The force of repulsion between two point charges is F, when they are d distance apart. If the point charges are replaced by conducted sphereas each of radius r and the charges remain same. The separation between the centre of sphere isd, then force of repulsion between them is

Options

A. A. Equal to F
B. B. Less than F
C. C. Greater than F
D. D. Cannot be sait

Answer

To solve the problem, we need to analyze the situation step by step:

### Step 1: Understand the Initial Condition
We start with two point charges, each with charge \( Q \), separated by a distance \( d \). The force of repulsion between these two point charges is given as \( F \).

**Hint:** Recall Coulomb's Law, which states that the force \( F \) between two point charges is given by:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
where \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.

### Step 2: Transition to Conducting Spheres
Now, we replace the point charges with conducting spheres, each of radius \( r \). The charges on the spheres remain the same, and the distance between the centers of the spheres is still \( d \).

**Hint:** Remember that for conducting spheres, the charges will distribute uniformly on the surface of the spheres.

### Step 3: Calculate the Effective Distance
When dealing with conducting spheres, the effective distance between the charges is not simply \( d \) but rather the distance between the surfaces of the spheres. Since each sphere has a radius \( r \), the distance between the surfaces of the spheres is:
\[ d' = d - 2r \]
where \( d' \) is the distance between the surfaces of the spheres.

**Hint:** This adjustment is necessary because the charges are now distributed over the surfaces of the spheres, and we need to account for their radii.

### Step 4: Apply Coulomb's Law Again
Using Coulomb's Law again for the conducting spheres, the new force of repulsion \( F' \) can be expressed as:
\[ F' = k \frac{Q^2}{(d - 2r)^2} \]

**Hint:** The formula remains the same, but ensure you substitute the correct effective distance.

### Step 5: Compare the Forces
Now, we need to compare \( F' \) with the original force \( F \):
- The original force was \( F = k \frac{Q^2}{d^2} \).
- The new force is \( F' = k \frac{Q^2}{(d - 2r)^2} \).

Since \( d - 2r < d \) (as long as \( r > 0 \)), it follows that:
\[ (d - 2r)^2 < d^2 \]
Thus, we can conclude:
\[ F' > F \]

### Conclusion
The force of repulsion between the two conducting spheres is greater than the force of repulsion between the two point charges.

**Final Answer:** The force of repulsion between the conducting spheres is greater than \( F \).