Two similarly charged bodies are kept 5 cm apart in air. If the secnd body is shifted away from the first by another 5 cm, their force of repulsion will be

Options

A. doubled
B. halved
C. quadrupled
D. reduced to one-fourth

Explanation

To solve the problem, we will use Coulomb's Law, which states that the force \( F \) between two charged bodies is given by:

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

where:
- \( F \) is the force between the charges,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and
- \( r \) is the distance between the charges.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The initial distance \( r_1 \) between the two similarly charged bodies is 5 cm.

2. **Calculate the Initial Force**:
- Let’s denote the initial force as \( F_1 \). According to Coulomb's Law:
\[ F_1 = k \frac{|q_1 q_2|}{(5 \text{ cm})^2} \]

3. **Determine the New Distance**:
- The second body is shifted away by another 5 cm, making the new distance \( r_2 = 5 \text{ cm} + 5 \text{ cm} = 10 \text{ cm} \).

4. **Calculate the New Force**:
- Let’s denote the new force as \( F_2 \). Using Coulomb's Law again:
\[ F_2 = k \frac{|q_1 q_2|}{(10 \text{ cm})^2} \]

5. **Relate the Two Forces**:
- Now, we can relate \( F_1 \) and \( F_2 \):
\[ \frac{F_1}{F_2} = \frac{(5 \text{ cm})^2}{(10 \text{ cm})^2} = \frac{25}{100} = \frac{1}{4} \]
- This implies:
\[ F_2 = 4 F_1 \]

6. **Conclusion**:
- The force of repulsion \( F_2 \) when the second body is shifted away to 10 cm will be one-fourth of the original force \( F_1 \).

### Final Answer:
The force of repulsion will be reduced to one-fourth of the original force.