Two charged particles placed at a separation of 20 cm exert 20 N of coulomb force on each other. What will be the force if the separation is increased to 25 cm?

Explanation

To solve the problem, we will use Coulomb's Law, which states that the force between two charged particles is inversely proportional to the square of the distance between them.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial separation, \( R_1 = 20 \) cm
- Initial force, \( F_1 = 20 \) N
- New separation, \( R_2 = 25 \) cm

2. **Coulomb's Law**:
The formula for the force between two charges is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
where \( F \) is the force, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the separation distance, and \( \epsilon_0 \) is the permittivity of free space.

3. **Set Up the Ratios**:
Since the charges remain constant, we can set up the ratio of the forces at the two separations:
\[ \frac{F_2}{F_1} = \frac{R_1^2}{R_2^2} \]

4. **Substitute Known Values**:
We know \( F_1 = 20 \) N, \( R_1 = 20 \) cm, and \( R_2 = 25 \) cm. Plugging these values into the equation gives:
\[ \frac{F_2}{20} = \frac{(20)^2}{(25)^2} \]

5. **Calculate the Right Side**:
Calculate \( (20)^2 = 400 \) and \( (25)^2 = 625 \):
\[ \frac{F_2}{20} = \frac{400}{625} \]

6. **Simplify the Fraction**:
Simplifying \( \frac{400}{625} \):
\[ \frac{400}{625} = \frac{16}{25} \]

7. **Find \( F_2 \)**:
Now, multiply both sides by 20 N:
\[ F_2 = 20 \times \frac{16}{25} \]

8. **Calculate \( F_2 \)**:
\[ F_2 = \frac{320}{25} = 12.8 \text{ N} \]

### Final Answer:
The force when the separation is increased to 25 cm is \( F_2 = 12.8 \) N.

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