SI unit of `in_(0)` (permittivity of free space) is:

Options

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

Answer

To find the SI unit of \( \epsilon_0 \) (the permittivity of free space), we can derive it using Coulomb's Law. Here’s a step-by-step solution:

### Step 1: Write down Coulomb's Law
Coulomb's Law states that the force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Express Coulomb's constant in terms of \( \epsilon_0 \)
Coulomb's constant \( k \) can be expressed in terms of \( \epsilon_0 \):
\[ k = \frac{1}{4 \pi \epsilon_0} \]
Substituting this into Coulomb's Law gives:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]

### Step 3: Rearrange the equation to solve for \( \epsilon_0 \)
Rearranging the equation to isolate \( \epsilon_0 \):
\[ \epsilon_0 = \frac{Q_1 Q_2}{F \cdot 4 \pi r^2} \]
Since \( 4 \pi \) is a constant and does not have any dimensions, we can ignore it for the purpose of finding the SI unit.

### Step 4: Identify the SI units of each quantity
- The unit of charge \( Q \) is Coulombs (C).
- The unit of force \( F \) is Newtons (N).
- The unit of distance \( r \) is meters (m).

### Step 5: Substitute the units into the equation
Substituting the units into the rearranged equation:
\[ \epsilon_0 = \frac{C^2}{N \cdot m^2} \]

### Step 6: Write the final expression for the SI unit of \( \epsilon_0 \)
Thus, the SI unit of \( \epsilon_0 \) (permittivity of free space) is:
\[ \epsilon_0 \text{ has units of } \frac{C^2}{N \cdot m^2} \]

### Summary
The SI unit of \( \epsilon_0 \) is \( \frac{C^2}{N \cdot m^2} \).

---