Two positive point charges are `3 m` apart their combined charge is `20 mu C`. If the force between them is `0.075 N`, the charges are

Options

A. `10 mu C , 10 mu C`
B. `15 mu C , 5 mu C`
C. `12 mu C , 8 mu C`
D. `14 mu C , 6 mu C`

Explanation

To solve the problem, we need to find the values of the two positive point charges, \( Q_1 \) and \( Q_2 \), given that they are 3 meters apart, their combined charge is \( 20 \, \mu C \), and the force between them is \( 0.075 \, N \).

### Step 1: Set up the equations
We know that:
1. The combined charge \( Q_1 + Q_2 = 20 \, \mu C = 20 \times 10^{-6} \, C \) (Equation 1)
2. The force between the charges is given by Coulomb's law:
\[ F = k \frac{Q_1 Q_2}{R^2} \]
where \( k = 9 \times 10^9 \, N m^2/C^2 \) and \( R = 3 \, m \).

### Step 2: Substitute values into Coulomb's law
Substituting the known values into the equation:
\[ 0.075 = \frac{9 \times 10^9 \cdot Q_1 Q_2}{(3)^2} \]
\[ 0.075 = \frac{9 \times 10^9 \cdot Q_1 Q_2}{9} \]
\[ 0.075 = 10^9 \cdot Q_1 Q_2 \]
From this, we can express \( Q_1 Q_2 \):
\[ Q_1 Q_2 = 0.075 \times 10^{-9} = 75 \times 10^{-12} \, C^2 \quad (Equation 2) \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( Q_1 + Q_2 = 20 \times 10^{-6} \) (Equation 1)
2. \( Q_1 Q_2 = 75 \times 10^{-12} \) (Equation 2)

We can express \( Q_2 \) in terms of \( Q_1 \) from Equation 1:
\[ Q_2 = 20 \times 10^{-6} - Q_1 \]

### Step 4: Substitute \( Q_2 \) into Equation 2
Substituting \( Q_2 \) into Equation 2:
\[ Q_1 \left(20 \times 10^{-6} - Q_1\right) = 75 \times 10^{-12} \]
Expanding this gives:
\[ 20 \times 10^{-6} Q_1 - Q_1^2 = 75 \times 10^{-12} \]
Rearranging gives us a quadratic equation:
\[ Q_1^2 - 20 \times 10^{-6} Q_1 + 75 \times 10^{-12} = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( Q_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -20 \times 10^{-6} \), and \( c = 75 \times 10^{-12} \):
\[ Q_1 = \frac{20 \times 10^{-6} \pm \sqrt{(20 \times 10^{-6})^2 - 4 \cdot 1 \cdot 75 \times 10^{-12}}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (20 \times 10^{-6})^2 = 400 \times 10^{-12} \]
\[ 4 \cdot 75 \times 10^{-12} = 300 \times 10^{-12} \]
\[ \text{Discriminant} = 400 \times 10^{-12} - 300 \times 10^{-12} = 100 \times 10^{-12} \]
Thus:
\[ Q_1 = \frac{20 \times 10^{-6} \pm 10 \times 10^{-6}}{2} \]
Calculating the two possible values:
1. \( Q_1 = \frac{30 \times 10^{-6}}{2} = 15 \times 10^{-6} \, C = 15 \, \mu C \)
2. \( Q_1 = \frac{10 \times 10^{-6}}{2} = 5 \times 10^{-6} \, C = 5 \, \mu C \)

### Step 6: Find \( Q_2 \)
Using \( Q_1 + Q_2 = 20 \, \mu C \):
1. If \( Q_1 = 15 \, \mu C \), then \( Q_2 = 20 - 15 = 5 \, \mu C \)
2. If \( Q_1 = 5 \, \mu C \), then \( Q_2 = 20 - 5 = 15 \, \mu C \)

Thus, the charges are:
- \( Q_1 = 15 \, \mu C \) and \( Q_2 = 5 \, \mu C \)

### Final Answer
The two positive point charges are \( Q_1 = 15 \, \mu C \) and \( Q_2 = 5 \, \mu C \).

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