Two point charges separated by a distance d repel each other with a force of 9N. If the separation between them becomes 3d, the force of repulsion will be

Options

A. 1N
B. 3N
C. 6N
D. 27N

Answer

To solve the problem, we will use Coulomb's law, which states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by the formula:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

### Step 1: Identify the initial conditions
We know that the initial force \( F \) between the charges separated by a distance \( d \) is 9 N. Therefore, we can write:

\[ F = 9 \, \text{N} \]

### Step 2: Write the expression for the initial force
Using Coulomb's law for the initial distance \( d \):

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2} \]

### Step 3: Determine the new distance
The new distance between the charges is given as \( 3d \).

### Step 4: Write the expression for the new force
Using Coulomb's law for the new distance \( 3d \):

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(3d)^2} \]

### Step 5: Simplify the expression for the new force
Substituting \( (3d)^2 \) into the equation:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{9d^2} \]

### Step 6: Relate the new force to the initial force
Now, we can relate \( F' \) to \( F \):

\[ F' = \frac{1}{9} \cdot \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2} = \frac{F}{9} \]

### Step 7: Substitute the value of the initial force
Since \( F = 9 \, \text{N} \):

\[ F' = \frac{9}{9} = 1 \, \text{N} \]

### Conclusion
The new force of repulsion when the distance becomes \( 3d \) is:

\[ F' = 1 \, \text{N} \]

### Final Answer
The force of repulsion will be **1 N**.

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