Two identical particles are charged and held at a distance of 1m form each pther. They are found to be attraction each other with a force of 0.027N . Now they are connected by a conducting wire so that charge folws between them. Whe the charge flow stop, they are found to be repelling each other with a force of 0.009N. Find the initial charge on each particle.

Explanation

To solve the problem, we need to determine the initial charges on two identical particles that initially attract each other and then repel after being connected by a conducting wire. We will use Coulomb's law to find the charges.

### Step 1: Understand the Initial Conditions
Initially, the two particles attract each other with a force of \( F_1 = 0.027 \, \text{N} \). Since they attract, one charge must be positive and the other must be negative. Let’s denote the charges as \( Q_1 \) and \( Q_2 \).

### Step 2: Apply Coulomb's Law for Attraction
Coulomb's law states that the force between two charges is given by:

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

Here, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( r = 1 \, \text{m} \). Since the force is attractive, we can write:

\[ 0.027 = 9 \times 10^9 \frac{|Q_1 Q_2|}{1^2} \]

This simplifies to:

\[ |Q_1 Q_2| = \frac{0.027}{9 \times 10^9} \]

Calculating this gives:

\[ |Q_1 Q_2| = 3 \times 10^{-12} \, \text{C}^2 \tag{1} \]

### Step 3: Understand the Conditions After Connecting the Wire
When the particles are connected by a conducting wire, charge flows until both particles have the same charge. Let’s denote the final charge on each particle as \( Q_f \). The total charge is conserved, so:

\[ Q_f = \frac{Q_1 + Q_2}{2} \]

### Step 4: Apply Coulomb's Law for Repulsion
After the charge has equalized, the particles repel each other with a force of \( F_2 = 0.009 \, \text{N} \). Now both charges are of the same sign (either both positive or both negative), so we can write:

\[ 0.009 = 9 \times 10^9 \frac{(Q_f)^2}{1^2} \]

This simplifies to:

\[ (Q_f)^2 = \frac{0.009}{9 \times 10^9} \]

Calculating this gives:

\[ (Q_f)^2 = 1 \times 10^{-12} \, \text{C}^2 \tag{2} \]

Taking the square root, we find:

\[ Q_f = 1 \times 10^{-6} \, \text{C} = 1 \, \mu\text{C} \]

### Step 5: Relate the Charges
From equation (1), we have:

\[ |Q_1 Q_2| = 3 \times 10^{-12} \]

And from the equalized charge:

\[ Q_f = \frac{Q_1 + Q_2}{2} = 1 \times 10^{-6} \]

Thus,

\[ Q_1 + Q_2 = 2 \times 10^{-6} \tag{3} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( Q_1 + Q_2 = 2 \times 10^{-6} \) (from equation 3)
2. \( Q_1 Q_2 = -3 \times 10^{-12} \) (from equation 1)

Let \( Q_1 = x \) and \( Q_2 = 2 \times 10^{-6} - x \). Substituting into the second equation gives:

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

Expanding this:

\[ 2 \times 10^{-6}x - x^2 = -3 \times 10^{-12} \]

Rearranging gives:

\[ x^2 - 2 \times 10^{-6}x - 3 \times 10^{-12} = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = -2 \times 10^{-6} \), and \( c = -3 \times 10^{-12} \).

Calculating the discriminant:

\[ b^2 - 4ac = (2 \times 10^{-6})^2 - 4(1)(-3 \times 10^{-12}) = 4 \times 10^{-12} + 12 \times 10^{-12} = 16 \times 10^{-12} \]

Now substituting back into the formula:

\[ x = \frac{2 \times 10^{-6} \pm \sqrt{16 \times 10^{-12}}}{2} \]
\[ x = \frac{2 \times 10^{-6} \pm 4 \times 10^{-6}}{2} \]

This gives two solutions:

1. \( x = 3 \times 10^{-6} \, \text{C} \) (which is not possible since it exceeds the total charge)
2. \( x = -1 \times 10^{-6} \, \text{C} \)

Thus, the charges are:

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

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