Two identical conducting sphere carrying different charges attact each other with a force F when placed in air medium at a distance `d` apart. The spheres are brought into contact and then taken to their original positions. Now, the two sphere repel each other with a force whole magnitude is equal to the initial attractive force. The ratio between initial charges on the spheres is

Options

A. `- (3 + sqrt8)` only
B. `-3 + sqrt8` only
C. `-(3 + sqrt8)` or `(-3 + sqrt8)`
D. `+ sqrt3`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the Initial Situation
We have two identical conducting spheres with charges \( Q_1 \) and \( Q_2 \) that attract each other with a force \( F \) when placed at a distance \( d \) apart. Since they attract each other, we can assume one charge is positive and the other is negative.

### Step 2: Apply Coulomb's Law
The electrostatic force of attraction between the two charges can be expressed using Coulomb's law:
\[ F = k \frac{|Q_1 Q_2|}{d^2} \]
where \( k \) is Coulomb's constant. This is our first equation.

### Step 3: Bring the Spheres into Contact
When the spheres are brought into contact, they share their charges. The total charge after contact is:
\[ Q_{total} = Q_1 + Q_2 \]
After contact, each sphere will have the same charge, which is:
\[ Q' = \frac{Q_1 + Q_2}{2} \]

### Step 4: Calculate the New Force After Contact
After being separated back to distance \( d \), both spheres now have the same charge \( Q' \). The force of repulsion between them can be expressed as:
\[ F' = k \frac{(Q')^2}{d^2} = k \frac{\left(\frac{Q_1 + Q_2}{2}\right)^2}{d^2} \]
This is our second equation.

### Step 5: Set the Forces Equal
According to the problem, the magnitude of the repulsive force \( F' \) is equal to the initial attractive force \( F \):
\[ F' = F \]
Substituting the expressions for \( F \) and \( F' \):
\[ k \frac{\left(\frac{Q_1 + Q_2}{2}\right)^2}{d^2} = k \frac{|Q_1 Q_2|}{d^2} \]
We can cancel \( k \) and \( d^2 \) from both sides:
\[ \left(\frac{Q_1 + Q_2}{2}\right)^2 = |Q_1 Q_2| \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ \frac{(Q_1 + Q_2)^2}{4} = |Q_1 Q_2| \]
This leads to:
\[ (Q_1 + Q_2)^2 = 4 |Q_1 Q_2| \]

### Step 7: Substitute \( Q_2 = -k Q_1 \)
Assuming \( Q_2 = -k Q_1 \) (since one charge is negative), we can substitute:
\[ (Q_1 - k Q_1)^2 = 4 |Q_1 (-k Q_1)| \]
This simplifies to:
\[ (1 - k)^2 Q_1^2 = 4k Q_1^2 \]
Dividing by \( Q_1^2 \) (assuming \( Q_1 \neq 0 \)):
\[ (1 - k)^2 = 4k \]

### Step 8: Solve for \( k \)
Expanding and rearranging gives:
\[ 1 - 2k + k^2 = 4k \]
\[ k^2 - 6k + 1 = 0 \]
Using the quadratic formula:
\[ k = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2} \]

### Step 9: Find the Ratio of Charges
The ratio of the initial charges \( \frac{Q_1}{Q_2} \) can be expressed as:
\[ \frac{Q_1}{Q_2} = \frac{1}{-k} = \frac{1}{-(3 \pm 2\sqrt{2})} \]

### Final Answer
Thus, the ratio of the initial charges on the spheres is:
\[ \frac{Q_1}{Q_2} = -\frac{1}{3 \pm 2\sqrt{2}} \]