Basic Physics

# 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

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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 10-18 Joule

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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 ?

Known :

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)

Solution :

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