In MKS system of units `epsilon_(0)` equals -

Options

A. `9xx10^(9) N-m^(2)//C^(2)`
B. `(1)/(9xx10^(9)) C^(2)//N -m^(2)`
C. `(1)/(4pi xx 9 xx 10^(9)) C^(2)//N-m^(2)`
D. `1C^(2)//N-m^(2)`

Explanation

To determine the value of \( \epsilon_0 \) in the MKS system of units, we start from the formula for the electrostatic force between two point charges.

### Step-by-Step Solution:

1. **Understand the Electrostatic Force Formula**:
The electrostatic force \( F \) between two charges \( Q_1 \) and \( Q_2 \) separated by a distance \( R \) is given by:
\[ F = \frac{k Q_1 Q_2}{R^2} \]
where \( k \) is Coulomb's constant.

2. **Relate Coulomb's Constant to \( \epsilon_0 \)**:
Coulomb's constant \( k \) can be expressed in terms of \( \epsilon_0 \):
\[ k = \frac{1}{4 \pi \epsilon_0} \]
Substituting this into the force equation gives:
\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1 Q_2}{R^2} \]

3. **Rearranging to Find \( \epsilon_0 \)**:
Rearranging the equation to isolate \( \epsilon_0 \):
\[ \epsilon_0 = \frac{Q_1 Q_2}{4 \pi F R^2} \]

4. **Identifying Units**:
In the MKS system:
- The unit of charge \( Q \) is Coulombs (C).
- The unit of force \( F \) is Newtons (N).
- The unit of distance \( R \) is meters (m).

5. **Substituting Units**:
Substituting the units into the equation for \( \epsilon_0 \):
\[ \epsilon_0 = \frac{(C^2)}{(N)(m^2)} \]

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

7. **Final Expression**:
Thus, the unit of \( \epsilon_0 \) in the MKS system is:
\[ \epsilon_0 = \frac{C^2}{N \cdot m^2} \]

### Conclusion:
The value of \( \epsilon_0 \) in the MKS system of units is expressed as \( \frac{C^2}{N \cdot m^2} \).

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