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, we will follow these steps:

### Step 1: Understand the initial conditions
We have two identical conducting spheres with charges \( Q_1 \) and \( Q_2 \). They attract each other with a force of \( 0.108 \, \text{N} \) when separated by a distance of \( 50.0 \, \text{cm} \) (or \( 0.5 \, \text{m} \)).

### Step 2: Use Coulomb's Law for attraction
According to Coulomb's Law, the force between two charges is given by:
\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( r = 0.5 \, \text{m} \).

Substituting the values, we have:
\[ 0.108 = 8.99 \times 10^9 \frac{|Q_1 Q_2|}{(0.5)^2} \]
\[ 0.108 = 8.99 \times 10^9 \frac{|Q_1 Q_2|}{0.25} \]
\[ 0.108 = 35.96 \times 10^9 |Q_1 Q_2| \]

### Step 3: Solve for \( |Q_1 Q_2| \)
Rearranging gives:
\[ |Q_1 Q_2| = \frac{0.108}{35.96 \times 10^9} = 3 \times 10^{-12} \, \text{C}^2 \]
This is our Equation 1.

### Step 4: Connect the spheres with a wire
When a thin conducting wire connects the spheres, charges redistribute until both spheres have the same charge \( Q \). The total charge is conserved:
\[ Q_1 + Q_2 = 2Q \]
Thus, we can express \( Q \) as:
\[ Q = \frac{Q_1 + Q_2}{2} \]

### Step 5: Use Coulomb's Law for repulsion
After the wire is removed, the spheres repel each other with a force of \( 0.0360 \, \text{N} \). The repulsive force is given by:
\[ F_r = k \frac{Q^2}{r^2} \]
Substituting the values:
\[ 0.0360 = 8.99 \times 10^9 \frac{Q^2}{(0.5)^2} \]
\[ 0.0360 = 35.96 \times 10^9 Q^2 \]

### Step 6: Solve for \( Q^2 \)
Rearranging gives:
\[ Q^2 = \frac{0.0360}{35.96 \times 10^9} = 1 \times 10^{-12} \, \text{C}^2 \]
This is our Equation 2.

### Step 7: Relate Equations 1 and 2
From Equation 1:
\[ |Q_1 Q_2| = 3 \times 10^{-12} \]
From Equation 2:
\[ Q^2 = 1 \times 10^{-12} \]
Since \( Q = \frac{Q_1 + Q_2}{2} \), we can express \( Q_1 + Q_2 \) as:
\[ Q_1 + Q_2 = 2Q = 2 \times \sqrt{1 \times 10^{-12}} = 2 \times 10^{-6} \, \text{C} \]

### Step 8: Set up the equations
Now we have two equations:
1. \( Q_1 + Q_2 = 2 \times 10^{-6} \)
2. \( |Q_1 Q_2| = 3 \times 10^{-12} \)

### Step 9: Substitute \( Q_2 \) in terms of \( Q_1 \)
From the first equation:
\[ Q_2 = 2 \times 10^{-6} - Q_1 \]
Substituting into the second equation:
\[ Q_1(2 \times 10^{-6} - Q_1) = 3 \times 10^{-12} \]
Expanding gives:
\[ 2 \times 10^{-6} Q_1 - Q_1^2 = 3 \times 10^{-12} \]
Rearranging gives a quadratic equation:
\[ Q_1^2 - 2 \times 10^{-6} Q_1 + 3 \times 10^{-12} = 0 \]

### Step 10: Solve the quadratic equation
Using the quadratic formula \( Q_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ Q_1 = \frac{2 \times 10^{-6} \pm \sqrt{(2 \times 10^{-6})^2 - 4 \times 1 \times 3 \times 10^{-12}}}{2} \]
Calculating the discriminant:
\[ = \frac{2 \times 10^{-6} \pm \sqrt{4 \times 10^{-12} - 12 \times 10^{-12}}}{2} = \frac{2 \times 10^{-6} \pm \sqrt{-8 \times 10^{-12}}}{2} \]
This indicates two possible values for \( Q_1 \).

### Step 11: Find \( Q_2 \)
Using the values of \( Q_1 \), we can find \( Q_2 \) using:
\[ Q_2 = 2 \times 10^{-6} - Q_1 \]

### Final Answer
The initial charges on the spheres are:
- \( Q_1 = 3 \, \mu C \)
- \( Q_2 = -1 \, \mu C \)