Two positively charged particles each of mass is `9 xx 10^(-30)kg` and carrying a charge of `1.6 xx 10^(-19)C` are placed at a distance .r. apart. If each experiences a force equal to its weight, the value of r is `(g = 10 ms^(-2))`

Options

A. 1.6m
B. 0.16m
C. 0.116m
D. 0.8m

Answer

To solve the problem, we need to find the distance \( r \) between two positively charged particles that experience a force equal to their weight. Here’s a step-by-step solution:

### Step 1: Understand the Forces Involved
The electrostatic force \( F \) between two charges is given by Coulomb's law:
\[ F = k \frac{q_1 q_2}{r^2} \]
where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( q_1 \) and \( q_2 \) are the charges,
- \( r \) is the distance between the charges.

### Step 2: Set Up the Weight of the Particles
The weight \( W \) of each particle is given by:
\[ W = mg \]
where:
- \( m = 9 \times 10^{-30} \, \text{kg} \) (mass of each particle),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 3: Equate the Electrostatic Force to the Weight
According to the problem, the electrostatic force is equal to the weight of each particle:
\[ k \frac{q^2}{r^2} = mg \]
Here, \( q = 1.6 \times 10^{-19} \, \text{C} \) is the charge of each particle.

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 9 \times 10^9 \frac{(1.6 \times 10^{-19})^2}{r^2} = (9 \times 10^{-30})(10) \]

### Step 5: Simplify the Equation
Calculating the right side:
\[ 9 \times 10^{-30} \times 10 = 9 \times 10^{-29} \]
Now, substituting this back into the equation:
\[ 9 \times 10^9 \frac{(1.6 \times 10^{-19})^2}{r^2} = 9 \times 10^{-29} \]

### Step 6: Calculate \( (1.6 \times 10^{-19})^2 \)
Calculating \( (1.6 \times 10^{-19})^2 \):
\[ (1.6)^2 = 2.56 \quad \text{and} \quad (10^{-19})^2 = 10^{-38} \]
Thus,
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]

### Step 7: Substitute Back and Solve for \( r^2 \)
Substituting this into the equation:
\[ 9 \times 10^9 \frac{2.56 \times 10^{-38}}{r^2} = 9 \times 10^{-29} \]
Now, simplifying:
\[ \frac{2.56 \times 9 \times 10^9}{9 \times 10^{-29}} = r^2 \]
Cancelling \( 9 \):
\[ \frac{2.56 \times 10^9}{10^{-29}} = r^2 \]
This simplifies to:
\[ 2.56 \times 10^{38} = r^2 \]

### Step 8: Solve for \( r \)
Taking the square root:
\[ r = \sqrt{2.56 \times 10^{38}} = 1.6 \times 10^{19} \, \text{m} \]

### Final Answer
Thus, the distance \( r \) is:
\[ r = 1.6 \, \text{m} \]