$6.6.15.$ 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.\
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<b>a.</b> To which electrode — to the wire or to the tube — are dust particles attracted?\
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<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.\
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<b>c.</b> How does the force of attraction depend on the potential difference? On the distance to the wire?\
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<b>d*.</b> 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
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### Statement
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$6.6.15.$ 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.
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<b>a.</b> To which electrode — to the wire or to the tube — are dust particles attracted?
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<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.
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<b>c.</b> How does the force of attraction depend on the potential difference? On the distance to the wire?
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$d^\ast$. 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.
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### Solution
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<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:
<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:
$$F = p \frac{\partial E}{\partial x} = \alpha E \frac{\partial E}{\partial x} = \frac{1}{2}\alpha\frac{\partial (E^2)}{\partial x}$$
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:
$$F \propto V^2; \quad F \propto\frac{1}{x^3}$$
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<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):
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<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):
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:
$6.6.15.$ 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?\
<b>d*.</b> 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
### 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.
<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:
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:
$$F_x = p \frac{\partial E}{\partial x}$$
$$F_x = p \frac{\partial E}{\partial x}$$
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>.
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:
<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:
<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 $R_0$:
<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 $R_0$:
$$E(x) = \frac{V}{x \ln(R_0/r_0)}$$
$$E(x) = \frac{V}{x \ln(R_0/r_0)}$$
Since $p = \alpha E$ (where $\alpha$ is a constant of the dust particle found in part b), the force can be rewritten as:
Since $p = \alpha E$ (where $\alpha$ is a constant of the dust particle found in part b), the force can be rewritten as:
$$F = p \frac{\partial E}{\partial x} = \alpha E \frac{\partial E}{\partial x} = \frac{1}{2}\alpha\frac{\partial (E^2)}{\partial x}$$
$$F = p \frac{\partial E}{\partial x} = \alpha E \frac{\partial E}{\partial x} = \frac{1}{2}\alpha\frac{\partial (E^2)}{\partial x}$$
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:
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:
$$F \propto V^2; \quad F \propto\frac{1}{x^3}$$
$$F \propto V^2; \quad F \propto\frac{1}{x^3}$$
<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):
<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):
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:
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:
$$F \propto p \propto R^3$$
$$F \propto p \propto R^3$$
Consequently, the ratio of the forces is equal to the ratio of the cubes of the dust particles' radii:
Consequently, the ratio of the forces is equal to the ratio of the cubes of the dust particles' radii:
