In coulomb's law,the magnitude of K in `Nm ^(2) //C ^(2)` is

Options

A. `9xx 10 ^(5)`
B. `9xx 10 ^(11)`
C. `9xx 10 ^(3)`
D. none of these

Answer

To find the magnitude of \( K \) in Coulomb's law, we can follow these steps:

### Step 1: Understand 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 the proportionality constant.

### Step 2: Express \( K \) in terms of \( \epsilon_0 \)
The constant \( K \) can be expressed as:
\[ K = \frac{1}{4 \pi \epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Substitute the value of \( \epsilon_0 \)
The value of \( \epsilon_0 \) is approximately:
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \]

### Step 4: Calculate \( K \)
Now substituting the value of \( \epsilon_0 \) into the equation for \( K \):
\[ K = \frac{1}{4 \pi (8.85 \times 10^{-12})} \]
Using \( \pi \approx 3.14 \):
\[ K = \frac{1}{4 \times 3.14 \times 8.85 \times 10^{-12}} \]
Calculating the denominator:
\[ 4 \times 3.14 \approx 12.56 \]
\[ 12.56 \times 8.85 \approx 111.3 \times 10^{-12} \]
Thus,
\[ K \approx \frac{1}{111.3 \times 10^{-12}} \approx 9 \times 10^{11} \, \text{N m}^2/\text{C}^2 \]

### Step 5: Conclusion
Therefore, the magnitude of \( K \) in \( \text{N m}^2/\text{C}^2 \) is:
\[ K \approx 9 \times 10^{11} \, \text{N m}^2/\text{C}^2 \]

### Final Answer
The magnitude of \( K \) is \( 9 \times 10^{11} \, \text{N m}^2/\text{C}^2 \).

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