Write down the value of obsolute permittivity of free space.

Options

A. `9 xx 10^(9) NC^(2)//m^(2)`
B. `8.85 xx 10^(-12) N-m^(2)//C^(2)s`
C. `8.85 xx 10^(-12) C^(2)//N-m^(2)`
D. `9 xx 10^(9)C^(2)//N - m^(2)`

Answer

To find the value of the absolute permittivity of free space (ε₀), we can follow these steps:

### Step 1: Understand the Concept
The absolute permittivity of free space (ε₀) is a fundamental physical constant that describes how electric fields interact with a vacuum. It is also related to Coulomb's law and the electrostatic force between charges.

### Step 2: Recall the Relationship with Coulomb's Law
Coulomb's law states that the force (F) between two point charges (Q₁ and Q₂) separated by a distance (r) is given by:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
where \( k \) is the Coulomb's constant.

### Step 3: Express Coulomb's Constant in Terms of ε₀
The Coulomb's constant \( k \) can also be expressed in terms of the permittivity of free space:
\[ k = \frac{1}{4 \pi \epsilon_0} \]

### Step 4: Substitute k into the Equation
From the above relationship, we can rearrange it to find ε₀:
\[ \epsilon_0 = \frac{1}{4 \pi k} \]

### Step 5: Use the Value of k
The value of Coulomb's constant \( k \) is approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \). Plugging this value into the equation gives:
\[ \epsilon_0 = \frac{1}{4 \pi \times 9 \times 10^9} \]

### Step 6: Calculate ε₀
Calculating this gives:
\[ \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/(\text{N m}^2) \]

### Step 7: Write the Final Answer
Thus, the value of the absolute permittivity of free space is:
\[ \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \]