1. Electron e potlakiswa ho tloha phomolong ka phapang e ka bang teng ea 12 V. Phetoho ke efe ho matla a motlakase ea elektrone?
Tse tsejoang:
The tefiso hodima elektrone (e) = -1.60 x 10-19 Coulomb
Bokhoni ba motlakase = Palo ea li-volts (V) = 12 Volt
Oa Batla: Phetoho ea matla a motlakase a elektrone (ΔPE)
Tharollo:
ΔPE = q V = (-1.60 x 10-19 C(12 V) = -19.2 x 10-19 Joule
Letšoao la minus le bontša hore matla a ka bang teng aa fokotseha.
2. Lipoleiti tse peli tse bapileng li tjhajwa. Karohano pakeng tsa lipoleiti ke 2 cm mme boholo ba tšimo ea motlakase Pakeng tsa lipoleiti ke 500 Volt/meter. Phetoho ea matla a ka bang teng a proton ke efe ha e potlakisoa ho tloha poleiting e nang le tjhaja e ntle ho ea poleiting e nang le tjhaja e mpe?
Tse tsejoang:
Boholo ba tšimo ea motlakase pakeng tsa lipoleiti (E) = 500 Volt/meter
The distanta pakeng tsa lipoleiti (li) = 2 cm = 0,02 m
Tefiso ho proton = +1.60 x 10-19 Coulomb
Ho batloa: Phetoho ea matla a motlakase (ΔPE)
Tharollo:
Bokhoni ba motlakase:
V = E s
V = (500 Volt/m)(0.02 m)
V = 10 Volt
Phetoho ea matla a motlakase:
ΔPE = q V
ΔPE = (1,60 x 10-19 C(10 V)
ΔPE = 16 x 10-19 Joule
ΔPE = 1.6 x 10-18 Joule
3. Litefiso tse peli tsa lintlha li arotsoe ka sebaka sa 10 cm. Litefiso ntlheng ea A =+9 μC le tefiso ntlheng ea B = -4 μC. k = 9 x 109 Nm2C-2, 1 μC = 10-6 C. Phetoho ea matla a motlakase a tjhaja ntlheng ea B ke efe haeba e potlakisetsoa ntlheng ea A?
Tse tsejoang:
Tefiso A (q)1) = +9 μC = +9 x 10-6 C
lefisa B (q1) = -4 μC = -4 x 10-6 C
k = 9 x 109 Nm2C-2
Sebaka se pakeng tsa litefiso A le B (r) = 10 cm = 0.1 m = 10-1 m
Ho batloa: Phetoho ea matla a motlakase (ΔEP)
Tharollo:
