SI unit of permittivity is

Options

A. `C^(2)m^(2)N^(-1)`
B. `C^(-1)m^(2)N^(-2)`
C. `C^(2)m^(2)N^(2)`
D. `C^(2)m^(-2)N^(-1)`

Answer

To find the SI unit of permittivity, we can follow these steps:

### Step 1: Understand the Concept of Permittivity
Permittivity is a measure of how much electric field is generated per unit charge in a medium. The permittivity of free space is denoted as \( \varepsilon_0 \).

### Step 2: Relate Permittivity to Electrostatic Force
The electrostatic force \( F \) between two charges \( Q \) separated by a distance \( R \) is given by Coulomb's law:
\[ F = \frac{k Q^2}{R^2} \]
where \( k \) is Coulomb's constant.

### Step 3: Express Coulomb's Constant in Terms of Permittivity
Coulomb's constant \( k \) can be expressed in terms of permittivity:
\[ k = \frac{1}{4 \pi \varepsilon_0} \]
Substituting this into the equation for force gives:
\[ F = \frac{Q^2}{4 \pi \varepsilon_0 R^2} \]

### Step 4: Rearranging for Permittivity
From the rearranged equation, we can express \( \varepsilon_0 \):
\[ \varepsilon_0 = \frac{Q^2}{4 \pi F R^2} \]

### Step 5: Identify the Units of Each Quantity
- The unit of charge \( Q \) is Coulombs (C).
- The unit of distance \( R \) is meters (m).
- The unit of force \( F \) is Newtons (N).

### Step 6: Substitute the Units into the Permittivity Equation
Substituting the units into the equation for permittivity:
\[ \varepsilon_0 = \frac{(C)^2}{N \cdot (m^2)} \]

### Step 7: Express Newton in Terms of Base SI Units
Recall that \( 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \). Therefore:
\[ \varepsilon_0 = \frac{C^2}{(kg \cdot m/s^2) \cdot m^2} = \frac{C^2 \cdot s^2}{kg \cdot m^3} \]

### Final Step: Write the SI Unit of Permittivity
Thus, the SI unit of permittivity \( \varepsilon_0 \) is:
\[ \text{C}^2 \cdot \text{m}^{-2} \cdot \text{N}^{-1} \]

### Conclusion
The SI unit of permittivity is \( \text{C}^2 \cdot \text{m}^{-2} \cdot \text{N}^{-1} \).

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