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| ### Statement |
| ### Statement |
| $6.6.22.$ One plate of an uncharged capacitor is made of a fine grid and lies on the surface of a liquid of density $\rho$ and dielectric permittivity $\varepsilon$. The area of each plate is $S$. To what height will the liquid level rise in the capacitor if a charge $Q$ is given to it? |
| $6.6.22.$ One plate of an uncharged capacitor is made of a fine grid and lies on the surface of a liquid of density $\rho$ and dielectric permittivity $\varepsilon$. The area of each plate is $S$. To what height will the liquid level rise in the capacitor if a charge $Q$ is given to it? |
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| ### Solution |
| ### Solution |
| The capacitance of the capacitor (which can be viewed as two capacitors connected in series) is: |
| The capacitance of the capacitor (which can be viewed as two capacitors connected in series) is: |
| $$ | | $$ |
| C = \left( \frac{d-h}{\varepsilon_0 S} + \frac{h}{\varepsilon \varepsilon_0 S} \right)^{-1} | | C = \left( \frac{d-h}{\varepsilon_0 S} + \frac{h}{\varepsilon \varepsilon_0 S} \right)^{-1} |
| = \frac{\varepsilon_0 S}{d - h + h/\varepsilon}. | | = \frac{\varepsilon_0 S}{d - h + h/\varepsilon}. |
| $$ | | $$ |
| The field energy at fixed charge is: | | The field energy at fixed charge is: |
| $$ | | $$ |
| W_c = \frac{Q^2}{2C} = \frac{Q^2}{2\varepsilon_0 S} \left( d - h + \frac{h}{\varepsilon} \right). | | W_c = \frac{Q^2}{2C} = \frac{Q^2}{2\varepsilon_0 S} \left( d - h + \frac{h}{\varepsilon} \right). |
| $$ | | $$ |
| The potential energy of the liquid is: | | The potential energy of the liquid is: |
| $$ | | $$ |
| W_p = \frac{h}{2} m g = \frac{1}{2} S h^2 \rho g. | | W_p = \frac{h}{2} m g = \frac{1}{2} S h^2 \rho g. |
| $$ | | $$ |
| Equilibrium corresponds to a minimum of the total energy; we find this position using the derivative: | | Equilibrium corresponds to a minimum of the total energy; we find this position using the derivative: |
| $$ | | $$ |
| \frac{d}{dh} \bigl[ W_p - W_c \bigr] = 0 | | \frac{d}{dh} \bigl[ W_p - W_c \bigr] = 0 |
| \quad\Rightarrow\quad | | \quad\Rightarrow\quad |
| S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) = 0. | | S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) = 0. |
| $$ | | $$ |
| The second derivative: | | The second derivative: |
| $$ | | $$ |
| \frac{d}{dh} \left[ S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) \right] > 0 | | \frac{d}{dh} \left[ S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) \right] > 0 |
| \quad\Rightarrow\quad h > 0 \text{ – obviously.} | | \quad\Rightarrow\quad h > 0 \text{ – obviously.} |
| $$ | | $$ |
| Hence the only equilibrium height of rise is: | | Hence the only equilibrium height of rise is: |
| $$ | | $$ |
| h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}. | | h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}. |
| $$ | | $$ |
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| #### Answer | | #### Answer |
| $$ | | $$ |
| \boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}. | | \boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}. |
| $$ | | $$ |
