3 ngā pātai mō te whārite rerekētanga pūmanawa
1. Ka nekehia tētahi utu hiko i roto i tētahi papa hiko ōrite me te kaha o te 2√3 N i te tawhiti o te 20 cm. Mena kei te koki o te 30 o te nekehanga o te utu hiko te ahunga o te kaha , he aha te rerekētanga o te kaha pūmanawa hiko i ngā tūranga tīmatanga me ngā tūranga whakamutunga o te utu hiko.
Mōhiotia:
Te kaha (F) = 2√3 N
Tawhiti (s) = 20 cm = 0.2 m
Koki (θ) = 30 o
E hiahiatia ana: Te rerekētanga pūmanawa hiko
Rongoā:
Ko te mahi o te whakawhiti i te utu +q mai i a ki b he ōrite ki te rerekētanga pūngao pūmanawa hiko i ngā pūwāhi a me b.
ΔEP = ΔW
Mena ko te ahunga o te kaha F e pā ana ki te ahunga o te whakawhiti utu +q he koki θ, ko te mahi o te whakawhiti utu +q mai i a ki b ko:
ΔW = F Δs cos θ = ( 2 √ 3 )(0,2)(cos 30) = (0,4 √ 3 )(0,5 √ 3 ) = (0,2)(3) = 0,6 Joule
2. Mena e mōhiotia ana ko te utu me te kaha o ngā pūnga iahiko he 5 µC me te 20 µF, ia, whakatauhia te rerekētanga pūmanawa o ngā pūnga iahiko.
Mōhiotia:
Utu hiko (q) = 5 µC = 5 x 10 -6 C
Te kaha o te pūnga (C) = 20 µF = 20 x 10 -6 F
E hiahiatia ana: Te rerekētanga pūmanawa o te pūnga (V)
Rongoā:
Ko te tātai mō te rerekētanga pūmanawa o te pūnga, te kaha o te pūnga me te utu:
V = q / C = (5 x 10 -6 ) / 20 x 10 -6 = 5/20 = 0,25 Ngā Waohiko
3. E 16 cm te tawhiti o ngā utu pūwāhi e rua, ko QA = -4 µC me QB = 8 µC. Tātaihia te pūmanawa hiko i waenganui i ngā utu e rua.
Mōhiotia:
Utu hiko 1 (Q A ) = 4 x 10 -6 C
Utu hiko 2 (Q B ) = 8 x 10 -6 C
Tawhiti (r) = 8 henimita = 0,08 m
Pūmau Coulomb (k) = 9 × 109 Nm2 / C2
E hiahiatia ana: Te pūmanawa hiko i C (V C )
Rongoā:
V A = k Q / r = (9 × 10 9 )(4 x 10 -6 ) / 0,08 = (36 × 10 3 ) / 0,08 = -450 Ngā Waeine
V B = k Q / r = (9 × 10 9 )(8 x 10 -6 ) / 0,08 = (72 × 10 3 ) / 0,08 = 900 Ngā Waeine
Ko te pūmanawa hiko i C (V C ) = 900 – 450 = 450 Volts
