Understanding Permittivity of Free Space SI Unit

The question asks us to find the SI unit of the permittivity of free space, denoted by the symbol $\(\epsilon_0\)$, using the fundamental SI units: kilogram (kg), meter (m), second (s), and ampere (A). Permittivity of free space is a fundamental physical constant that describes how an electric field affects, and is affected by, a vacuum. It essentially measures the capability of a vacuum to permit electric field lines.

Deriving the SI Unit of Permittivity of Free Space ($\\epsilon_0$)

To determine the SI unit of $\\epsilon_0$, we can use Coulomb's Law, which describes the electrostatic force (F) between two point charges ($q_1$ and $q_2$) separated by a distance ($r$):

$ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} $

We need to rearrange this formula to solve for $\\epsilon_0$:

$ \epsilon_0 = \frac{q_1 q_2}{4 \pi F r^2} $

Now, let's determine the SI units for each term in the rearranged equation:

• Force ($F$): The SI unit of force is the Newton (N). In terms of fundamental units, $1 \text{ N} = 1 \text{ kg} \cdot \text{m} \cdot \text{s}^{-2}\).
• Charge ($q_1, q_2$): The SI unit of electric charge is the Coulomb (C). The Ampere (A) is the SI unit of electric current, and it's defined such that $1 \text{ C} = 1 \text{ A} \cdot \text{s}\). Therefore, the unit of charge squared ($q_1 q_2$) is \( \text{C}^2 = (\text{A} \cdot \text{s})^2 = \text{A}^2 \text{s}^2 \).
• Distance ($r$): The SI unit of distance is the meter (m). Therefore, the unit of distance squared ($r^2$) is \( \text{m}^2 \).
• \( 4 \pi \): This is a dimensionless constant and does not affect the units.

Substitute these units back into the formula for $\\epsilon_0$:

$ \text{Unit}(\epsilon_0) = \frac{\text{Unit}(q_1 q_2)}{\text{Unit}(F) \cdot \text{Unit}(r^2)} $

$ \text{Unit}(\epsilon_0) = \frac{\text{A}^2 \text{s}^2}{(\text{kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot (\text{m}^2)} $

Now, let's simplify the expression by combining the units:

$ \text{Unit}(\epsilon_0) = \frac{\text{A}^2 \text{s}^2}{\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2}} $

To simplify further, move the terms from the denominator to the numerator by inverting their signs:

$ \text{Unit}(\epsilon_0) = \text{A}^2 \cdot \text{s}^2 \cdot \text{kg}^{-1} \cdot \text{m}^{-3} \cdot \text{s}^{2} $

Combine the powers of second (s):

$ \text{Unit}(\epsilon_0) = \text{A}^2 \cdot \text{kg}^{-1} \cdot \text{m}^{-3} \cdot \text{s}^{(2+2)} $

$ \text{Unit}(\epsilon_0) = \text{A}^2 \text{kg}^{-1} \text{m}^{-3} \text{s}^{4} $

Conclusion: SI Unit of Permittivity of Free Space

Based on the derivation using Coulomb's Law and the fundamental SI units, the SI unit for the permittivity of free space ($\\epsilon_0$) is \( \text{A}^2 \text{kg}^{-1} \text{m}^{-3} \text{s}^{4} \).

Comparing this result with the given options:

• Option 1: \( \text{A}^2 \text{kg}^{-1} \text{m}^{-3} \text{s}^{4} \) - Matches our derived unit.
• Option 2: \( \text{A}^{-2} \text{kg}^1 \text{m}^3 \text{s}^{-4} \)
• Option 3: \( \text{A}^2 \text{kg}^{-1} \text{m}^{-2} \text{s}^{3} \)
• Option 4: \( \text{A}^2 \text{kg}^1 \text{m}^{-3} \text{s}^{-4} \)

Therefore, the correct option representing the SI unit of permittivity of free space in terms of kg, m, s, and A is the first one.