Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of `0.108 N` when separated by `50.0 cm`, centre-to-centre. A thin conducting wire then connects the spheres. When the wire is removed, the spheres repel each other with an electrostatic force of `0.0360 N`. What were the initial charges on the spheres?

Explanation

To solve the problem step by step, we will use Coulomb's law and the principle of conservation of charge.

### Step 1: Understand the Initial Attraction
We know that two identical conducting spheres attract each other with an electrostatic force of \( F_1 = 0.108 \, \text{N} \) when separated by a distance \( r = 50.0 \, \text{cm} = 0.5 \, \text{m} \). Let the charges on the spheres be \( Q_1 \) and \( Q_2 \). According to Coulomb's law, the force between two charges is given by:

\[ F = k \frac{|Q_1 Q_2|}{r^2} \]

Where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 2: Set Up the Equation for Initial Attraction
Substituting the known values into Coulomb's law:

\[ 0.108 = k \frac{|Q_1 Q_2|}{(0.5)^2} \]

This simplifies to:

\[ 0.108 = k \frac{|Q_1 Q_2|}{0.25} \]

Rearranging gives:

\[ |Q_1 Q_2| = 0.108 \times 0.25 / k \]

### Step 3: Calculate \( Q_1 Q_2 \)
Substituting \( k \):

\[ |Q_1 Q_2| = \frac{0.108 \times 0.25}{8.99 \times 10^9} \]

Calculating this gives:

\[ |Q_1 Q_2| = \frac{0.027}{8.99 \times 10^9} \approx 3 \times 10^{-12} \, \text{C}^2 \]

### Step 4: Understand the Effect of Shorting the Spheres
When the spheres are connected by a wire, they will share charge and reach the same potential. Let the new charge on each sphere be \( Q' \). The total charge is conserved:

\[ Q' + Q' = Q_1 + Q_2 \implies 2Q' = Q_1 + Q_2 \implies Q' = \frac{Q_1 + Q_2}{2} \]

### Step 5: Set Up the Equation for Repulsion After Shorting
After the wire is removed, the spheres repel each other with a force of \( F_2 = 0.0360 \, \text{N} \). Using Coulomb's law again:

\[ 0.0360 = k \frac{(Q')^2}{(0.5)^2} \]

This simplifies to:

\[ 0.0360 = k \frac{(Q')^2}{0.25} \]

Rearranging gives:

\[ (Q')^2 = 0.0360 \times 0.25 / k \]

### Step 6: Calculate \( (Q')^2 \)
Substituting \( k \):

\[ (Q')^2 = \frac{0.0360 \times 0.25}{8.99 \times 10^9} \]

Calculating this gives:

\[ (Q')^2 = \frac{0.009}{8.99 \times 10^9} \approx 1 \times 10^{-12} \, \text{C}^2 \]

### Step 7: Solve for \( Q' \)
Taking the square root gives:

\[ Q' = \sqrt{1 \times 10^{-12}} = 1 \times 10^{-6} \, \text{C} \]

### Step 8: Use Charge Conservation to Find \( Q_1 \) and \( Q_2 \)
From the charge conservation equation:

\[ Q_1 + Q_2 = 2Q' = 2 \times 1 \times 10^{-6} = 2 \times 10^{-6} \, \text{C} \]

### Step 9: Solve the System of Equations
We have two equations:

1. \( |Q_1 Q_2| = 3 \times 10^{-12} \)
2. \( Q_1 + Q_2 = 2 \times 10^{-6} \)

Let \( Q_2 = 2 \times 10^{-6} - Q_1 \) and substitute into the first equation:

\[ |Q_1(2 \times 10^{-6} - Q_1)| = 3 \times 10^{-12} \]

This is a quadratic equation in terms of \( Q_1 \). Solving this gives:

\[ Q_1 = 3 \times 10^{-6} \, \text{C} \quad \text{and} \quad Q_2 = -1 \times 10^{-6} \, \text{C} \]

### Final Result
The initial charges on the spheres are:

- \( Q_1 = 3 \times 10^{-6} \, \text{C} \)
- \( Q_2 = -1 \times 10^{-6} \, \text{C} \)