Electric potential energy – problems and solutions
1. An electron is accelerated from rest through a potential difference 12 V. What is the change in electric potential energy of the electron?
The charge on an electron (e) = -1.60 x 10-19 Coulomb
Wanted: The change in electric potential energy of the electron (ΔPE)
ΔPE = q V = (-1.60 x 10-19 C)(12 V) = -19.2 x 10-19 Joule
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.
The magnitude of the electric field between the plates (E) = 500 Volt/meter
The distance between the plates (s) = 2 cm = 0,02 m
The charge on an proton = +1.60 x 10-19 Coulomb
Wanted : The change in electric potential energy (ΔPE)
Electric potential :
V = E s
V = (500 Volt/m)(0.02 m)
V = 10 Volt
The change in electric potential energy :
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. 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 ?
Charge A (q1) = +9 μC = +9 x 10−6 C
Charge 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
Wanted : The change in electric potential energy (ΔEP)