Electric Potential Energy Formula | Problems And Solutions

by
Electric Potential Energy Formula

Electric potential energy (UE) depends upon the coulomb’s constant (k), quantity of charge (q) and the distance of separation (r). Using the formula of electric potential energy: UE = k [q1 q2] ÷ r, the value of electric potential energy can be calculated.

Let’s solve some problems based on this formula, so you’ll get a clear idea.

Electric Potential Energy Practice Problems

Problem 1: Two balls 1 and 2 with charges 15 µC and 35 µC are separated by a distance of 2 m. Calculate the electric potential energy of this system of charges. (Take the value of coulomb’s constant, k = 8.98 × 109 N m2/C2)

Solution:

Given data:
Quantity of charge on ball 1, q1 = 15 µC = 15 × 10-6 C
Quantity of charge on ball 2, q2 = 35 µC = 35 × 10-6 C
Distance between the two charged balls, r = 2 m
Coulomb’s constant, k = 8.98 × 109 N m2/C2
Electric potential energy of the system, UE = ?

Using the formula of electric potential energy,
UE = k [q1 q2] ÷ r
UE = [(8.98 × 109) × (15 × 10-6) × (35 × 10-6)] ÷ 2
UE = [8.98 × 15 × 35 × 10-3] ÷ 2
UE = 4.7145 ÷ 2
UE = 2.35 N

Therefore, the electric potential energy of the system is 2.35 N.

Problem 2: Two spheres 1 and 2 with charges q1 = 12 µC and q2 = 24 µC are separated by a distance of 1 m. If the value of coulomb’s constant is k = 8.98 × 109 N m2/C2, then find the electric potential energy of the system.

Solution:

Given data:
Quantity of charge on sphere 1, q1 = 12 µC = 12 × 10-6 C
Quantity of charge on sphere 2, q2 = 24 µC = 24 × 10-6 C
Distance between the two charged spheres, r = 1 m
Coulomb’s constant, k = 8.98 × 109 N m2/C2
Electric potential energy of the system, UE = ?

Using the formula of electric potential energy,
UE = k [q1 q2] ÷ r
UE = [(8.98 × 109) × (12 × 10-6) × (24 × 10-6)] ÷ 1
UE = [8.98 × 12 × 24 × 10-3] ÷ 1
UE = 2.58 N

Therefore, the electric potential energy of the system is 2.58 N.

Problem 3: Calculate the electric potential energy of the system in which two particles A and B with charges 14 µC and 18 µC are separated by a distance of 3 m. (Take the value of coulomb’s constant, k = 8.98 × 109 N m2/C2)

Solution:

Given data:
Electric potential energy of the system, UE = ?
Quantity of charge on particle 1, q1 = 14 µC = 14 × 10-6 C
Quantity of charge on particle 2, q2 = 18 µC = 18 × 10-6 C
Distance between the two charged particles, r = 3 m
Coulomb’s constant, k = 8.98 × 109 N m2/C2

Using the formula of electric potential energy,
UE = k [q1 q2] ÷ r
UE = [(8.98 × 109) × (14 × 10-6) × (18 × 10-6)] ÷ 3
UE = [8.98 × 14 × 18 × 10-3] ÷ 3
UE = 2.2629 ÷ 3
UE = 0.75 N

Therefore, the electric potential energy of the system is 0.75 N.

Problem 4: Two masses m1 and m2 are separated by a distance of 1.5 m. Calculate the electric potential energy of the system, if the masses m1 and m2 have the charges 13 µC and 16 µC respectively. (Take the value of coulomb’s constant, k = 8.98 × 109 N m2/C2)

Solution:

Given data:
Distance between the two charged masses, r = 1.5 m
Electric potential energy of the system, UE = ?
Quantity of charge on mass 1, q1 = 13 µC = 13 × 10-6 C
Quantity of charge on mass 2, q2 = 16 µC = 16 × 10-6 C
Coulomb’s constant, k = 8.98 × 109 N m2/C2

Using the formula of electric potential energy,
UE = k [q1 q2] ÷ r
UE = [(8.98 × 109) × (13 × 10-6) × (16 × 10-6)] ÷ 1.5
UE = [8.98 × 13 × 16 × 10-3] ÷ 1.5
UE = 1.8678 ÷ 1.5
UE = 1.24 N

Therefore, the electric potential energy of the system is 1.24 N.

.
.
.

Related:

Also Read:

Leave a Comment