**Electric potential energy** (U_{E}) depends upon the **coulomb’s constant** (k), **quantity of charge** (q) and the **distance of separation **(r). Using the formula of electric potential energy: **U**_{E}** = k [q**_{1}** q**_{2}**] ÷ 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 × 10^{9} N m^{2}/C^{2})

Solution:

Given data:

Quantity of charge on ball 1, q_{1} = 15 µC = 15 × 10^{-6} C

Quantity of charge on ball 2, q_{2} = 35 µC = 35 × 10^{-6} C

Distance between the two charged balls, r = 2 m

Coulomb’s constant, k = 8.98 × 10^{9} N m^{2}/C^{2}

Electric potential energy of the system, U_{E} = ?

Using the formula of electric potential energy,

U_{E} = k [q_{1} q_{2}] ÷ r

U_{E} = [(8.98 × 10^{9}) × (15 × 10^{-6}) × (35 × 10^{-6})] ÷ 2

U_{E} = [8.98 × 15 × 35 × 10^{-3}] ÷ 2

U_{E} = 4.7145 ÷ 2

U_{E} = 2.35 N

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

**Problem 2:** Two spheres 1 and 2 with charges q_{1} = 12 µC and q_{2} = 24 µC are separated by a distance of 1 m. If the value of coulomb’s constant is k = 8.98 × 10^{9} N m^{2}/C^{2}, then find the electric potential energy of the system.

Solution:

Given data:

Quantity of charge on sphere 1, q_{1} = 12 µC = 12 × 10^{-6} C

Quantity of charge on sphere 2, q_{2} = 24 µC = 24 × 10^{-6} C

Distance between the two charged spheres, r = 1 m

Coulomb’s constant, k = 8.98 × 10^{9} N m^{2}/C^{2}

Electric potential energy of the system, U_{E} = ?

Using the formula of electric potential energy,

U_{E} = k [q_{1} q_{2}] ÷ r

U_{E} = [(8.98 × 10^{9}) × (12 × 10^{-6}) × (24 × 10^{-6})] ÷ 1

U_{E} = [8.98 × 12 × 24 × 10^{-3}] ÷ 1

U_{E} = 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 × 10^{9} N m^{2}/C^{2})

Solution:

Given data:

Electric potential energy of the system, U_{E} = ?

Quantity of charge on particle 1, q_{1} = 14 µC = 14 × 10^{-6} C

Quantity of charge on particle 2, q_{2} = 18 µC = 18 × 10^{-6} C

Distance between the two charged particles, r = 3 m

Coulomb’s constant, k = 8.98 × 10^{9} N m^{2}/C^{2}

Using the formula of electric potential energy,

U_{E} = k [q_{1} q_{2}] ÷ r

U_{E} = [(8.98 × 10^{9}) × (14 × 10^{-6}) × (18 × 10^{-6})] ÷ 3

U_{E} = [8.98 × 14 × 18 × 10^{-3}] ÷ 3

U_{E} = 2.2629 ÷ 3

U_{E} = 0.75 N

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

**Problem 4:** Two masses m_{1} and m_{2} are separated by a distance of 1.5 m. Calculate the electric potential energy of the system, if the masses m_{1} and m_{2} have the charges 13 µC and 16 µC respectively. (Take the value of coulomb’s constant, k = 8.98 × 10^{9} N m^{2}/C^{2})

Solution:

Given data:

Distance between the two charged masses, r = 1.5 m

Electric potential energy of the system, U_{E} = ?

Quantity of charge on mass 1, q_{1} = 13 µC = 13 × 10^{-6} C

Quantity of charge on mass 2, q_{2} = 16 µC = 16 × 10^{-6} C

Coulomb’s constant, k = 8.98 × 10^{9} N m^{2}/C^{2}

Using the formula of electric potential energy,

U_{E} = k [q_{1} q_{2}] ÷ r

U_{E} = [(8.98 × 10^{9}) × (13 × 10^{-6}) × (16 × 10^{-6})] ÷ 1.5

U_{E} = [8.98 × 13 × 16 × 10^{-3}] ÷ 1.5

U_{E} = 1.8678 ÷ 1.5

U_{E} = 1.24 N

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

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**Related:**

- Thermal Energy Equation
- Potential Energy Formula
- Gravitational Potential Energy Formula
- Elastic Potential Energy Formula
- Kinetic Energy Formula
- Rotational Kinetic Energy Formula
- Electrical Energy Equation
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