Math Problem Statement

One model for a water molecule is to have a point charge of magnitude βˆ’2π‘’βˆ’ at point (π‘₯, π‘¦, π‘§) = (0, 0, βˆ’π‘‘/2) and two point charges of magnitude +π‘’βˆ’ at points (±𝑑/4, 0, π‘‘/2). Verify that far from the molecule the fields for this model give a dipole distribution. How far away (relative to π‘‘) does the observation need to be for this to be true? 

Solution

To verify that the far-field distribution of the water molecule model corresponds to a dipole distribution, let's consider the arrangement of charges and calculate the dipole moment. Then, we'll determine the condition under which the far-field electric field resembles that of a dipole.

Step 1: Calculate the Dipole Moment

The water molecule model consists of three point charges:

  • A negative charge of βˆ’2e-2e at (0,0,βˆ’d/2)(0, 0, -d/2).
  • Two positive charges of +e+e at (d/4,0,d/2)(d/4, 0, d/2) and (βˆ’d/4,0,d/2)(-d/4, 0, d/2).

The electric dipole moment p\mathbf{p} is defined as:

p=βˆ‘iqiri\mathbf{p} = \sum_i q_i \mathbf{r}_i

where qiq_i is the charge and ri\mathbf{r}_i is its position vector.

For this system:

  • Negative charge contribution: pneg=βˆ’2eβ‹…(0,0,βˆ’d/2)=(0,0,ed)\mathbf{p}_\text{neg} = -2e \cdot (0, 0, -d/2) = (0, 0, ed)
  • Positive charge contribution (since the two charges are symmetric and only contribute along the z-axis): ppos=eβ‹…(d4,0,d2)+eβ‹…(βˆ’d4,0,d2)=(0,0,ed)\mathbf{p}_\text{pos} = e \cdot \left(\frac{d}{4}, 0, \frac{d}{2}\right) + e \cdot \left(-\frac{d}{4}, 0, \frac{d}{2}\right) = (0, 0, ed)

Thus, the total dipole moment is:

p=(0,0,ed)+(0,0,ed)=(0,0,2ed)\mathbf{p} = (0, 0, ed) + (0, 0, ed) = (0, 0, 2ed)

Step 2: Far-Field Condition for Dipole Distribution

In the far-field region, the distance rr from the dipole to the observation point should be much larger than the separation between the charges dd. For the far-field approximation to hold, the distance rr must satisfy:

r≫dr \gg d

This ensures that the electric field behaves like that of a pure dipole, which falls off as 1/r31/r^3 rather than the individual fields of the point charges.

Conclusion

To observe a dipole distribution, the observation point needs to be at a distance rr from the molecule such that rr is significantly larger than the separation distance dd. Typically, this is interpreted as rβ‰₯10dr \geq 10d to ensure the dominance of the dipole term in the field.

Would you like more details or have any questions?

Related Questions:

  1. How does the field of a dipole differ from that of a monopole?
  2. How is the potential field calculated for a dipole at far distances?
  3. What is the significance of the dipole moment in molecular physics?
  4. How would the electric field behave if the charges were not symmetrically placed?
  5. What are real-world applications of understanding dipole fields?

Tip:

When dealing with multipole expansions, remember that higher-order terms become negligible faster as the distance from the source increases.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Electric Fields
Point Charges
Dipole Moment

Formulas

Electric dipole moment formula

Theorems

-

Suitable Grade Level

Advanced Undergraduate