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en/6.6.14.md
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| + | ### Statement | ||
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| + | $6.6.14.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | ### Statement | ||
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| + | $6.6.14.$ Charges $\pm q$ are placed on the plates of a flat capacitor. The gap between the plates is filled with a substance whose dielectric permittivity varies in the direction perpendicular to the plates according to the law $\varepsilon = \varepsilon_0(1 + x/d)^{-1}$, where $x$ is the distance to the positive plate, and $d$ is the distance between the plates. Find the volume charge density as a function of $x$. The area of the plates is $S$. | ||
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| + | ### Solution | ||
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| + | <b>1.</b> The electric displacement vector $\vec{D}$ inside the dielectric is determined only by the free charges on the metal plates of the capacitor. Since there is no free volume charge in the dielectric itself ($\nabla \cdot \vec{D} = 0$), the displacement field $\vec{D}$ is uniform throughout the volume and directed from the positive plate to the negative one (along the $x$-axis). Its magnitude equals the surface free charge density: | ||
| + | $$D = \frac{q}{S}$$ | ||
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| + | <b>2.</b> The electric field strength $\vec{E}$ is related to the displacement vector by the equation $\vec{D} = \varepsilon_{v} \varepsilon(x) \vec{E}$ (in SI units, where $\varepsilon_{v}$ is the vacuum permittivity and $\varepsilon(x)$ is the relative permittivity of the material). We express the electric field $E(x)$ as: | ||
| + | $$E(x) = \frac{D}{\varepsilon_{v} \varepsilon(x)} = \frac{q}{S \varepsilon_{v} \varepsilon_0 (1 + x/d)^{-1}} = \frac{q}{S \varepsilon_{v} \varepsilon_0} \left(1 + \frac{x}{d}\right)$$ | ||
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| + | <b>3.</b> The volume density of all charges in the dielectric (which consists only of the required polarization/bound charges $\rho'$) is related to the field $\vec{E}$ by the local Gauss's theorem: $\nabla \cdot \vec{E} = \rho' / \varepsilon_{v}$. | ||
| + | Since the field depends only on the $x$ coordinate, the divergence reduces to a simple derivative: | ||
| + | $$\rho' = \varepsilon_{v} \frac{d E(x)}{dx}$$ | ||
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| + | <b>4.</b> Substitute the function $E(x)$ into the derivative: | ||
| + | $$\rho' = \varepsilon_{v} \frac{d}{dx} \left[ \frac{q}{S \varepsilon_{v} \varepsilon_0} \left(1 + \frac{x}{d}\right) \right] = \varepsilon_{v} \frac{q}{S \varepsilon_{v} \varepsilon_0} \cdot \frac{1}{d} = \frac{q}{S d \varepsilon_0}$$ | ||
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| + | <i>Note:</i> The dielectric permittivity decreases with distance, so the polarization weakens, causing positive bound charge to accumulate in the volume. The minus sign in the textbook's official answer is a mathematical error (the minus sign was lost in the relation $\rho' = -\nabla \cdot \vec{P}$). Also, in some editions, the parameter $\varepsilon_0$ is mistakenly printed as $\varepsilon_1$ in the answer key. | ||
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| + | #### Answer | ||
| + | $$\rho' = \frac{q}{S d \varepsilon_0}$$ | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $6.6.14.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $6.6.14.$ Charges $\pm q$ are placed on the plates of a flat capacitor. The gap between the plates is filled with a substance whose dielectric permittivity varies in the direction perpendicular to the plates according to the law $\varepsilon = \varepsilon_0(1 + x/d)^{-1}$, where $x$ is the distance to the positive plate, and $d$ is the distance between the plates. Find the volume charge density as a function of $x$. The area of the plates is $S$. | |||
| ### Solution | |||
| <b>1.</b> The electric displacement vector $\vec{D}$ inside the dielectric is determined only by the free charges on the metal plates of the capacitor. Since there is no free volume charge in the dielectric itself ($\nabla \cdot \vec{D} = 0$), the displacement field $\vec{D}$ is uniform throughout the volume and directed from the positive plate to the negative one (along the $x$-axis). Its magnitude equals the surface free charge density: | |||
| $$D = \frac{q}{S}$$ | |||
| <b>2.</b> The electric field strength $\vec{E}$ is related to the displacement vector by the equation $\vec{D} = \varepsilon_{v} \varepsilon(x) \vec{E}$ (in SI units, where $\varepsilon_{v}$ is the vacuum permittivity and $\varepsilon(x)$ is the relative permittivity of the material). We express the electric field $E(x)$ as: | |||
| $$E(x) = \frac{D}{\varepsilon_{v} \varepsilon(x)} = \frac{q}{S \varepsilon_{v} \varepsilon_0 (1 + x/d)^{-1}} = \frac{q}{S \varepsilon_{v} \varepsilon_0} \left(1 + \frac{x}{d}\right)$$ | |||
| <b>3.</b> The volume density of all charges in the dielectric (which consists only of the required polarization/bound charges $\rho'$) is related to the field $\vec{E}$ by the local Gauss's theorem: $\nabla \cdot \vec{E} = \rho' / \varepsilon_{v}$. | |||
| Since the field depends only on the $x$ coordinate, the divergence reduces to a simple derivative: | |||
| $$\rho' = \varepsilon_{v} \frac{d E(x)}{dx}$$ | |||
| <b>4.</b> Substitute the function $E(x)$ into the derivative: | |||
| $$\rho' = \varepsilon_{v} \frac{d}{dx} \left[ \frac{q}{S \varepsilon_{v} \varepsilon_0} \left(1 + \frac{x}{d}\right) \right] = \varepsilon_{v} \frac{q}{S \varepsilon_{v} \varepsilon_0} \cdot \frac{1}{d} = \frac{q}{S d \varepsilon_0}$$ | |||
| <i>Note:</i> The dielectric permittivity decreases with distance, so the polarization weakens, causing positive bound charge to accumulate in the volume. The minus sign in the textbook's official answer is a mathematical error (the minus sign was lost in the relation $\rho' = -\nabla \cdot \vec{P}$). Also, in some editions, the parameter $\varepsilon_0$ is mistakenly printed as $\varepsilon_1$ in the answer key. | |||
| #### Answer | |||
| $$\rho' = \frac{q}{S d \varepsilon_0}$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||