**Electric force** is the force of attraction or repulsion between charged particles due to their electric fields. It is one of the fundamental forces of nature and is described by Coulomb’s law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. For instance, when you rub a balloon on your hair, electrons transfer from your hair to the balloon, creating a static electric charge. This charge allows the balloon to stick to a wall because of the electric force between the charged balloon and the neutral wall. Electric force is crucial in a wide range of phenomena, from the behavior of atoms and molecules to the operation of electrical devices and the transmission of electrical power.

## 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 × 10^{9} N m^{2}/C^{2}.

**Solution**

Given data:

- Electric force acting between two charged objects, F
_{e}= ? - Quantity of charge on object 1, q
_{1}= 20 µC = 20 × 10^{-6}C - Quantity of charge on object 2, q
_{2}= 15 µC = 15 × 10^{-6}C - Distance between the two charged objects, r = 1 m
- Proportionality constant, k = 8.98 × 10
^{9}N m^{2}/C^{2}

Applying the formula:

- F
_{e}= k [q_{1}q_{2}] ÷ r^{2} - F
_{e}= [(8.98 × 10^{9}) × (20 × 10^{-6}) × (15 × 10^{-6})] ÷ (1)^{2} - F
_{e}= [8.98 × 20 × 15 × 10^{-3}] ÷ 1 - F
_{e}= 8.98 × 20 × 15 × 10^{-3} - F
_{e}= 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 × 10^{9} N m^{2}/C^{2}. The charges of the balls are q_{1} = 16 µC and q_{2} = 8 µC.

**Solution**

Given data:

- Electric force acting between two charged plastic balls, F
_{e}= ? - Distance between the two charged plastic balls, r = 150 cm = 1.5 m
- Proportionality constant, k = 8.98 × 10
^{9}N m^{2}/C^{2} - Quantity of charge on 1
^{st}plastic ball, q_{1}= 16 µC = 16 × 10^{-6}C - Quantity of charge on 2
^{nd}plastic ball, q_{2}= 8 µC = 8 × 10^{-6}C

Applying the formula:

- F
_{e}= k [q_{1}q_{2}] ÷ r^{2} - F
_{e}= [(8.98 × 10^{9}) × (16 × 10^{-6}) × (8 × 10^{-6})] ÷ (1.5)^{2} - F
_{e}= [8.98 × 16 × 8 × 10^{-3}] ÷ 2.25 - F
_{e}= 1.1494 ÷ 2.25 - F
_{e}= 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 × 10^{9} N m^{2}/C^{2}.

**Solution**

Given data:

- Quantity of charge on sphere 1, q
_{1}= 30 µC = 30 × 10^{-6}C - Quantity of charge on sphere 2, q
_{2}= 7 µC = 7 × 10^{-6}C - Distance between the two charged spheres, r = 2.1 m
- Electric force acting between two charged spheres, F
_{e}= ? - Proportionality constant, k = 8.98 × 10
^{9}N m^{2}/C^{2}

Applying the formula:

- F
_{e}= k [q_{1}q_{2}] ÷ r^{2} - F
_{e}= [(8.98 × 10^{9}) × (30 × 10^{-6}) × (7 × 10^{-6})] ÷ (2.1)^{2} - F
_{e}= [8.98 × 30 × 7 × 10^{-3}] ÷ 4.41 - F
_{e}= 1.8858 ÷ 4.41 - F
_{e}= 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 × 10^{9} N m^{2}/C^{2}. The charges of the balloons are q_{1} = -25 µC and q_{2} = 5 µC.

**Solution**

Given data:

- Electric force acting between two charged balloons, F
_{e}= ? - Distance between the two charged sphere, r = 250 cm = 2.5 m
- Proportionality constant, k = 8.98 × 10
^{9}N m^{2}/C^{2} - Quantity of charge on balloon 1, q
_{1}= -25 µC = -25 × 10^{-6}C - Quantity of charge on balloon 2, q
_{2}= 5 µC = 5 × 10^{-6}C

Applying the formula:

- F
_{e}= k [q_{1}q_{2}] ÷ r^{2} - F
_{e}= [(8.98 × 10^{9}) × (-25 × 10^{-6}) × (5 × 10^{-6})] ÷ (2.5)^{2} - F
_{e}= [8.98 × (-25) × 5 × 10^{-3}] ÷ 6.25 - F
_{e}= (-1.1225) ÷ 4.41 - F
_{e}= -0.25 N - |F
_{e}| = |-0.25| = 0.25 N

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

## More topics

- Compression (physics)
- Net force
- Magnetism
- Centripetal force
- Centrifugal force
- Spring force
- Tension (physics)
**Electric force**

## External links

- Electric Forces, Fields & Potential | Equation & Relationship – Study.com
- Physics Tutorial: Coulomb’s Law – The Physics Classroom
- 2.8 Electric Field: Concept of a Field Revisited – BCcampus Pressbooks
- Coulomb’s Law Calculator – Omni Calculator
- 18.3 Electric Field – Physics – OpenStax
- Electric force (article) | Electrostatics – Khan Academy
- Coulomb’s law – Wikipedia
- What is the equation for calculating the electrical force, Fe, between two charges? – Quora
- 18.3: Coulomb’s Law – Physics LibreTexts
- Electric forces – HyperPhysics Concepts
- Coulomb’s Law: Calculating the Electrostatic Force – StickMan Physics
- Electric Force: Definition, Equation & Examples – Vaia
- Electric Field: Concept of a Field Revisited | Physics – Lumen Learning
- Q: How does electrical force equation related to Coulomb’s Law? – CK-12 Foundation
- Electric Charge And Electric Force Study Guide – Inspirit VR
- Electric Force – Physics – Socratic
- How to Calculate Electrostatic Force – Sciencing

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.