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 find the value of \( \epsilon_0 \) in the MKS (Meter-Kilogram-Second) system of units, we can follow these steps:

### Step 1: Understand the relationship between force, charge, and distance
The electric force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's law:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Substitute known values
We are given that both charges \( q_1 \) and \( q_2 \) are equal to 1 coulomb, and the distance \( r \) is 1 meter. Thus, we can substitute these values into the equation:
\[ F = k \frac{(1 \, \text{C})(1 \, \text{C})}{(1 \, \text{m})^2} \]
This simplifies to:
\[ F = k \]

### Step 3: Use the known value of the force
It is known that the force between two 1 coulomb charges separated by 1 meter is approximately \( 9 \times 10^9 \, \text{N} \). Therefore, we can set:
\[ k = 9 \times 10^9 \, \text{N} \]

### Step 4: Relate \( k \) to \( \epsilon_0 \)
Coulomb's constant \( k \) is also related to \( \epsilon_0 \) by the equation:
\[ k = \frac{1}{4 \pi \epsilon_0} \]
Substituting the value of \( k \) we found:
\[ 9 \times 10^9 = \frac{1}{4 \pi \epsilon_0} \]

### Step 5: Solve for \( \epsilon_0 \)
Rearranging the equation to solve for \( \epsilon_0 \):
\[ \epsilon_0 = \frac{1}{4 \pi (9 \times 10^9)} \]

### Step 6: Calculate the value
Now we can calculate \( \epsilon_0 \):
\[ \epsilon_0 = \frac{1}{36 \pi \times 10^9} \]

### Final Answer
Thus, the value of \( \epsilon_0 \) in the MKS system is:
\[ \epsilon_0 = \frac{1}{36 \pi \times 10^9} \, \text{F/m} \]