The distance between `H^(+) and Cl^(-)` ions in HCl molecules is `1.38Å.` The potential due to this dipole at a distance of `10Å` on the axis of dipole is

Options

A. `2.1` V
B. `1.8` V
C. `0.2` V
D. `1.2` V

Explanation

To find the electric potential \( V \) due to a dipole at a point along its axis, we can use the formula:

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{p \cdot \cos(\theta)}{r^2} \]

Where:
- \( p \) is the dipole moment,
- \( \epsilon_0 \) is the permittivity of free space,
- \( r \) is the distance from the dipole to the point where the potential is being calculated,
- \( \theta \) is the angle between the dipole moment and the line joining the dipole to the point.

### Step 1: Calculate the dipole moment \( p \)

The dipole moment \( p \) is given by:

\[ p = q \cdot d \]

Where:
- \( q \) is the charge of the ion (for \( H^+ \) or \( Cl^- \), \( q = 1.6 \times 10^{-19} \, C \)),
- \( d \) is the distance between the charges (given as \( 1.38 \, \text{Å} = 1.38 \times 10^{-10} \, m \)).

Calculating \( p \):

\[ p = (1.6 \times 10^{-19} \, C) \cdot (1.38 \times 10^{-10} \, m) = 2.208 \times 10^{-29} \, C \cdot m \]

### Step 2: Calculate the potential \( V \)

We are given that the distance \( r = 10 \, \text{Å} = 10 \times 10^{-10} \, m \). Since we are on the axis of the dipole, \( \theta = 0^\circ \) and \( \cos(0) = 1 \).

Substituting the values into the potential formula:

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{p}{r^2} \]

Using \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \):

\[ V = \frac{1}{4\pi (8.85 \times 10^{-12})} \cdot \frac{2.208 \times 10^{-29}}{(10 \times 10^{-10})^2} \]

Calculating \( r^2 \):

\[ r^2 = (10 \times 10^{-10})^2 = 100 \times 10^{-20} = 10^{-18} \, m^2 \]

Now substituting \( r^2 \) into the potential formula:

\[ V = \frac{1}{4\pi (8.85 \times 10^{-12})} \cdot \frac{2.208 \times 10^{-29}}{10^{-18}} \]

Calculating the fraction:

\[ V = \frac{1}{4\pi (8.85 \times 10^{-12})} \cdot 2.208 \times 10^{-11} \]

Calculating \( 4\pi (8.85 \times 10^{-12}) \):

\[ 4\pi (8.85 \times 10^{-12}) \approx 1.11 \times 10^{-10} \]

Thus:

\[ V \approx \frac{2.208 \times 10^{-11}}{1.11 \times 10^{-10}} \approx 0.198 \, V \]

### Final Answer

The potential due to the dipole at a distance of \( 10 \, \text{Å} \) on the axis of the dipole is approximately \( 0.198 \, V \).