3 su'aalood oo ku saabsan isle'egyada goobta korantada
1. Kubbad qabsata oo leh gacan 10 cm ah waxay leedahay koronto koronto oo ah 500 μC. Dhibcaha A, B, iyo C waxay la siman yihiin bartamaha kubadda masaafo dhan 12 cm, 10 cm iyo 8 cm siday u kala horreeyaan bartamaha kubadda. Xisaabi xoogga goobta korantada ee dhibcaha A, B, iyo C!
Known:
Gacanka kubbadda hagaya (R) = 10 cm = 0.1 m
Dakhli koronto (q) = 500 μC = 500 x 10 -6 C
r A = 12 cm = 0,12 m
r B = 10 cm = 0,1 m
r C = 8 cm = 0,08 m
Joogtada Coulomb (k) = 9 x 10 9
La Rabay: Xoogga goobta korantada ee barta A (E A ), barta B (E B ) iyo barta C (E C )
Solution:
a) Xoogga goobta korantada ee barta A
E A = kq / r A 2 = (9 x 10 9 )(500 x 10 -6 ) / (0,12) 2 = (4500 x 10 3 ) / 0,0144 = 312500 x 10 3 = 3,125 x 10 8 N/C
b) Xoogga goobta korantada ee barta B
E B = kq / r B 2 = (9 x 10 9 )(500 x 10 -6 ) / (0,1) 2 = (4500 x 10 3 ) / 0,01 = 450.000 x 10 3 = 4,5 x 10 8 N/C
c) Xoogga goobta korantada ee barta C
E C = 0 marka aad kubadda ku jirto.
2. Haddii tijaabo lagu sameeyo 4 nC meel, dalacku wuxuu la kulmaa xoog dhan 5 × 10 – 4 N. Waa maxay baaxadda goobta korantada E ee markaas?
La yaqaan
Tijaabi dallacaadda korantada (q) = 4 nC = 4 x 10 -9 Coulomb
Xoogga korontada (F) = 5 × 10 -4 N
La Doonayo: Cabbirka goobta korantada (E)
Solution:
E = F / q = (5 × 10 -4 ) / (4 x 10 -9 ) = 1,25 x 10 5 N/C
3. Laba dallacaad q B = 12 μC iyo q C = 9 μC ayaa la dhigayaa geesaha saddexagalka midig sida ku cad Jaantuska. Go'aami xoogga goobta korantada ee la dareemay barta A!
La yaqaan
Dalacaadda barta B (qB) = 12 μC = 12 x 10 -6 C
Dalacaadda barta C (qC) = 9 μC = 9 x 10 -6 C
Joogtada Coulomb (k) = 9 x 10 9
r AC = 4 cm = 0,04 m
r AB = 3 cm = 0,03 m
SE buska: xoogga goobta korantada ee barta A
Solution:
E AC = kq / r 2 = (9 x 10 9 )(9 x 10 -6 ) / (0,04) 2 = 81 x 10 3 / 0,0016 = 5,0 x 10 7 N/C
E AB = kq / r 2 = (9 x 10 9 )(9 x 10 -6 ) / (0,03) 2 = 81 x 10 3 / 0,0009 = 9,0 x 10 7 N/C
