The force of repulsion between two point charges is F, when these are at a distance of 1 m apart Now the point charges are replaced by spheres of radii 25 cm having the same charge as that of point charge and same distance apart. Then the new force of repulsion will

Options

A. Increase
B. decrease
C. Remain same
D. First increase then decrease

Answer

To solve the problem, we need to analyze the change in the force of repulsion when point charges are replaced by charged spheres.

### Step-by-step Solution:

1. **Understanding the Initial Force**:
The initial force of repulsion \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r = 1 \, \text{m} \) is given by Coulomb's law:
\[ F = \frac{k \cdot Q_1 \cdot Q_2}{r^2} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \).

2. **Replacing Point Charges with Spheres**:
The point charges are replaced by spheres of radius \( R = 25 \, \text{cm} = 0.25 \, \text{m} \). The distance between the centers of the spheres when they are 1 m apart is now the distance from the center of one sphere to the center of the other sphere.

3. **Calculating the New Distance**:
The new distance \( d \) between the centers of the two spheres is:
\[ d = 1 \, \text{m} + 0.25 \, \text{m} + 0.25 \, \text{m} = 1.5 \, \text{m} \]

4. **Calculating the New Force**:
The new force of repulsion \( F' \) between the two charged spheres is given by:
\[ F' = \frac{k \cdot Q_1 \cdot Q_2}{d^2} \]
Substituting \( d = 1.5 \, \text{m} \):
\[ F' = \frac{k \cdot Q_1 \cdot Q_2}{(1.5)^2} = \frac{k \cdot Q_1 \cdot Q_2}{2.25} \]

5. **Comparing the Forces**:
Now we compare \( F \) and \( F' \):
\[ F' = \frac{F}{2.25} \]
Since \( 2.25 > 1 \), we conclude that:
\[ F' < F \]
Therefore, the new force of repulsion \( F' \) is less than the original force \( F \).

### Conclusion:
The new force of repulsion will decrease when the point charges are replaced by spheres of the same charge.

### Final Answer:
The new force of repulsion will be \( F' < F \).

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