# 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?

__Known :__

The charge on an electron (e) = -1.60 x 10^{-19} Coulomb

Electric potential = voltage (V) = 12 Volt

__Wanted:__ The change in electric potential energy of the electron (ΔPE)

__Solution :__

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.

__Known :__

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)

__Solution :__

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^{-1}

^{8}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 10^{9 }Nm^{2}C^{−2}, 1 μC = 10^{−6} C. What is the change in electric potential energy of charge on point B if accelerated to point A ?

__Known :__

Charge A (q_{1}) = +9 μC = +9 x 10^{−6} C

Charge B (q_{1}) = -4 μC = -4 x 10^{−6} C

k = 9 x 10^{9} Nm^{2}C^{−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)

__Solution :__