The unit of permittivity of free space `epsilon_(0)` is:

Options

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

Answer

To find the unit of permittivity of free space, denoted as \( \epsilon_0 \), we can start from the formula for the electrostatic force between two point charges. The formula is given by:

\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1 Q_2}{R^2} \]

Where:
- \( F \) is the electrostatic force,
- \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
- \( R \) is the distance between the charges,
- \( \epsilon_0 \) is the permittivity of free space.

### Step 1: Rearranging the formula to solve for \( \epsilon_0 \)

To isolate \( \epsilon_0 \), we can rearrange the formula:

\[ \epsilon_0 = \frac{Q_1 Q_2}{F} \cdot \frac{1}{4 \pi} \cdot R^2 \]

Since \( 4 \pi \) is a constant and does not have any units, we can ignore it for our unit calculation:

\[ \epsilon_0 = \frac{Q_1 Q_2}{F} \cdot R^2 \]

### Step 2: Identifying the units of each variable

- The unit of charge \( Q \) is the coulomb (C).
- The unit of force \( F \) is the newton (N).
- The unit of distance \( R \) is the meter (m).

### Step 3: Substituting the units into the equation

Substituting the units into the rearranged formula gives:

\[ \text{Unit of } \epsilon_0 = \frac{(C)(C)}{N} \cdot (m^2) \]

This simplifies to:

\[ \text{Unit of } \epsilon_0 = \frac{C^2}{N \cdot m^2} \]

### Step 4: Expressing newton in terms of base units

Recall that:

\[ 1 \text{ N} = 1 \text{ kg} \cdot \text{m/s}^2 \]

Thus, we can rewrite the unit of \( \epsilon_0 \):

\[ \text{Unit of } \epsilon_0 = \frac{C^2}{\text{kg} \cdot \text{m/s}^2 \cdot m^2} \]

This simplifies to:

\[ \text{Unit of } \epsilon_0 = \frac{C^2 \cdot s^2}{\text{kg} \cdot m^3} \]

### Final Answer

Therefore, the unit of permittivity of free space \( \epsilon_0 \) is:

\[ \epsilon_0 = \frac{C^2 \cdot s^2}{\text{kg} \cdot m^3} \]