Edits to “Statement”, “Solution”, “Answer”
en/11.2.21.md
+6 −7
| @@ -1,7 +1,11 @@ | |||
| ### Statement | |||
| − | $11.2.21.$ | ||
| + | $11.2.21.$ | ||
| + | Experiments on colliding electron-electron beams have shown that the elec- | ||
| + | tron charge is distributed in a region whose dimensions are less than 10−18m. | ||
| + | Estimate the upper limit of the electromagnetic mass of an electron. | ||
| + | |||
| ### Solution | |||
| We model the electron as a sphere of radius R with charge e distributed on its surface. The electrostatic potential energy in SI is: | |||
| $U = \frac{1}{2} \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{R}$ | |||
| The equivalent electromagnetic mass, according to $E = mc^2 $ is | |||
| $m_{\text{em}} = \frac{U}{c^2}$ | |||
| Data | |||
| Electron charge: $e = 1.60 \times 10^{-19} \text{ C}$ | |||
| Coulomb constant: $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2$ | |||
| Radius (experimental limit): $R = 10^{-18} \text{ m}$ | |||
| Speed of light: $c = 3.0 \times 10^8 \text{ m/s}$ | |||
| Calculation | |||
| $e^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 | |||
| \frac{1}{4\pi\varepsilon_0} e^2 = (9.0 \times 10^9) \times (2.56 \times 10^{-38}) = 2.304 \times 10^{-28} \, \text{J·m}$ | |||
| Potential energy U: | |||
| $U = \frac{1}{2} \times \frac{2.304 \times 10^{-28}}{10^{-18}} = \frac{1}{2} \times 2.304 \times 10^{-10} = 1.152 \times 10^{-10} \, \text{J}$ | |||
| Electromagnetic mass: | |||
| @@ -34,11 +38,6 @@Solution | |||
| $m_{\text{em}} = \frac{U}{c^2} = \frac{1.152 \times 10^{-10}}{(3.0 \times 10^8)^2} = \frac{1.152 \times 10^{-10}}{9.0 \times 10^{16}} = 1.28 \times 10^{-27} \, \text{kg}$ | |||
| − | Result | ||
| − | |||
| − | $\boxed{m_{\text{em}} \approx 1.3 \times 10^{-27} \text{ kg}}$ | ||
| − | |||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{m_{\text{em}} \approx 1.3 \times 10^{-27} \text{ kg}}$ | ||
| @@ -1,7 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $11.2.21.$ |
$11.2.21.$ | ||
| Experiments on colliding electron-electron beams have shown that the elec- | |||
| tron charge is distributed in a region whose dimensions are less than 10−18m. | |||
| Estimate the upper limit of the electromagnetic mass of an electron. | |||
| ### Solution | ### Solution | ||
| We model the electron as a sphere of radius R with charge e distributed on its surface. The electrostatic potential energy in SI is: | We model the electron as a sphere of radius R with charge e distributed on its surface. The electrostatic potential energy in SI is: | ||
| $U = \frac{1}{2} \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{R}$ | $U = \frac{1}{2} \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{R}$ | ||
| The equivalent electromagnetic mass, according to $E = mc^2 $ is | The equivalent electromagnetic mass, according to $E = mc^2 $ is | ||
| $m_{\text{em}} = \frac{U}{c^2}$ | $m_{\text{em}} = \frac{U}{c^2}$ | ||
| Data | Data | ||
| Electron charge: $e = 1.60 \times 10^{-19} \text{ C}$ | Electron charge: $e = 1.60 \times 10^{-19} \text{ C}$ | ||
| Coulomb constant: $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2$ | Coulomb constant: $\frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2$ | ||
| Radius (experimental limit): $R = 10^{-18} \text{ m}$ | Radius (experimental limit): $R = 10^{-18} \text{ m}$ | ||
| Speed of light: $c = 3.0 \times 10^8 \text{ m/s}$ | Speed of light: $c = 3.0 \times 10^8 \text{ m/s}$ | ||
| Calculation | Calculation | ||
| $e^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 | $e^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 | ||
| \frac{1}{4\pi\varepsilon_0} e^2 = (9.0 \times 10^9) \times (2.56 \times 10^{-38}) = 2.304 \times 10^{-28} \, \text{J·m}$ | \frac{1}{4\pi\varepsilon_0} e^2 = (9.0 \times 10^9) \times (2.56 \times 10^{-38}) = 2.304 \times 10^{-28} \, \text{J·m}$ | ||
| Potential energy U: | Potential energy U: | ||
| $U = \frac{1}{2} \times \frac{2.304 \times 10^{-28}}{10^{-18}} = \frac{1}{2} \times 2.304 \times 10^{-10} = 1.152 \times 10^{-10} \, \text{J}$ | $U = \frac{1}{2} \times \frac{2.304 \times 10^{-28}}{10^{-18}} = \frac{1}{2} \times 2.304 \times 10^{-10} = 1.152 \times 10^{-10} \, \text{J}$ | ||
| Electromagnetic mass: | Electromagnetic mass: | ||
| @@ -34,11 +38,6 @@Solution | |||
| $m_{\text{em}} = \frac{U}{c^2} = \frac{1.152 \times 10^{-10}}{(3.0 \times 10^8)^2} = \frac{1.152 \times 10^{-10}}{9.0 \times 10^{16}} = 1.28 \times 10^{-27} \, \text{kg}$ | $m_{\text{em}} = \frac{U}{c^2} = \frac{1.152 \times 10^{-10}}{(3.0 \times 10^8)^2} = \frac{1.152 \times 10^{-10}}{9.0 \times 10^{16}} = 1.28 \times 10^{-27} \, \text{kg}$ | ||
| Result | |||
| $\boxed{m_{\text{em}} \approx 1.3 \times 10^{-27} \text{ kg}}$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{m_{\text{em}} \approx 1.3 \times 10^{-27} \text{ kg}}$ | ||