Misalai 10 na Dokar Coulomb
1. An shirya caji guda biyu kamar yadda aka nuna a hoton da ke ƙasa. Cajin da ke A shine +8 micro Coulombs kuma cajin da ke B shine -5 micro Coulombs. Girman ƙarfin Coulomb da ke aiki akan caji biyu shine… (k = 9 x 10 9 Nm 2 C −2 , micro Coulombs 1 = 10 −6 C)
Tattaunawa
An sani :
An tambaya : Girman ƙarfin wutar lantarki da ke aiki akan caji biyu
Jawab :
Tsarin dokar Coulomb :
Girman ƙarfin wutar lantarki da ke aiki akan caji biyu:
2. Cajin wutar lantarki P = +10 micro Coulombs da cajin wutar lantarki Q = +20 micro Coulombs an raba su kamar yadda aka nuna a cikin hoton. Girman ƙarfin Coulomb da ke aiki akan duka caji shine…
Tattaunawa
An sani :
An tambaya Girman rundunar Coulomb da ke aiki a kan dukkan tuhume-tuhumen
Jawab :
Girman ƙarfin Coulomb ko ƙarfin lantarki da ke aiki akan duka caji shine 125 Newtons.
3. An sanya cajin lantarki guda uku daban-daban kamar yadda aka nuna a hoton! Cajin A = -5 micro Coulombs, cajin B = +10 micro Coulombs, da cajin C = -12 micro Coulombs. Girma da alkiblar ƙarfin lantarki akan cajin B sune…
Tattaunawa
An sani :
An tambaya : Girma da alkiblar ƙarfin lantarki akan caji B
Jawab :
Ƙarfin wutar lantarki da ke kan caji B shine sakamakon ƙarfin wutar lantarki tsakanin caji A da B da kuma ƙarfin wutar lantarki tsakanin caji B da C.
Ƙarfin wutar lantarki tsakanin caji A da B :
Cajin A yana da negative kuma chajin B yana da positive don haka alkiblar ƙarfin Coulomb tana zuwa ga chajin A kuma tana nesa da chajin B (a hagu).
Ƙarfin wutar lantarki tsakanin caji B da C :
Caji B yana da kyau kuma chaji C yana da negative don haka alkiblar ƙarfin Coulomb tana zuwa ga chaji C kuma tana nesa da chaji B (dama).
Sakamakon ƙarfin wutar lantarki akan caji B :
Alkiblar FAB zuwa hagu da kuma alkiblar FBC Zuwa hannun dama.
FB = FAB - FBC = 675 N – 125 N = 550 Newtons.
Alkiblar ƙarfin wutar lantarki da aka samu a kan caji B (F)B) = alkiblar ƙarfin lantarki FAB, wato zuwa ga cajin C (a dama).
4. An raba cajin wutar lantarki +Q 1 = ƙananan Coulombs 10, +Q 2 = ƙananan Coulombs 50 da Q 3 kamar yadda aka nuna a cikin hoton. Don haka ƙarfin wutar lantarki da ke aiki akan cajin Q 2 = sifili to cajin Q 3 shine…
Tattaunawa
An sani :
Caji 1 (q1) = +10 μC = +10 x 10-6 Coulomb
Caji 2 (q2) = +50 μC = +50 x 10-6 Coulomb
Nisa tsakanin caji 1 da 2 (r12) = 2 cm = 0,02 m = 2 x 10-2 m
Nisa tsakanin caji 2 da 3 (r23) = 6 cm = 0,06 m = 6 x 10-2 m
Ƙarfin wutar lantarki da aka samu a caji 2 (F)2) = 0
An tambaya : caji 3 (q3)
Jawab :
Ƙarfin wutar lantarki da ke kan caji 2 shine sakamakon ƙarfin wutar lantarki tsakanin caji 1 da 2 tare da ƙarfin wutar lantarki tsakanin caji 2 da 3.
Ƙarfin wutar lantarki tsakanin caji 1 da 2 :
Caji na 1 yana da kyau kuma chaji na 2 yana da kyau, don haka alkiblar ita ce F12 yana zuwa ga caji 3 (a hannun dama).
Ƙarfin wutar lantarki tsakanin caji 2 da 3 :
Caji na 2 yana da kyau kuma chaji na 3 yana da kyau, don haka alkiblar ita ce F23 yana zuwa ga caji 1 (a hagu).
Ƙarfin wutar lantarki da aka samu a caji 2 = 0 :
Alkiblar F12 zuwa dama da alkiblar F23 Zuwa hannun hagu.
5. A cikin hoton da ke gefe, ƙarfin da ke tattare da cajin biyu shine 144 N, don haka nisan da ke tsakanin cajin biyu shine… (1 μC = 10 -6 C da k = 9.10 9 Nm 2 .C -2 )
A. 4 cm
B. 5 cm
C. 8 cm
D. 10 cm
E. 12 cm
Tattaunawa
An san cewa:
Ƙarfin wutar lantarki (F 12 ) = 144 N
Caji 1 (q 1 ) = 10 μC = 10 x 10 -6 C
Caji 2 (q 2 ) = 4 μC = 4 x 10 -6 C
Madaidaitan Coulomb (k) = 9 x 10 9 Nm 2 .C -2
C = Coulomb, N = Newton, m = mita
Tambaya: Nisa tsakanin caji biyu (r)
Amsa:
Tsarin dokar Coulomb:
Bayani: F = ƙarfin lantarki, k = Madaidaitan Coulomb, q = cajin lantarki, r = nisan da ke tsakanin cajin lantarki guda biyu.
Nisa tsakanin caji biyu (r) = mita 5 x 10 -2 = 5 x 10 -2 x 10 santimita 2 = santimita 5.
Amsar da ta dace ita ce B.
6. Cajin wutar lantarki guda biyu kamar yadda aka nuna a hoton da ke ƙasa. Idan ƙarfin Coulomb da aka samu a duka caji guda biyu shine 3,6 N, to nisan da ke tsakanin caji biyu shine… (1 μC = 10 -6 C da k = 9.10 9 Nm 2 .C -2 )
A. 30,0 cm
B. 27,0 cm
C. 9,0 cm
D. 3,0 cm
E. 0,3 cm
E. 12 cm
Tattaunawa
An san cewa:
Ƙarfin wutar lantarki (F 12 ) = 3,6 N
Caji 1 (q 1 ) = 9 μC = 9 x 10 -6 C
Caji 2 (q 2 ) = 4 μC = 4 x 10 -6 C
Madaidaitan Coulomb (k) = 9 x 10 9 Nm 2 .C -2
C = Coulomb, N = Newton, m = mita
Tambaya: Nisa tsakanin caji biyu (r)
Amsa:
Tsarin dokar Coulomb:
Bayani: F = ƙarfin lantarki, k = Madaidaitan Coulomb, q = cajin lantarki, r = nisan da ke tsakanin cajin lantarki guda biyu.
Nisa tsakanin caji biyu (r) = mita 0,3 = santimita 0,3 x 100 = santimita 30.
Amsar da ta dace ita ce A.
7. Ana samun caji mai maki uku a kusurwoyin alwatika, kamar yadda aka nuna a cikin hoton. Girman ƙarfin Coulomb a wurin A shine… (k = 9.10 9 Nm 2 .C -2 ; 1 μ = 10 -6 )
A. 10 N
B. 60 N
C. 90 N
D. 90 √2 N
E. 100 √2 N
Tattaunawa
An san cewa:
Cajin A (qA ) = -2 μC = -2 x 10 -6 C
Cajin B (qB ) = 8 μC = 8 x 10 -6 C
Cajin C (qC ) = 4,5 μC = 4,5 x 10 -6 C
Madaidaitan Coulomb (k) = 9 x 10 9 Nm 2 .C -2
Nisa tsakanin caji A da B (r AB ) = 4 cm = mita 4 x 10 -2
Nisa tsakanin caji A da C (r AC ) = 3 cm = mita 3 x 10 -2
C = Coulomb, N = Newton, m = mita
Tambaya: Ƙarfin wutar lantarki a wuri na A
Amsa:
Tsarin dokar Coulomb:
Bayani: F = ƙarfin lantarki, k = Madaidaitan Coulomb, q = cajin lantarki, r = nisan da ke tsakanin cajin lantarki guda biyu.
Ƙarfin wutar lantarki a wuri na A:
F AB da F AC suna samar da kusurwar dama don haka ana ƙididdige ƙarfin sakamakon ta amfani da dabarar Pythagorean.
Amsar da ta dace ita ce D.
8. Abubuwa biyu masu cajin lantarki Q1 da Q2 a nisan r cm, yana haifar da ƙarfin 10 N. Sannan cajin Q1 an canza shi ta yadda ƙarfin da ke tasowa zai zama N40. Daidaiton k = 9 x 109 Nm2.C-2, sannan cajin Q1 dole ne a motsa ta…
A. 1/2 r nesa da Q2
B. 1/2 r ya kusanci Q2
C. 1 r ya ƙaura daga Q2
D. 2 r ya kusanci Q2
E. 2 r ya ƙaura daga Q2
Tattaunawa
An san cewa:
Nisa tsakanin Q 1 da Q 2 = r cm
Ƙarfin da ke juyewa (F) = 10 N
Idan aka canza caji Q1 , ƙarfin ƙin yarda (F) = 40 N
Daidaitacce (k) = 9 x 10 9 Nm 2 .C -2
Tambaya: dole ne a canza caji Q1 ta hanyar...
Amsa:
Tsarin ƙarfin lantarki:
F = k Q 1 Q 2 / r 2
Nisa (r) daidai yake da ƙarfi (F). Wannan yana nufin cewa idan ƙarfin ya fi girma, nisan da ke tsakanin cajin zai yi ƙanƙanta. Don haka, dole ne a matsar da caji Q1 kusa da caji Q2.
Amsar da ta dace ita ce B.
9. Cajin wutar lantarki + q3 = 20 μC, +q2 = 10 μC da q1 an raba shi kamar yadda aka nuna a hoton. Don haka rundunar Coulomb da ke aiki a kan cajin q2 = sifili sannan cajin q1 shine…
A. +2 μC
B. –2 μC
C. +5 μC
D. –5 μC
E. +8 μC
Tattaunawa:
An sani :
+q3 = 20 μC
+q2 = 10 μC
An tambaya :
q1 = ?
Jawab :
Mai caji q2 ma'aikacin F23 da kuma F21, inda F23 an nufi hagu. Don haka ƙarfin da ya biyo baya yana aiki akan q2 daidai da sifili idan aka kwatanta da q1 dole ne ya zama tabbatacce don ƙarfin F ya kasance21 zuwa dama.
Cajin q1 = +5 μC.
Amsar da ta dace ita ce C.
10. Cajin wutar lantarki guda uku +q 1 = 5 μC, +q 2 = 10 μC da +q 3 = 10 μC suna cikin layi. Idan a = 10 cm to girma da alkiblar ƙarfin wutar lantarki da ke aiki akan caji q 2 shine… (k = 9 x 10 9 Nm 2 C -2 )
A. 7,5 N zuwa q1
B. 11,25 N zuwa q2
C. 15 N zuwa q1
D. 22,5 N zuwa q1
E. 78,75 N zuwa q1
Tattaunawa:
An sani :
q1= 5 μC, q2 = 10 μC da q3 = 10 μC
a = 0,1 m
An tambaya :
Girma da alkiblar ƙarfin Coulomb da aka samu akan q2 ?
Jawab :
Akwai ƙarfin Coulomb guda biyu da ke aiki akan q2, wato F21 da kuma F23, inda alkiblar waɗannan ƙarfin biyu take akasin haka. q1 tabbatacce da q2 tabbatacce don haka ƙarfin F21 zuwa dama (zuwa q)3) q2 tabbatacce da q3 tabbatacce don haka ƙarfin F23 zuwa hagu (zuwa q)1).
Girman ƙarfin wutar lantarki da aka samu = 90 N – 11,25 N = 78,75 N.
Alkiblar ƙarfin da aka samu iri ɗaya ce da alkiblar ƙarfin F23, zuwa ga q1.
Amsar da ta dace ita ce E.
