The centres of two identical small conducting sphere are 1 m apart. They carry charge of opposite kind and attract each other with a force F. when they connected by conducting thin wire they repel each other with a force `F//3`. The ratio of magnitude of charges carried by the spheres initially in `n : 1`. Find value of n

Answer

To solve the problem, we will follow these steps systematically:

### Step 1: Understand the Initial Setup
We have two identical small conducting spheres with charges \( +q_1 \) and \( -q_2 \) respectively, separated by a distance of 1 meter. They attract each other with a force \( F \).

### Step 2: Write the Expression for the Attractive Force
The attractive force \( F \) between the two charges can be expressed using Coulomb's law:
\[ F = k \frac{q_1 q_2}{r^2} \]
Given that \( r = 1 \) meter, we can simplify this to:
\[ F = k q_1 q_2 \quad \text{(Equation 1)} \]

### Step 3: Connect the Spheres with a Wire
When the spheres are connected by a conducting wire, they share their charges equally because they are identical spheres. The total charge after connecting is:
\[ Q_{\text{total}} = q_1 - q_2 \]
Each sphere will then have a charge of:
\[ q' = \frac{q_1 - q_2}{2} \]

### Step 4: Write the Expression for the Repulsive Force
After connecting the spheres, they repel each other with a force of \( \frac{F}{3} \). The repulsive force can also be expressed using Coulomb's law:
\[ \frac{F}{3} = k \frac{(q')^2}{r^2} \]
Substituting \( r = 1 \) meter and \( q' = \frac{q_1 - q_2}{2} \), we get:
\[ \frac{F}{3} = k \left(\frac{q_1 - q_2}{2}\right)^2 \]
This simplifies to:
\[ \frac{F}{3} = k \frac{(q_1 - q_2)^2}{4} \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 1 into Equation 2
From Equation 1, we have:
\[ F = k q_1 q_2 \]
Substituting this into Equation 2 gives:
\[ \frac{k q_1 q_2}{3} = k \frac{(q_1 - q_2)^2}{4} \]
We can cancel \( k \) from both sides (assuming \( k \neq 0 \)):
\[ \frac{q_1 q_2}{3} = \frac{(q_1 - q_2)^2}{4} \]

### Step 6: Cross Multiply and Rearrange
Cross multiplying gives:
\[ 4 q_1 q_2 = 3 (q_1 - q_2)^2 \]
Expanding the right side:
\[ 4 q_1 q_2 = 3 (q_1^2 - 2q_1 q_2 + q_2^2) \]
This simplifies to:
\[ 4 q_1 q_2 = 3 q_1^2 - 6 q_1 q_2 + 3 q_2^2 \]
Rearranging gives:
\[ 3 q_1^2 + 3 q_2^2 - 10 q_1 q_2 = 0 \]

### Step 7: Divide by \( q_2^2 \)
Let \( x = \frac{q_1}{q_2} \). Then we can rewrite the equation as:
\[ 3 x^2 - 10 x + 3 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
Calculating the discriminant:
\[ x = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \]
This gives us two possible solutions:
\[ x = \frac{18}{6} = 3 \quad \text{or} \quad x = \frac{2}{6} = \frac{1}{3} \]

### Step 9: Determine the Value of \( n \)
Since the problem states that the ratio of the charges is \( n:1 \), we take the integer solution:
\[ n = 3 \]

### Final Answer
The value of \( n \) is \( 3 \).

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