Новое решение
en/6.6.15.md
+63 −0
| @@ -0,0 +1,63 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $6.6.15.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $6.6.15.$ An electrostatic precipitator consists of a long metal tube and a wire directed along the axis. A potential difference $V$ is created between them. Air with dust is passed through the tube. | ||
| + | <b>a.</b> To which electrode — to the wire or to the tube — are dust particles attracted? | ||
| + | <b>b.</b> What is the force acting on a dust particle with dielectric permittivity $\varepsilon_2$, if the force acting on a dust particle of the same radius, but with dielectric permittivity $\varepsilon_1$, is equal to $F_1$? Both dust particles are equally distant from the wire. | ||
| + | <b>c.</b> How does the force of attraction depend on the potential difference? On the distance to the wire? | ||
| + | $d^\ast$. How many times is the force acting on a dust particle of radius $R$ greater than the force acting on a dust particle of radius $r < R$? The dielectric permittivity of the dust particles is the same, and they are located at the same distance from the wire. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | <b>a)</b> Let us find the distribution of the electric field inside the electrostatic precipitator. The system is a cylindrical capacitor. According to Gauss's law, the electric field strength $E$ at a distance $x$ from the wire axis is inversely proportional to this distance: $E(x) \propto 1/x$. Thus, the field is maximum near the wire and decreases as it approaches the tube. | ||
| + | The dust particle is initially electrically neutral, but in an external field it polarizes, acquiring an induced dipole moment $\vec{p}$ co-directed with the field. The force acting on a dipole in a non-uniform field is: | ||
| + | $$F_x = p \frac{\partial E}{\partial x}$$ | ||
| + | Since the field decreases with distance, the gradient $\partial E / \partial x < 0$. This means that the force vector is directed opposite to the $x$-axis, i.e., it pulls the dust particle into the region of a stronger field. Consequently, the dust particles are always attracted <b>to the wire</b>. | ||
| + | |||
| + | <b>b)</b> Let us find the dependence of the induced dipole moment on the dielectric permittivity. Consider a dielectric sphere (dust particle) of radius $R$ placed in a uniform external field $\vec{E}_0$. Polarization leads to the appearance of bound charges on the surface, which create a depolarizing field inside the sphere: | ||
| + | $$\vec{E}_{\text{dep}} = -\frac{\vec{P}}{3\varepsilon_0}$$ | ||
| + | where $\vec{P}$ is the polarization vector. The true field inside the sphere is the sum of these fields: | ||
| + | $$\vec{E}_{\text{in}} = \vec{E}_0 - \frac{\vec{P}}{3\varepsilon_0}$$ | ||
| + | By definition, the polarization of a linear dielectric is: | ||
| + | $$\vec{P} = \varepsilon_0(\varepsilon - 1)\vec{E}_{\text{in}} = \varepsilon_0(\varepsilon - 1)\left(\vec{E}_0 - \frac{\vec{P}}{3\varepsilon_0}\right)$$ | ||
| + | Let us express the polarization vector from this equation: | ||
| + | $$\vec{P} \left( 1 + \frac{\varepsilon - 1}{3} \right) = \varepsilon_0(\varepsilon - 1)\vec{E}_0 \implies \vec{P} = 3\varepsilon_0 \frac{\varepsilon - 1}{\varepsilon + 2} \vec{E}_0$$ | ||
| + | The total dipole moment of the dust particle $p$ is equal to the product of the polarization and the volume of the sphere $V = \frac{4}{3}\pi R^3$: | ||
| + | $$p = P \cdot V = 4\pi\varepsilon_0 R^3 \frac{\varepsilon - 1}{\varepsilon + 2} E_0$$ | ||
| + | Since the attractive force is proportional to the dipole moment ($F \propto p$), it depends on the permittivity as: | ||
| + | $$F \propto \frac{\varepsilon - 1}{\varepsilon + 2}$$ | ||
| + | Let us write the ratio of forces for two dust particles: | ||
| + | $$\frac{F_2}{F_1} = \frac{\frac{\varepsilon_2 - 1}{\varepsilon_2 + 2}}{\frac{\varepsilon_1 - 1}{\varepsilon_1 + 2}}$$ | ||
| + | From this, we find the required force: | ||
| + | $$F_2 = F_1 \frac{(\varepsilon_2 - 1)(\varepsilon_1 + 2)}{(\varepsilon_1 - 1)(\varepsilon_2 + 2)}$$ | ||
| + | |||
| + | <b>c)</b> The field strength of a cylindrical capacitor is expressed through the potential difference $V$, the radius of the wire $r_0$, and the radius of the tube $R_0$: | ||
| + | $$E(x) = \frac{V}{x \ln(R_0/r_0)}$$ | ||
| + | Since $p = \alpha E$ (where $\alpha$ is a constant of the dust particle found in part b), the force can be rewritten as: | ||
| + | $$F = p \frac{\partial E}{\partial x} = \alpha E \frac{\partial E}{\partial x} = \frac{1}{2}\alpha \frac{\partial (E^2)}{\partial x}$$ | ||
| + | Substitute the field function $E(x)$: | ||
| + | $$E^2 = \frac{V^2}{x^2 \ln^2(R_0/r_0)} \implies \frac{\partial (E^2)}{\partial x} = -\frac{2V^2}{x^3 \ln^2(R_0/r_0)}$$ | ||
| + | It can be seen that the magnitude of the attractive force is directly proportional to the square of the voltage and inversely proportional to the cube of the distance: | ||
| + | $$F \propto V^2; \quad F \propto \frac{1}{x^3}$$ | ||
| + | |||
| + | <b>d)</b> As rigorously derived in part (b), the induced dipole moment of a sphere is directly proportional to its volume (the cube of its radius): | ||
| + | $$p = 4\pi\varepsilon_0 R^3 \frac{\varepsilon - 1}{\varepsilon + 2} E_0 \implies p \propto R^3$$ | ||
| + | For the same material ($\varepsilon$) and the same position in space ($E$ and $\partial E/\partial x$ are equal), the attractive force depends exclusively on the dipole moment: | ||
| + | $$F \propto p \propto R^3$$ | ||
| + | Consequently, the ratio of the forces is equal to the ratio of the cubes of the dust particles' radii: | ||
| + | $$\frac{F_R}{F_r} = \frac{R^3}{r^3}$$ | ||
| + | |||
| + | #### Answer | ||
| + | a. To the wire | ||
| + | b. $F_2 = F_1 \frac{(\varepsilon_2 - 1)(\varepsilon_1 + 2)}{(\varepsilon_1 - 1)(\varepsilon_2 + 2)}$ | ||
| + | c. $F \propto V^2$, $F \propto 1/x^3$ | ||
| + | d. By a factor of $R^3 / r^3$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,63 @@ | |||
| ### Statement | |||
| $6.6.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $6.6.15.$ An electrostatic precipitator consists of a long metal tube and a wire directed along the axis. A potential difference $V$ is created between them. Air with dust is passed through the tube. | |||
| <b>a.</b> To which electrode — to the wire or to the tube — are dust particles attracted? | |||
| <b>b.</b> What is the force acting on a dust particle with dielectric permittivity $\varepsilon_2$, if the force acting on a dust particle of the same radius, but with dielectric permittivity $\varepsilon_1$, is equal to $F_1$? Both dust particles are equally distant from the wire. | |||
| <b>c.</b> How does the force of attraction depend on the potential difference? On the distance to the wire? | |||
| $d^\ast$. How many times is the force acting on a dust particle of radius $R$ greater than the force acting on a dust particle of radius $r < R$? The dielectric permittivity of the dust particles is the same, and they are located at the same distance from the wire. | |||
| ### Solution | |||
| <b>a)</b> Let us find the distribution of the electric field inside the electrostatic precipitator. The system is a cylindrical capacitor. According to Gauss's law, the electric field strength $E$ at a distance $x$ from the wire axis is inversely proportional to this distance: $E(x) \propto 1/x$. Thus, the field is maximum near the wire and decreases as it approaches the tube. | |||
| The dust particle is initially electrically neutral, but in an external field it polarizes, acquiring an induced dipole moment $\vec{p}$ co-directed with the field. The force acting on a dipole in a non-uniform field is: | |||
| $$F_x = p \frac{\partial E}{\partial x}$$ | |||
| Since the field decreases with distance, the gradient $\partial E / \partial x < 0$. This means that the force vector is directed opposite to the $x$-axis, i.e., it pulls the dust particle into the region of a stronger field. Consequently, the dust particles are always attracted <b>to the wire</b>. | |||
| <b>b)</b> Let us find the dependence of the induced dipole moment on the dielectric permittivity. Consider a dielectric sphere (dust particle) of radius $R$ placed in a uniform external field $\vec{E}_0$. Polarization leads to the appearance of bound charges on the surface, which create a depolarizing field inside the sphere: | |||
| $$\vec{E}_{\text{dep}} = -\frac{\vec{P}}{3\varepsilon_0}$$ | |||
| where $\vec{P}$ is the polarization vector. The true field inside the sphere is the sum of these fields: | |||
| $$\vec{E}_{\text{in}} = \vec{E}_0 - \frac{\vec{P}}{3\varepsilon_0}$$ | |||
| By definition, the polarization of a linear dielectric is: | |||
| $$\vec{P} = \varepsilon_0(\varepsilon - 1)\vec{E}_{\text{in}} = \varepsilon_0(\varepsilon - 1)\left(\vec{E}_0 - \frac{\vec{P}}{3\varepsilon_0}\right)$$ | |||
| Let us express the polarization vector from this equation: | |||
| $$\vec{P} \left( 1 + \frac{\varepsilon - 1}{3} \right) = \varepsilon_0(\varepsilon - 1)\vec{E}_0 \implies \vec{P} = 3\varepsilon_0 \frac{\varepsilon - 1}{\varepsilon + 2} \vec{E}_0$$ | |||
| The total dipole moment of the dust particle $p$ is equal to the product of the polarization and the volume of the sphere $V = \frac{4}{3}\pi R^3$: | |||
| $$p = P \cdot V = 4\pi\varepsilon_0 R^3 \frac{\varepsilon - 1}{\varepsilon + 2} E_0$$ | |||
| Since the attractive force is proportional to the dipole moment ($F \propto p$), it depends on the permittivity as: | |||
| $$F \propto \frac{\varepsilon - 1}{\varepsilon + 2}$$ | |||
| Let us write the ratio of forces for two dust particles: | |||
| $$\frac{F_2}{F_1} = \frac{\frac{\varepsilon_2 - 1}{\varepsilon_2 + 2}}{\frac{\varepsilon_1 - 1}{\varepsilon_1 + 2}}$$ | |||
| From this, we find the required force: | |||
| $$F_2 = F_1 \frac{(\varepsilon_2 - 1)(\varepsilon_1 + 2)}{(\varepsilon_1 - 1)(\varepsilon_2 + 2)}$$ | |||
| <b>c)</b> The field strength of a cylindrical capacitor is expressed through the potential difference $V$, the radius of the wire $r_0$, and the radius of the tube $R_0$: | |||
| $$E(x) = \frac{V}{x \ln(R_0/r_0)}$$ | |||
| Since $p = \alpha E$ (where $\alpha$ is a constant of the dust particle found in part b), the force can be rewritten as: | |||
| $$F = p \frac{\partial E}{\partial x} = \alpha E \frac{\partial E}{\partial x} = \frac{1}{2}\alpha \frac{\partial (E^2)}{\partial x}$$ | |||
| Substitute the field function $E(x)$: | |||
| $$E^2 = \frac{V^2}{x^2 \ln^2(R_0/r_0)} \implies \frac{\partial (E^2)}{\partial x} = -\frac{2V^2}{x^3 \ln^2(R_0/r_0)}$$ | |||
| It can be seen that the magnitude of the attractive force is directly proportional to the square of the voltage and inversely proportional to the cube of the distance: | |||
| $$F \propto V^2; \quad F \propto \frac{1}{x^3}$$ | |||
| <b>d)</b> As rigorously derived in part (b), the induced dipole moment of a sphere is directly proportional to its volume (the cube of its radius): | |||
| $$p = 4\pi\varepsilon_0 R^3 \frac{\varepsilon - 1}{\varepsilon + 2} E_0 \implies p \propto R^3$$ | |||
| For the same material ($\varepsilon$) and the same position in space ($E$ and $\partial E/\partial x$ are equal), the attractive force depends exclusively on the dipole moment: | |||
| $$F \propto p \propto R^3$$ | |||
| Consequently, the ratio of the forces is equal to the ratio of the cubes of the dust particles' radii: | |||
| $$\frac{F_R}{F_r} = \frac{R^3}{r^3}$$ | |||
| #### Answer | |||
| a. To the wire | |||
| b. $F_2 = F_1 \frac{(\varepsilon_2 - 1)(\varepsilon_1 + 2)}{(\varepsilon_1 - 1)(\varepsilon_2 + 2)}$ | |||
| c. $F \propto V^2$, $F \propto 1/x^3$ | |||
| d. By a factor of $R^3 / r^3$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||