Edits to “Problem”, “1st Solution Option”, “0. Write down what we have”

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revision #18983 parent #18980 ← older
en/6.6.2.md +40 −22
@@ -1,12 +1,13 @@
+
### Problem
+
6.6.2. The dielectric constant of helium at temperature $0^\circ C$ and pressure 1 atm is $1.000074$. Find the dipole moment of a helium atom in a uniform electric field of strength 300 V/cm.
### Solution
+### 1st Solution Option.
−### 1st Solution Option
−
−#### 0. Write down what we have.
+### 0. Write down what we have.
\[
\varepsilon = 1.000074
\]
@@ -40,9 +41,8 @@2. Relationship between $\varepsilon$ and polarization $P_{\text{pol}}$
#### 3. Dipole moment of one atom.
−By definition:
\[
−P_{\text{pol}} := n \cdot p
+P_{\text{pol}} := n \cdot p \quad ((:=) - \text{equal by definition})
\]
\[
@@ -55,83 +55,101 @@3. Dipole moment of one atom.
= \frac{8.85 \times 10^{-12} \cdot (1.000074 - 1) \cdot 3 \times 10^{4} \cdot 1.38 \times 10^{-23} \cdot 273}{1.01 \times 10^{5}}
\approx 7.3 \times 10^{-37}\ \text{C} \cdot \text{m}
\]
−### Answer
+
+### Answer:
\[
\boxed{p \approx 7.3 \times 10^{-37}\ \text{C} \cdot \text{m}}
\]
+#### 2nd Solution Option.
−### 2nd Solution Option
−
−
−#### 0. Write down what we have
+#### 0. Write down what we have.
\[
\varepsilon = 1.000074
\]
+
\[
t = 0^\circ C \Leftrightarrow T = 273\ \text{K}
\]
+
\[
E = 300\ \text{V/cm} \Rightarrow 3 \times 10^{4}\ \text{V/m}
\]
+
\[
P = 1\ \text{atm} \approx 1.01 \times 10^{5}\ \text{Pa}
\]
#### 1. First, write down what we need to find.
+
\[
−p = \alpha E
+p = \alpha E_{\text{loc}}
\]
−where $\alpha$ is the polarizability of the atom, which can be found using the Clausius-Mossotti formula (an explanation of this formula will be given before the final answer, in case you are encountering it for the first time).
+where $\alpha$ is the polarizability of the atom, which can be found using the Clausius-Mossotti formula, and
+\[
+E_{\text{loc}} = E + E_{\text{l}} = \frac{\varepsilon + 2}{3}E
+\]
+Since the fraction $\approx 1$, we have $E_{\text{loc}} \approx E$.
+$E_{\text{l}}$ is the electric field of the surroundings, created by polarization outside the Lorentz sphere. (An explanation of this formula will be given before the final answer, in case you are encountering it for the first time.)
+
#### 2. Finding the polarizability of the atom.
−Write down the Clausius--Mossotti formula, then express $\alpha$:
+Write down the Clausius-Mossotti formula, then express $\alpha$:
+
\[
\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n\alpha}{3\varepsilon_{0}}}
\]
+
\[
\alpha = \frac{3\varepsilon_{0}}{n}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right)
\]
−where $n$ is the concentration of helium atoms, which we can find from the ideal gas equation of state:
+where $n$ is the concentration of helium atoms, which we can find from the ideal gas equation of state.
+
\[
P = nKT \Rightarrow n = \frac{P}{KT}
\]
where $K$ is the Boltzmann constant.
Now we have what we were looking for:
+
\[
\alpha = \frac{3KT\varepsilon_{0}}{P}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right)
\]
#### 3. Substitute $\alpha$ into the dipole moment formula and find the dipole moment of the helium atom.
+
\[
\boxed{p = \frac{3KT\varepsilon_{0}}{P}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) \cdot E}
\]
+
\[
p = \frac{3 \cdot 1.38 \times 10^{-23} \cdot 273 \cdot 8.85 \times 10^{-12}}{1.01 \times 10^{5}}
\times \left(\frac{1.000074 - 1}{1.000074 + 2}\right)
\times 3 \times 10^{4}
−\approx 7.3 \times 10^{-38}\ \text{C} \cdot \text{m}
+\approx 7.3 \times 10^{-37}\ \text{C} \cdot \text{m}
\]
−#### Theory on the Clausius- Mossotti Formula
+#### Theory on the Clausius-Mossotti Formula
−The Clausius- Mossotti formula describes the relationship between the static dielectric constant of a dielectric and the polarizability of its constituent particles. It was derived independently by Ottaviano F. Mossotti in 1850 and by Rudolf J. E. Clausius in 1879. In cases where the substance consists of particles of one kind, in the Gaussian system of units the formula is:
+The Clausius--Mossotti formula describes the relationship between the static dielectric constant of a dielectric and the polarizability of its constituent particles. It was derived independently by Ottaviano F. Mossotti in 1850 and by Rudolf J. E. Clausius in 1879. In cases where the substance consists of particles of one kind, in the Gaussian system of units the formula is:
+
\[
−\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3} N \alpha}
+\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3}N\alpha}
\]
+
where $\varepsilon$ is the dielectric constant, $N$ is the number of particles per unit volume, and $\alpha$ is their polarizability.
If the substance consists of particles of different kinds, the formula takes the following form:
+
\[
−\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3} \sum_{i=1}^{j} N_{i} \alpha_{i}}
+\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3}\sum_{i=1}^{j} N_{i}\alpha_{i}}
\]
−For more details, you can read here: (https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation)
+For more details, you can read here: https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation
−#### Answer
+### Answer:
\[
−\boxed{p \approx 7.3 \times 10^{-38}\ \text{C} \cdot \text{m}}
+\boxed{p \approx 7.3 \times 10^{-37}\ \text{C} \cdot \text{m}}
\]