1. Ka whakaterea te tere o tētahi irahiko mai i te okioki mā roto i tētahi rerekētanga pūmanawa o te 12 V. He aha te panonitanga o te pūngao pūmanawa hiko o te irahiko?
Mōhiotia:
Ko te utu i runga i te irahiko (e) = -1.60 x 10 -19 Coulomb
Pūmanawa hiko = ngaohiko (V) = 12 Volts
E hiahiatia ana: Te huringa o te pūngao pūmanawa hiko o te irahiko (ΔPE)
Rongoā:
ΔPE = q V = (-1.60 x 10 -19 C)(12 V) = -19.2 x 10 -19 Joules
Ko te tohu tango e tohu ana ka heke te pūngao pūmanawa.
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2. E rua ngā pereti whakarara e utaina ana. Ko te wehenga i waenga i ngā pereti he 2 cm, ā, ko te rahi o te āpure hiko i waenga i ngā pereti he 500 Volt/mita. He aha te huringa o te pūngao pūmanawa o te pūrotene ina whakaterea mai i te pereti utu pai ki te pereti utu kino?
Mōhiotia:
Ko te rahi o te papa hiko i waenganui i ngā pereti (E) = 500 Volt/mita
Ko te tawhiti i waenganui i ngā pereti (s) = 2 cm = 0,02 m
Ko te utu i runga i te porotini = +1.60 x 10 -19 Coulombs
E hiahiatia ana: Te huringa o te pūngao pūmanawa hiko (ΔPE)
Rongoā:
Pūmanawa hiko:
V = E s
V = (500 Ngaohiko/m)(0.02 m)
V = 10 Ngāohiko
Te huringa o te pūngao pūmanawa hiko:
ΔPE = q V
ΔPE = (1,60 x 10 -19 C)(10 V)
ΔPE = 16 x 10 -19 Joule
ΔPE = 1.6 x 10 -1 8 Joule
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3. E wehea ana ngā utu pūwāhi e rua e te tawhiti o te 10 cm. Ko te utu i runga i te pūwāhi A = +9 μC me te utu i runga i te pūwāhi B = -4 μC. k = 9 x 10 9 Nm 2 C −2 , 1 μC = 10 −6 C. He aha te huringa o te kaha hiko o te utu i te pūwāhi B mēnā ka whakaterea ki te pūwāhi A?
Mōhiotia:
Utu A (q 1 ) = +9 μC = +9 x 10 −6 C
Utu B (q 1 ) = -4 μC = -4 x 10 −6 C
k = 9 x 10 9 Nm 2 C −2
Ko te tawhiti i waenganui i ngā utu A me B (r) = 10 cm = 0.1 m = 10 -1 m
E hiahiatia ana: Te huringa o te pūngao pūmanawa hiko (ΔEP)
Rongoā:
