Question

A neutral water molecule is placed in an electric field $ E = 2.5 \times 10^4 \, \text{N/C} $. The work done to rotate it by 180° is $ 5 \times 10^{-25} \, \text{J} $. Find the approximate separation of the center of charges.

• A. \( 1.25 \times 10^{-10} \, \text{m} \)
• B. \( 0.625 \times 10^{-10} \, \text{m} \)
• C. \( 0.625 \times 10^{-9} \, \text{m} \)
• D. \( 0.998 \times 10^{-10} \, \text{m} \)

Hint:

In problems involving the rotation of a dipole in an electric field, use the work formula \( W = pE(1 - \cos \theta) \) to find the dipole moment and then use \( p = qd \) to determine the separation between the charges.

Answer 1

The Correct Option is B

The work done to rotate the water molecule in an electric field is given by the formula: \[ W = pE \left( 1 - \cos \theta \right) \] where: - \( p \) is the dipole moment of the water molecule, - \( E \) is the electric field, - \( \theta \) is the angle of rotation (which is \( 180^\circ \), so \( \cos 180^\circ = -1 \)). The work done to rotate the molecule is: \[ W = pE \left( 1 - (-1) \right) = 2pE \] Therefore: \[ p = \frac{W}{2E} \] Substitute the given values: \[ p = \frac{5 \times 10^{-25}}{2 \times 2.5 \times 10^4} = \frac{5 \times 10^{-25}}{5 \times 10^4} = 10^{-29} \, \text{C} \cdot \text{m} \] Now, the dipole moment \( p \) is related to the separation between the charges \( d \) by the formula: \[ p = qd \] where \( q \) is the charge on the water molecule and \( d \) is the separation between the charges. For a water molecule, the charge is approximately the charge of an electron, \( q \approx 1.6 \times 10^{-19} \, \text{C} \).
Thus, we have: \[ d = \frac{p}{q} = \frac{10^{-29}}{1.6 \times 10^{-19}} = 0.625 \times 10^{-10} \, \text{m} \]
Thus, the approximate separation of the charges is \( 0.625 \times 10^{-10 \, \text{m}} \).