Question

S.I. unit of permittivity is

• A. N\(^{-1}\) C\(^{-1}\) m\(^{2}\)
• B. NC\(^{2}\) m\(^{2}\)
• C. NC\(^{-2}\) m\(^{2}\)
• D. C\(^{2}\) N\(^{-1}\) m\(^{-2}\)

Hint:

Another common unit for permittivity is Farads per meter (F/m). You can verify that this is equivalent: \(1 \text{ F} = 1 \text{ C/V}\) and \(1 \text{ V} = 1 \text{ J/C} = 1 \text{ N} \cdot \text{m/C}\). Therefore, \(1 \text{ F} = 1 \text{ C}^2 / (\text{N} \cdot \text{m})\). Dividing by meters gives \(1 \text{ F/m} = 1 \text{ C}^2 / (\text{N} \cdot \text{m}^2)\). Remembering both units (F/m and C\(^2\) N\(^{-1}\) m\(^{-2}\)) can be helpful.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Permittivity (\(\epsilon\)) is a measure of how an electric field affects, and is affected by, a dielectric medium. The question asks for the SI unit of permittivity, often referring to the permittivity of free space, \(\epsilon_0\).
Step 2: Key Formula or Approach:
We can derive the unit of permittivity from Coulomb's Law, which describes the electrostatic force (\(F\)) between two point charges (\(q_1\) and \(q_2\)) separated by a distance (\(r\)).
\[ F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \] Step 3: Detailed Explanation:
First, we rearrange Coulomb's Law to solve for \(\epsilon_0\):
\[ \epsilon_0 = \frac{1}{4\pi F} \frac{q_1 q_2}{r^2} \] Now, we substitute the SI units for each physical quantity into the equation. The term \(4\pi\) is a dimensionless constant.
\begin{itemize} \item Unit of Force (\(F\)): Newton (N) \item Unit of Charge (\(q_1, q_2\)): Coulomb (C) \item Unit of Distance (\(r\)): meter (m) \end{itemize} Substituting these units:
\[ \text{Unit of } \epsilon_0 = \frac{1}{\text{N}} \cdot \frac{\text{C} \cdot \text{C}}{\text{m}^2} = \frac{\text{C}^2}{\text{N} \cdot \text{m}^2} \] This unit can be written in exponent notation as:
\[ C^2 N^{-1} m^{-2} \] Step 4: Final Answer:
The SI unit of permittivity is C\(^2\) N\(^{-1}\) m\(^{-2}\). This matches option (D).