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:

\[ F = k \frac{q_1 q_2}{r^2} \]

where \( k \) is Coulomb's constant.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- The initial force \( F_1 = 9 \, \text{N} \)
- The initial distance \( d \)

According to Coulomb's Law, we can write:
\[ F_1 = k \frac{q_1 q_2}{d^2} = 9 \, \text{N} \]

2. **Determine the new conditions:**
- The new distance is \( 3d \).
- We need to find the new force \( F_2 \) when the distance is increased to \( 3d \).

3. **Apply Coulomb's Law for the new distance:**
\[ F_2 = k \frac{q_1 q_2}{(3d)^2} \]

4. **Simplify the expression for \( F_2 \):**
\[ F_2 = k \frac{q_1 q_2}{9d^2} \]

5. **Relate \( F_2 \) to \( F_1 \):**
Since we know from the first condition that:
\[ F_1 = k \frac{q_1 q_2}{d^2} = 9 \, \text{N} \]
We can substitute this into the equation for \( F_2 \):
\[ F_2 = \frac{1}{9} \cdot k \frac{q_1 q_2}{d^2} = \frac{1}{9} \cdot F_1 \]

6. **Calculate \( F_2 \):**
\[ F_2 = \frac{1}{9} \cdot 9 \, \text{N} = 1 \, \text{N} \]

### Final Answer:
The force of repulsion when the distance becomes \( 3d \) is \( 1 \, \text{N} \).

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