Two positive charges separated by a distance 2m repel each other with a force of 0.36N. If the combined charge is `26muC`, the charges are

Options

A. `20muC, 6muC`
B. `16muC, 10 mu C`
C. `18 mu C, 8 mu C`
D. `13 mu C, 13 mu C`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two positive charges \( q_1 \) and \( q_2 \) separated by a distance of 2 meters. The force of repulsion between them is given as 0.36 N, and the combined charge is \( q_1 + q_2 = 26 \, \mu C \).

### Step 2: Use Coulomb's Law
According to Coulomb's Law, the force \( F \) between two charges is given by the formula:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
where \( \epsilon_0 \) is the permittivity of free space, and \( r \) is the distance between the charges.

### Step 3: Substitute known values into the equation
We know:
- \( F = 0.36 \, N \)
- \( r = 2 \, m \)
- \( \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, N \cdot m^2/C^2 \)

Substituting these values into the equation gives:
\[ 0.36 = 9 \times 10^9 \cdot \frac{q_1 q_2}{(2)^2} \]
This simplifies to:
\[ 0.36 = 9 \times 10^9 \cdot \frac{q_1 q_2}{4} \]
Multiplying both sides by 4:
\[ 1.44 = 9 \times 10^9 q_1 q_2 \]
Now, dividing both sides by \( 9 \times 10^9 \):
\[ q_1 q_2 = \frac{1.44}{9 \times 10^9} = 0.16 \times 10^{-9} \, C^2 \]

### Step 4: Set up the equations
We have two equations:
1. \( q_1 + q_2 = 26 \, \mu C \) (or \( 26 \times 10^{-6} \, C \))
2. \( q_1 q_2 = 0.16 \times 10^{-9} \, C^2 \)

### Step 5: Express \( q_1 \) in terms of \( q_2 \)
From the first equation:
\[ q_1 = 26 \times 10^{-6} - q_2 \]

### Step 6: Substitute \( q_1 \) into the second equation
Substituting \( q_1 \) into the second equation:
\[ (26 \times 10^{-6} - q_2) q_2 = 0.16 \times 10^{-9} \]
Expanding this gives:
\[ 26 \times 10^{-6} q_2 - q_2^2 = 0.16 \times 10^{-9} \]
Rearranging gives:
\[ q_2^2 - 26 \times 10^{-6} q_2 + 0.16 \times 10^{-9} = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( q_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -26 \times 10^{-6} \), and \( c = 0.16 \times 10^{-9} \).
Calculating the discriminant:
\[ D = b^2 - 4ac = (26 \times 10^{-6})^2 - 4 \times 1 \times 0.16 \times 10^{-9} \]
Calculating \( D \):
\[ D = 676 \times 10^{-12} - 0.64 \times 10^{-9} = 676 \times 10^{-12} - 640 \times 10^{-12} = 36 \times 10^{-12} \]
Now substituting back into the quadratic formula:
\[ q_2 = \frac{26 \times 10^{-6} \pm \sqrt{36 \times 10^{-12}}}{2} \]
Calculating the square root:
\[ \sqrt{36 \times 10^{-12}} = 6 \times 10^{-6} \]
Thus:
\[ q_2 = \frac{26 \times 10^{-6} \pm 6 \times 10^{-6}}{2} \]
Calculating the two possible values:
1. \( q_2 = \frac{32 \times 10^{-6}}{2} = 16 \times 10^{-6} \, C \)
2. \( q_2 = \frac{20 \times 10^{-6}}{2} = 10 \times 10^{-6} \, C \)

### Step 8: Find \( q_1 \)
Using \( q_1 + q_2 = 26 \, \mu C \):
1. If \( q_2 = 6 \, \mu C \), then \( q_1 = 26 - 6 = 20 \, \mu C \)
2. If \( q_2 = 16 \, \mu C \), then \( q_1 = 26 - 16 = 10 \, \mu C \)

### Final Answer
The individual charges are:
- \( q_1 = 20 \, \mu C \)
- \( q_2 = 6 \, \mu C \)