Find the dimensions and units of `epsilon_0`

Explanation

To find the dimensions and units of \( \epsilon_0 \) (the permittivity of free space), we can start from Coulomb's law, which relates the electrostatic force between two charges to their magnitudes and the distance between them.

### Step-by-Step Solution:

1. **Coulomb's Law**:
The formula for the electrostatic force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

2. **Rearranging for \( \epsilon_0 \)**:
We can rearrange this equation to express \( \epsilon_0 \):
\[ \epsilon_0 = \frac{q_1 q_2}{4 \pi F r^2} \]

3. **Identifying Units**:
- 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).

Substituting these units into the equation for \( \epsilon_0 \):
\[ \text{Units of } \epsilon_0 = \frac{(C)^2}{N \cdot m^2} \]

4. **Expressing Newton in Base Units**:
Recall that \( 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \). Thus, we can express the units of \( \epsilon_0 \) as:
\[ \text{Units of } \epsilon_0 = \frac{C^2}{\text{kg} \cdot \text{m/s}^2 \cdot m^2} = \frac{C^2 \cdot s^2}{\text{kg} \cdot m^3} \]

5. **Finding Dimensions**:
- The dimension of charge \( q \) is given by \( [I][T] \) where \( I \) is the current (Ampere) and \( T \) is time (seconds).
- Therefore, the dimension of charge squared is \( [I^2][T^2] \).
- The dimension of force \( F \) is \( [M][L][T^{-2}] \).
- The dimension of distance \( r \) is \( [L] \).

Now substituting these dimensions into the equation:
\[ \epsilon_0 = \frac{[I^2][T^2]}{[M][L][T^{-2}][L^2]} = \frac{[I^2][T^2]}{[M][L^3][T^{-2}]} \]

Simplifying this gives:
\[ \epsilon_0 = [M^{-1}][L^{-3}][T^4][I^2] \]

### Final Results:
- **Units of \( \epsilon_0 \)**: \( \frac{C^2}{N \cdot m^2} \) or equivalently \( \frac{C^2 \cdot s^2}{kg \cdot m^3} \)
- **Dimensions of \( \epsilon_0 \)**: \( [M^{-1}][L^{-3}][T^4][I^2] \)