Ƙarfin Electrostatic - Matsaloli da mafita
1. Idan ƙarfin wutar lantarki mai tsauri shine 144 N, menene tazara tsakanin caji biyu ... (1 μC = 10 -6 C da k = 9.10 9 Nm 2 .C -2 )
An sani:
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
Daidaitacce (k) = 9 x 10 9 Nm 2 .C -2
C = Coulomb, N = Newton, m = mita
Ana nema: Nisa tsakanin caji biyu (r)
Magani:
Daidaiton dokar Coulomb :
F = ƙarfin lantarki mai tsauri, k = Madaidaitan Coulomb, q = cajin lantarki, r = tazara tsakanin caji biyu
Nisa tsakanin caji biyu (r) = mita 5 x 10 -2 = mita 5 x 10 -2 x 10 2 cm = 5 cm.
2. Caji biyu da nisan r ya raba . Idan ƙarfin lantarki mai tsauri shine 3.6 N, menene nisan da ke tsakanin caji biyu (1 μC = 10 -6 C da k = 9.10 9 Nm 2 .C -2 )
An sani:
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
Ana nema: Nisa tsakanin caji biyu (r)
Magani:
Nisa tsakanin caji biyu (r) = 0.3 m = 0.3 x 100 cm = 30 cm.
3. Cajin maki uku kamar yadda aka nuna a cikin hoton da ke ƙasa. Menene girman ƙarfin wutar lantarki mai tsayawa a wurin A (k = 9.10 9 Nm 2 .C -2 ; 1 μ = 10 -6 )
An sani:
cajin A (q)A) = -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 = 4 x 10 -2 m
Nisa tsakanin caji A da C (r AC ) = 3 cm = 3 x 10 -2 m
C = Coulomb, N = Newton, m = mita
Ana so: Ƙarfin wutar lantarki mai motsi a wuri na A
Magani:
Daidaiton dokar Coulomb:
F = ƙarfin lantarki mai tsauri , k = Madaidaici na Coulomb , q = cajin lantarki , r = tazara tsakanin caji biyu
Ƙarfin wutar lantarki mai motsi a wuri na A:
4. Cajin maki uku kamar yadda aka nuna a cikin hoton da ke ƙasa. Menene girman ƙarfin Coulomb da Charge B ya fuskanta (k = 9.10 9 Nm 2 .C -2 ; 1 μC = 10 -6 C ).
An sani:
q A = 10 µC = 10 x 10 -6 C = 10 -5 Coulombs
q B = 10 µC = 10 x 10 -6 = 10 -5 Coulombs
q C = 20 µC = 20 x 10 -6 = 2 x 10 -5 Coulombs
r AB = mita 0.1 = mita 10 -1
r BC = mita 0.1 = mita 10 -1
k = 9 x 10 9 Nm 2 C −2
Ana Neman : Rundunar Coulomb da Charge B ya samu
Magani:
Akwai ƙarfin coulomb guda biyu da ke aiki akan cajin B, ƙarfin Coulomb tsakanin caji A da B (F AB ) da ƙarfin Coulomb tsakanin caji B da C (F BC ). Ƙarfin Coulomb da aka samu ta hanyar caji B shine ƙarfin F AB da F BC.
Ƙarfin Coulomb tsakanin caji A da B:
Caji A ya sanya hannu kan positive da caji B ya sanya hannu kan positive don haka F AB ya nuna positive zuwa C.
Ƙarfin Coulomb tsakanin caji na B da C:
Caji B ya nuna alamar positive da kuma alamar positive don haka F BC ya nuna alamar positive A.
Ƙarfin Coulomb wanda aka samu ta hanyar caji B:
F B = F BC – F AB = 180 – 90 = 90 N
Ƙarfin Coulomb da aka samu ta hanyar cajin B (F B ) shine Newtons 90. Alkiblar F B iri ɗaya ce da alkiblar maki F BC zuwa cajin A.
5. Kayyade girma da alkiblar ƙarfin Coulomb akan caji B ( k = 9 x 10 9 Nm 2 C −2 , 1 μC = 10 −6 C).
An sani:
Cajin A (qA ) = +Q
Cajin B (qB ) = -2Q
Cajin C (qC ) = -Q
Nisa tsakanin caji A da B (r AB ) = r
Nisa tsakanin caji B da C (r BC ) = 2r
k = 9 x 10 9 Nm 2 C −2
Ana Neman: Girma da alkiblar ƙarfin Coulomb akan caji B
Magani:
Ƙarfin Coulomb tsakanin caji A da B:
Cajin A yana da kyau kuma Cajin B yana da kyau, don haka alkiblar F AB tana nuna zuwa ga Cajin A.
Ƙarfin Coulomb tsakanin caji B da caji C:
Cajin B yana da negative kuma Cajin C yana da negative don haka alkiblar F BC tana nuna zuwa ga cajin A.
Dokar ƙarfi ta hanyar amfani da caji 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 maki masu ƙarfi don cajin A ko zuwa hagu.
