The speed of light in vacuum, c, depends on two fundamental constants, the permeability of free space, `mu_(0)` and the permittivity of free space, `epsilon_(0)`. The speed of light is given by `c=(1)/(sqrt(mu_(0)epsilon_(0)))`. The units for `mu_(0)` are

Options

A. `kg^(-1)m^(-1)C^(2)`
B. `kgmC^(-2)`
C. `kg ms^(-4)C^(-2)`
D. `kg^(-1)s^(-3)C^(-2)`

Answer

To find the units of the permeability of free space, \( \mu_0 \), we start from the relationship given for the speed of light in vacuum:

\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]

### Step 1: Rearranging the equation
First, we can rearrange the equation to express \( \mu_0 \) in terms of \( c \) and \( \epsilon_0 \):

\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]

This implies:

\[ \mu_0 = \frac{1}{c^2 \epsilon_0} \]

### Step 2: Finding the units of \( \epsilon_0 \)
Next, we need to find the units of \( \epsilon_0 \). We can use the formula from electrostatics:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{r^2} \]

Where:
- \( F \) is force (units: \( \text{kg} \cdot \text{m/s}^2 \))
- \( q \) is charge (units: coulombs, \( C \))
- \( r \) is distance (units: meters, \( m \))

Rearranging gives:

\[ \epsilon_0 = \frac{1}{4 \pi} \frac{q^2}{F r^2} \]

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

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

This simplifies to:

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

### Step 4: Substituting \( \epsilon_0 \) back into \( \mu_0 \)
Now we substitute this expression for \( \epsilon_0 \) back into the equation for \( \mu_0 \):

\[ \mu_0 = \frac{1}{c^2 \cdot \epsilon_0} = \frac{1}{c^2} \cdot \frac{\text{kg} \cdot m^3 \cdot s^{-2}}{C^2} \]

### Step 5: Finding the units of \( c \)
The speed of light \( c \) has units of:

\[ c = \text{m/s} \]

Thus,

\[ c^2 = \text{m}^2/\text{s}^2 \]

### Step 6: Final expression for \( \mu_0 \)
Substituting \( c^2 \) into the equation for \( \mu_0 \):

\[ \mu_0 = \frac{\text{kg} \cdot m^3 \cdot s^{-2}}{C^2 \cdot \text{m}^2/s^2} \]

This simplifies to:

\[ \mu_0 = \frac{\text{kg}}{\text{m} \cdot s^2} \cdot C^2 \]

### Final Units of \( \mu_0 \)
Thus, the units of the permeability of free space \( \mu_0 \) are:

\[ \mu_0 = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2} \cdot \text{C}^2 \]