A water molecule is a permanent dipole with a known dipole moment

A water molecule is a permanent dipole with a known dipole moment
There is a water molecule in the air a very short distance
from the midpoint of a long glass rod of length
carrying a uniformly distributed positive charge
(Figure 15.53 ). The axis of the dipole is perpendicular to the rod. Note that
. (The charged rod induces an increase in the dipole moment, but the induced portion of the dipole moment is completely negligible compared to
. It is convenient to use the “binomial expansion” that you may have learned in calculus, that
is
Note that
can be negative. (a) Find the magnitude and direction of the electric force acting on the water molecule. Your final result for the magnitude of the force must be expressed in terms of
and
You can use
and
in your calculations, but your final result must not include
or
since it is only their product
that is measurable for a water molecule. Explain your work carefully, including appropriate diagrams. (b) If the electric field of the rod has the magnitude
at the location of the water molecule,
from the rod, what is the magnitude of the horizontal component of the acceleration of the water molecule? The measured dipole moment for
is
and the mass of one mole is
Be sure to check units in your calculation!

The Correct Answer and Explanation is:

To solve this problem, we need to break it down into parts. We have a water molecule, which behaves as a permanent dipole, near a charged rod. The force and acceleration it experiences are due to the electric field generated by the rod.

Part (a): Magnitude and Direction of the Electric Force on the Water Molecule

Step 1: Electric Field of a Charged Rod

We assume the rod is long and uniformly charged. The electric field EEE created by a uniformly charged rod at a point near its midpoint (in the direction perpendicular to the rod) is given by the formula:E=keλrE = \frac{k_e \lambda}{r}E=rke​λ​

where:

• kek_eke​ is Coulomb’s constant (8.99×109 N\cdotpm2/C28.99 \times 10^9 \, \text{N·m}^2/\text{C}^28.99×109N\cdotpm2/C2),
• λ\lambdaλ is the charge per unit length of the rod,
• rrr is the distance of the water molecule from the rod.

This electric field is directed radially outward from the rod (assuming positive charge on the rod).

Step 2: Induced Dipole Moment

The water molecule itself is a dipole, so it has an electric dipole moment, p⃗\vec{p}p​. A dipole in an external electric field experiences a force given by:F⃗=(p⃗⋅∇)E⃗\vec{F} = (\vec{p} \cdot \nabla) \vec{E}F=(p​⋅∇)E

However, for a uniform electric field, the force on a dipole simplifies to:F=pEF = p EF=pE

where:

• ppp is the dipole moment of the water molecule, and
• EEE is the magnitude of the electric field at the location of the dipole.

Since the axis of the dipole is perpendicular to the rod, the force acting on the dipole is aligned with the electric field direction. Thus, the magnitude of the force is:F=p⋅keλrF = p \cdot \frac{k_e \lambda}{r}F=p⋅rke​λ​

The direction of the force will be aligned with the direction of the electric field, which is radially outward from the rod.

Step 3: Final Expression for the Force

The final result for the magnitude of the force on the water molecule is:F=keλprF = \frac{k_e \lambda p}{r}F=rke​λp​

where:

• kek_eke​ is Coulomb’s constant,
• λ\lambdaλ is the charge per unit length of the rod,
• ppp is the dipole moment of the water molecule, and
• rrr is the distance between the water molecule and the rod.

Part (b): Magnitude of the Horizontal Component of the Acceleration

To find the magnitude of the horizontal component of the acceleration of the water molecule, we first need to use Newton’s second law:F=maF = maF=ma

where:

• FFF is the force on the water molecule,
• mmm is the mass of the water molecule,
• aaa is the acceleration.

The force FFF has already been found as:F=keλprF = \frac{k_e \lambda p}{r}F=rke​λp​

The mass of a single water molecule is given by the molar mass of water (18.015 g/mol18.015 \, \text{g/mol}18.015g/mol) divided by Avogadro’s number:m=18.015 g/mol6.022×1023 molecules/mol=2.99×10−26 kgm = \frac{18.015 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} = 2.99 \times 10^{-26} \, \text{kg}m=6.022×1023molecules/mol18.015g/mol​=2.99×10−26kg

Now, using Newton’s second law, we find the acceleration:a=Fm=keλpr⋅ma = \frac{F}{m} = \frac{k_e \lambda p}{r \cdot m}a=mF​=r⋅mke​λp​

Conclusion

Thus, the magnitude of the horizontal component of the acceleration of the water molecule depends on the electric field, the dipole moment, the mass of the water molecule, and the distance from the rod. The units should check out correctly, given that kek_eke​ has units of N\cdotpm2/C2\text{N·m}^2/\text{C}^2N\cdotpm2/C2, λ\lambdaλ has units of C/m\text{C/m}C/m, and ppp has units of C\cdotpm\text{C·m}C\cdotpm.