Question

The unit of permittivity of free space ε₀ is

• A. Newton metre² / Coulomb²
• B. Coulomb² /Newton metre²
• C. Coulomb² / (Newton metre)²
• D. Coulomb/Newton metre

Answer 1

The Correct Option is B

The unit of permittivity of free space, denoted by \varepsilon_0, is crucial in electromagnetic theory. It appears in Coulomb's law and the equations governing electric fields and forces. Let's derive and determine the correct unit for \varepsilon_0:

Coulomb's law is given by:

F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r^2}

where:

• F is the force between two charges q_1 and q_2
• r is the distance between the charges
• \varepsilon_0 is the permittivity of free space

Rearranging to solve for \varepsilon_0, we have:

\varepsilon_0 = \frac{1}{4\pi} \cdot \frac{q_1 q_2}{F r^2}

Thus, the units of \varepsilon_0 can be deduced by analyzing the units on the right-hand side:

- The unit of force F is Newtons (\text{N})

- The unit of charge q is Coulombs (\text{C})

- The unit of distance r is meters (\text{m})

Therefore, the unit equation becomes:

\varepsilon_0 = \frac{\text{C} \cdot \text{C}}{\text{N} \cdot \text{m}^2} = \frac{\text{C}^2}{\text{N} \cdot \text{m}^2}

Hence, the unit of permittivity of free space \varepsilon_0 is:

Coulomb² / Newton metre²

This matches the given correct answer: Coulomb2 / Newton metre2, making it the right choice.