# Electric force equation

The electric force equation, based on Coulomb’s law, allows us to calculate the force between two charged objects. It is expressed as Fe = k [q1 q2] ÷ r2, where Fe represents the electric force, k is the Coulomb’s constant, q1 and q2 are the charges of the objects, and r is the distance between the two charged objects.

Contents

## Practice problems

### Problem #1

Calculate the electric force acting between two objects, object 1 and object 2, with charges of 20 µC and 15 µC, respectively, separated by a distance of 1 m. Take the value of the proportionality constant, k, as 8.98 × 109 N m2/C2.

Solution

Given data:

• Electric force acting between two charged objects, Fe = ?
• Quantity of charge on object 1, q1 = 20 µC = 20 × 10-6 C
• Quantity of charge on object 2, q2 = 15 µC = 15 × 10-6 C
• Distance between the two charged objects, r = 1 m
• Proportionality constant, k = 8.98 × 109 N m2/C2

Applying the formula:

• Fe = k [q1 q2] ÷ r2
• Fe = [(8.98 × 109) × (20 × 10-6) × (15 × 10-6)] ÷ (1)2
• Fe = [8.98 × 20 × 15 × 10-3] ÷ 1
• Fe = 8.98 × 20 × 15 × 10-3
• Fe = 2.69 N

Therefore, the electric force acting between two charged objects is 2.69 N.

### Problem #2

Determine the electric force acting between two charged plastic balls separated by a distance of 150 cm. The proportionality constant, k, is 8.98 × 109 N m2/C2. The charges of the balls are q1 = 16 µC and q2 = 8 µC.

Solution

Given data:

• Electric force acting between two charged plastic balls, Fe = ?
• Distance between the two charged plastic balls, r = 150 cm = 1.5 m
• Proportionality constant, k = 8.98 × 109 N m2/C2
• Quantity of charge on 1st plastic ball, q1 = 16 µC = 16 × 10-6 C
• Quantity of charge on 2nd plastic ball, q2 = 8 µC = 8 × 10-6 C

Applying the formula:

• Fe = k [q1 q2] ÷ r2
• Fe = [(8.98 × 109) × (16 × 10-6) × (8 × 10-6)] ÷ (1.5)2
• Fe = [8.98 × 16 × 8 × 10-3] ÷ 2.25
• Fe = 1.1494 ÷ 2.25
• Fe = 0.51 N

Therefore, the electric force acting between two charged plastic balls is 0.51 N.

### Problem #3

Given two spheres with charges of 30 µC and 7 µC, placed 2.1 m apart, find the value of the electric force acting between them. The proportionality constant, k, is 8.98 × 109 N m2/C2.

Solution

Given data:

• Quantity of charge on sphere 1, q1 = 30 µC = 30 × 10-6 C
• Quantity of charge on sphere 2, q2 = 7 µC = 7 × 10-6 C
• Distance between the two charged spheres, r = 2.1 m
• Electric force acting between two charged spheres, Fe = ?
• Proportionality constant, k = 8.98 × 109 N m2/C2

Applying the formula:

• Fe = k [q1 q2] ÷ r2
• Fe = [(8.98 × 109) × (30 × 10-6) × (7 × 10-6)] ÷ (2.1)2
• Fe = [8.98 × 30 × 7 × 10-3] ÷ 4.41
• Fe = 1.8858 ÷ 4.41
• Fe = 0.42 N

Therefore, the electric force acting between two charged spheres is 0.42 N.

### Problem #4

Calculate the magnitude of the electric force between two charged balloons, separated by a distance of 250 cm. The proportionality constant, k, is 8.98 × 109 N m2/C2. The charges of the balloons are q1 = -25 µC and q2 = 5 µC.

Solution

Given data:

• Electric force acting between two charged balloons, Fe = ?
• Distance between the two charged sphere, r = 250 cm = 2.5 m
• Proportionality constant, k = 8.98 × 109 N m2/C2
• Quantity of charge on balloon 1, q1 = -25 µC = -25 × 10-6 C
• Quantity of charge on balloon 2, q2 = 5 µC = 5 × 10-6 C

Applying the formula:

• Fe = k [q1 q2] ÷ r2
• Fe = [(8.98 × 109) × (-25 × 10-6) × (5 × 10-6)] ÷ (2.5)2
• Fe = [8.98 × (-25) × 5 × 10-3] ÷ 6.25
• Fe = (-1.1225) ÷ 4.41
• Fe = -0.25 N
• |Fe| = |-0.25| = 0.25 N

Therefore, the magnitude of electric force acting between the two charged balloons is 0.25 N.