A proton and an electron are placed 1.6cm apart in free space. Find the magnitude of electrostatic force between them. The nature of this force.

Options

A. `9 xx 10^(-25)` repulsion
B. `90 xx 10^(-25)` repulsion
C. `9 xx 10^(-25)`N, attractive
D. `9 xx 10^(25)` attractive

Answer

To find the magnitude of the electrostatic force between a proton and an electron placed 1.6 cm apart in free space, we can use Coulomb's Law, which is given by the formula:

\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]

Where:
- \( F \) is the magnitude of the electrostatic force,
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges of the proton and electron, respectively,
- \( r \) is the distance between the charges.

### Step 1: Identify the charges
The charge of a proton (\( q_1 \)) is \( +1.6 \times 10^{-19} \, \text{C} \) and the charge of an electron (\( q_2 \)) is \( -1.6 \times 10^{-19} \, \text{C} \).

### Step 2: Convert the distance to meters
The distance between the proton and electron is given as 1.6 cm. We need to convert this to meters:
\[ r = 1.6 \, \text{cm} = 0.016 \, \text{m} \]

### Step 3: Substitute the values into Coulomb's Law
Now, we can substitute the values into the formula:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
\[ F = (8.99 \times 10^9) \frac{|(1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})|}{(0.016)^2} \]

### Step 4: Calculate the product of the charges
Calculating the product of the charges:
\[ |q_1 \cdot q_2| = |(1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})| = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]

### Step 5: Calculate \( r^2 \)
Now calculate \( r^2 \):
\[ r^2 = (0.016)^2 = 0.000256 \, \text{m}^2 \]

### Step 6: Substitute and calculate the force
Now substituting back into the equation:
\[ F = (8.99 \times 10^9) \frac{2.56 \times 10^{-38}}{0.000256} \]
\[ F = (8.99 \times 10^9) \times (1.0 \times 10^{-34}) \]
\[ F = 8.99 \times 10^{-25} \, \text{N} \]

### Step 7: Determine the nature of the force
Since the proton has a positive charge and the electron has a negative charge, the force between them is attractive.

### Final Answer
The magnitude of the electrostatic force between the proton and electron is \( 8.99 \times 10^{-25} \, \text{N} \), and the nature of this force is attractive.

---