Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting spheres, C,is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to :

Options

A. F
B. `(3F)/(8)`
C. `(3F)/(4)`
D. `(F)/(2)`

Answer

To solve the problem step by step, we will analyze the situation involving the three identical conducting spheres A, B, and C.

### Step 1: Initial Charges on Spheres A and B
Let the charge on sphere A be \( Q \) and the charge on sphere B also be \( Q \). The force \( F \) between them can be expressed using Coulomb's law:
\[ F = k \frac{Q^2}{R^2} \]
where \( k \) is Coulomb's constant and \( R \) is the distance between the centers of the spheres.

**Hint for Step 1:** Remember that the force between two charges is given by Coulomb's law.

### Step 2: Touching Sphere C to Sphere A
Sphere C is initially uncharged. When it touches sphere A, the total charge \( Q \) on sphere A is shared equally between A and C. After touching:
- Charge on A becomes \( \frac{Q}{2} \)
- Charge on C becomes \( \frac{Q}{2} \)

**Hint for Step 2:** When two identical conductors touch, the charge distributes equally between them.

### Step 3: Touching Sphere C to Sphere B
Next, sphere C (which now has a charge of \( \frac{Q}{2} \)) is brought into contact with sphere B. The total charge on sphere B before contact is \( Q \). When they touch, the total charge \( Q + \frac{Q}{2} = \frac{3Q}{2} \) is shared equally:
- Charge on B becomes \( \frac{3Q}{4} \)
- Charge on C becomes \( \frac{3Q}{4} \)

**Hint for Step 3:** Again, remember that the charge distributes equally between two identical conductors when they touch.

### Step 4: Final Charges on Spheres A and B
After the two contacts, the charges on the spheres are:
- Charge on A: \( \frac{Q}{2} \)
- Charge on B: \( \frac{3Q}{4} \)

**Hint for Step 4:** Keep track of the charges after each contact to ensure accuracy.

### Step 5: Calculating the New Force Between A and B
Now, we can calculate the new force \( F' \) between spheres A and B using the new charges:
\[ F' = k \frac{\left(\frac{Q}{2}\right) \left(\frac{3Q}{4}\right)}{R^2} = k \frac{3Q^2}{8R^2} \]

### Step 6: Relating the New Force to the Original Force
From the original force \( F = k \frac{Q^2}{R^2} \), we can express the new force in terms of the original force:
\[ F' = \frac{3}{8} F \]

### Conclusion
Thus, the new force between spheres A and B after the interactions with sphere C is:
\[ F' = \frac{3}{8} F \]

**Final Answer:** The force between A and B would be equal to \( \frac{3}{8} F \).