1. An electron is accelerated from rest through a potential difference 12 V. What is the change in potensyal na enerhiyang elektrikal of the electron?
Kilala:
Ang bayad on an electron (e) = -1.60 x 10-19 Coulomb
Potensyal ng kuryente = boltahe (V) = 12 Boltahe
Wanted: The change in electric potential energy of the electron (ΔPE)
solusyon:
ΔPE = q V = (-1.60 x 10-19 C)(12 V) = -19.2 x 10-19 Dyul
The minus sign indicates that the potential energy decreases.
2. Two parallel plates are charged. The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate.
Kilala:
The magnitude of the electric field between the plates (E) = 500 Volt/meter
Ang layo between the plates (s) = 2 cm = 0,02 m
The charge on an proton = +1.60 x 10-19 Coulomb
Hinahanap: The change in electric potential energy (ΔPE)
solusyon:
Electric potential :
V = E s
V = (500 Volt/m)(0.02 m)
V = 10 Boltahe
The change in electric potential energy :
ΔPE = q V
ΔPE = (1,60 x 10-19 C)(10 V)
ΔPE = 16 x 10-19 Dyul
ΔPE = 1.6 x 10-18 Dyul
3. Two point charges are separated by a distance of 10 cm. Charge on point A =+9 μC and charge on point B = -4 μC. k = 9 x 109 Nm2C-2, 1 µC = 10-6 C. What is the change in electric potential energy of charge on point B if accelerated to point A ?
Kilala:
Singil A (q1) = +9 μC = +9 x 10-6 C
Bayad B (q1) = -4 μC = -4 x 10-6 C
k = 9 x 109 Nm2C-2
The distance between charge A and B (r) = 10 cm = 0.1 m = 10-1 m
Hinahanap: The change in electric potential energy (ΔEP)
solusyon:
