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, we need to analyze the situation step by step.

### Step 1: Understand the Initial Setup
We have two identical small conducting spheres, separated by a distance of 1 meter. Let the charges on the spheres be \( Q_1 \) and \( -Q_2 \) (since they carry opposite charges). The force of attraction between them is given as \( F \).

### Step 2: Apply Coulomb's Law
According to Coulomb's law, the force \( F \) between two charges is given by:
\[ F = \frac{k |Q_1 \cdot (-Q_2)|}{r^2} \]
where \( k \) is Coulomb's constant and \( r \) is the distance between the centers of the spheres (1 meter in this case). Thus, we can write:
\[ F = \frac{k Q_1 Q_2}{1^2} = k Q_1 Q_2 \quad \text{(Equation 1)} \]

### Step 3: Connect the Spheres with a Wire
When the two spheres are connected by a conducting wire, charges will redistribute until both spheres have the same potential. Let the final charges on the spheres be \( Q_1' \) and \( Q_2' \). Since they repel each other with a force of \( \frac{F}{3} \), we can express this force as:
\[ \frac{F}{3} = \frac{k Q_1' Q_2'}{1^2} = k Q_1' Q_2' \quad \text{(Equation 2)} \]

### Step 4: Conservation of Charge
The total charge before connecting the spheres is conserved. Thus, we have:
\[ Q_1' + Q_2' = Q_1 - Q_2 \quad \text{(Equation 3)} \]

### Step 5: Equal Potential Condition
Since the spheres are identical, the potentials must be equal after connecting them:
\[ \frac{k Q_1'}{r} = \frac{k Q_2'}{r} \]
This implies:
\[ Q_1' = Q_2' \quad \text{(Equation 4)} \]

### Step 6: Substitute Equation 4 into Equation 3
From Equation 4, we can substitute \( Q_1' \) with \( Q_2' \) in Equation 3:
\[ 2Q_1' = Q_1 - Q_2 \implies Q_1' = \frac{Q_1 - Q_2}{2} \]

### Step 7: Substitute \( Q_1' \) into Equation 2
Now substituting \( Q_1' \) into Equation 2:
\[ \frac{F}{3} = k \left(\frac{Q_1 - Q_2}{2}\right) \left(\frac{Q_1 - Q_2}{2}\right) = k \frac{(Q_1 - Q_2)^2}{4} \]
Thus:
\[ F = \frac{k (Q_1 - Q_2)^2}{4} \cdot 3 = \frac{3k (Q_1 - Q_2)^2}{4} \]

### Step 8: Relate Equations 1 and 7
From Equation 1, we have \( F = k Q_1 Q_2 \). Setting the two expressions for \( F \) equal gives:
\[ k Q_1 Q_2 = \frac{3k (Q_1 - Q_2)^2}{4} \]
We can cancel \( k \) from both sides (assuming \( k \neq 0 \)):
\[ Q_1 Q_2 = \frac{3 (Q_1 - Q_2)^2}{4} \]

### Step 9: Rearranging and Solving for the Ratio
Expanding and rearranging gives:
\[ 4Q_1 Q_2 = 3(Q_1^2 - 2Q_1 Q_2 + Q_2^2) \]
This simplifies to:
\[ 3Q_1^2 - 10Q_1 Q_2 + 3Q_2^2 = 0 \]

### Step 10: Solve the Quadratic Equation
Using the quadratic formula \( Q_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = -10Q_2 \), and \( c = 3Q_2^2 \).
\[ Q_1 = \frac{10Q_2 \pm \sqrt{(-10Q_2)^2 - 4 \cdot 3 \cdot 3Q_2^2}}{2 \cdot 3} \]
\[ Q_1 = \frac{10Q_2 \pm \sqrt{100Q_2^2 - 36Q_2^2}}{6} \]
\[ Q_1 = \frac{10Q_2 \pm \sqrt{64Q_2^2}}{6} \]
\[ Q_1 = \frac{10Q_2 \pm 8Q_2}{6} \]
This gives two possible values:
1. \( Q_1 = \frac{18Q_2}{6} = 3Q_2 \)
2. \( Q_1 = \frac{2Q_2}{6} = \frac{Q_2}{3} \)

### Step 11: Determine the Ratios
Thus, the ratios of \( Q_1 \) to \( Q_2 \) can be:
1. \( \frac{Q_1}{Q_2} = 3:1 \)
2. \( \frac{Q_1}{Q_2} = 1:3 \)

### Final Answer
The possible ratios of the magnitudes of charges carried by the spheres initially are \( 3:1 \) and \( 1:3 \).

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