattaunawa
An sani :
Nisa tsakanin caji P da Q (r)PQ) = 10 cm = 0,1 m = 1 x 10-1 m
Ƙarfin wutar lantarki tsakanin caji P da Q (F) = 8 N
Cajin wutar lantarki Q (q)Q) = 40 µC = 40 x 10-6 C
Daidaitacce (k) = 9 x 109 Nm2.C-2
An tambaya : Ƙarfin wutar lantarki tsakanin caji P da Q idan aka canza caji Q zuwa caji P 5 cm
Jawab :
Da farko a lissafta cajin lantarki na P, sannan a lissafta ƙarfin lantarki tsakanin cajin lantarki guda biyu, idan an canza cajin lantarki na Q zuwa ga cajin P zuwa ga cajin.
Cajin wutar lantarki 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
Ƙarfin wutar lantarki tsakanin cajin lantarki P da Q :
Idan aka canza cajin da ke Q zuwa hagu da 5 cm, nisan da ke tsakanin cajin biyu zai zama 5 cm = mita 0,05 = 5 x 10-2 mita
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
Amsar da ta dace ita ce C.
20. Kalli hoton da ke ƙasa na cajin lantarki. Ƙarfin wutar lantarki da cajin q ke fuskanta shineB shine 8 N (1 µC = 10-6 C) da (k = 9.109 Nm2.C-2) Idan cajin qB An canza shi zuwa 4 cm daga A, to ƙarfin wutar lantarki da aka fuskanta shine qB yanzu shine…
A. 2 N
B. 4 N
C. 6 N
D. 8 N
E. 10 N
Tattaunawa
An sani :
Nisa tsakanin caji A da B (r)AB) = 2 cm = 0,02 m = 2 x 10-2 m
Ƙarfin wutar lantarki tsakanin caji A da B (F) = 8 N
Cajin wutar lantarki A (q)A) = 2 µC = 2 x 10-6 C
Daidaitacce (k) = 9 x 109 Nm2.C-2
An tambaya : Ƙarfin wutar lantarki tsakanin caji A da B idan nisan da ke tsakanin caji biyu ya kai cm 4
Jawab :
Da farko a ƙididdige cajin lantarki na B, bayan haka a ƙididdige ƙarfin lantarki tsakanin cajin lantarki guda biyu idan nisan da ke tsakanin cajin lantarki guda biyu shine 4 cm = mita 0,04 = 4 x 10-2 Mita.
Cajin wutar lantarki 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
Ƙarfin wutar lantarki tsakanin caji A da 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
Amsar da ta dace ita ce A.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
