The permittivity of free space, often denoted as ε₀, is a fundamental physical constant that plays a crucial role in electromagnetism. To determine its S.I. unit, we need to understand its relationship with electric fields and forces. The correct answer to your question is D. C²N⁻¹m⁻².
Understanding Permittivity
Permittivity is a measure of how much electric field (E) is generated per unit charge (Q) in a medium. In free space, this constant helps us understand how electric fields interact with charges. The formula that relates these quantities is derived from Coulomb's law and the definition of electric field strength.
Breaking Down the Units
To derive the unit of permittivity, we can start with the relationship defined by Coulomb's law:
- The electric force (F) between two point charges is given by: F = k * (Q₁ * Q₂) / r², where k is Coulomb's constant.
- The electric field (E) created by a charge (Q) at a distance (r) is: E = F / Q = k * Q / r².
From these relationships, we can express the permittivity of free space as:
- ε₀ = Q² / (F * m²), where Q is charge, F is force, and m is distance.
Unit Analysis
Now, let's break down the units involved:
- Charge (Coulombs, C)
- Force (Newtons, N), where 1 N = 1 kg·m/s²
- Distance (meters, m)
Substituting these into the equation for ε₀, we get:
- ε₀ = (C²) / (N·m²)
- Since N can be expressed as kg·m/s², we can rewrite this as: ε₀ = (C²) / (kg·m/s²·m²) = (C²) / (kg·m³/s²)
To express this in terms of the S.I. units, we can rearrange it to show that:
Finalizing the Units
Now, we can relate this back to the options provided. The correct unit of permittivity of free space is:
- D. C²N⁻¹m⁻², which can be derived from the relationships discussed above.
In summary, the S.I. unit of permittivity of free space is indeed C²N⁻¹m⁻², reflecting how electric fields behave in a vacuum. This understanding is essential for further studies in electromagnetism and related fields.