The centres of two identical small conducting spheres are 1m apart. They carry charge of opposite kind and attract each other with a force F. When they are connected by a conducting thin wire they repel each other with a force `F//3`. What is the ratio of magnitude of charge carried by the sphere initially ?

Options

A. `1 : 1`
B. `2 : 1`
C. `3 : 1`
D. `4 : 1`

Answer

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

### Step 1: Define the Charges
Let the charges on the two identical small conducting spheres be \( Q_1 \) and \( Q_2 \). Since they carry opposite charges, we can assume:
- \( Q_1 = +q \) (positive charge)
- \( Q_2 = -p \) (negative charge)

### Step 2: Use Coulomb's Law for Initial Attraction
According to Coulomb's law, the force \( F \) between the two charges is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1 \cdot Q_2}{r^2} \]
Since the distance \( r = 1 \, \text{m} \), we can write:
\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q \cdot (-p)}{1^2} = -\frac{qp}{4 \pi \epsilon_0} \]
Since \( F \) is positive (attraction), we can ignore the negative sign:
\[ F = \frac{qp}{4 \pi \epsilon_0} \]

### Step 3: Connecting the Spheres with a Wire
When the spheres are connected by a conducting wire, charge redistributes equally because they are identical. The new charges on both spheres will be:
\[ Q_1' = \frac{q - p}{2}, \quad Q_2' = \frac{-p + q}{2} \]

### Step 4: Use Coulomb's Law for Repulsion
After connecting the spheres, they repel each other with a force of \( \frac{F}{3} \):
\[ \frac{F}{3} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1' \cdot Q_2'}{1^2} \]
Substituting \( Q_1' \) and \( Q_2' \):
\[ \frac{F}{3} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\left(\frac{q - p}{2}\right) \left(\frac{-p + q}{2}\right)}{1} \]
This simplifies to:
\[ \frac{F}{3} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(q - p)(q - p)}{4} \]
Thus:
\[ F = \frac{(q - p)^2}{12 \pi \epsilon_0} \]

### Step 5: Equate the Two Expressions for F
From the two expressions for \( F \):
1. \( F = \frac{qp}{4 \pi \epsilon_0} \)
2. \( F = \frac{(q - p)^2}{12 \pi \epsilon_0} \)

Setting them equal gives:
\[ \frac{qp}{4 \pi \epsilon_0} = \frac{(q - p)^2}{12 \pi \epsilon_0} \]
Multiplying through by \( 12 \pi \epsilon_0 \):
\[ 3qp = (q - p)^2 \]

### Step 6: Expand and Rearrange
Expanding the right-hand side:
\[ 3qp = q^2 - 2qp + p^2 \]
Rearranging gives:
\[ q^2 - 5qp + p^2 = 0 \]

### Step 7: Solve the Quadratic Equation
Let \( x = \frac{q}{p} \). Then \( q = xp \):
\[ (x^2 - 5x + 1) p^2 = 0 \]
This implies:
\[ x^2 - 5x + 1 = 0 \]

### Step 8: Use the Quadratic Formula
Using the quadratic formula:
\[ x = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2} \]

### Step 9: Find the Ratio
The possible values of \( x \) give us the ratio of the charges:
- The ratio \( \frac{q}{p} \) can be either \( \frac{5 + \sqrt{21}}{2} \) or \( \frac{5 - \sqrt{21}}{2} \).

### Final Answer
The ratio of the magnitudes of the charges carried by the spheres initially is:
\[ \text{Ratio} = 3:1 \text{ or } 1:3 \]