Solution to Question 19
Problem Analysis
A mica sheet of thickness $d/2$ is being pulled out of a capacitor of plate separation $d$ and area $l \times l$. Let $x$ be the length of the mica sheet currently inside the capacitor.
The capacitor can be modeled as two parallel regions:
- Region 1 (Length $x$): A composite capacitor with air ($d/2$) and mica ($d/2$) in series.
- Region 2 (Length $l-x$): A pure air capacitor of thickness $d$.
1. Equivalent Capacitance $C(x)$
Region 1 (Composite):
$$ \frac{1}{C_1} = \frac{1}{C_{air}’} + \frac{1}{C_{mica}’} = \frac{d/2}{\varepsilon_0 l x} + \frac{d/2}{\varepsilon_r \varepsilon_0 l x} $$ $$ \frac{1}{C_1} = \frac{d}{2 \varepsilon_0 l x} \left( 1 + \frac{1}{\varepsilon_r} \right) = \frac{d (\varepsilon_r + 1)}{2 \varepsilon_0 \varepsilon_r l x} $$ $$ C_1 = \frac{2 \varepsilon_0 \varepsilon_r l x}{d (\varepsilon_r + 1)} $$Region 2 (Air):
$$ C_2 = \frac{\varepsilon_0 l (l-x)}{d} $$Total Capacitance:
$$ C(x) = C_1 + C_2 = \frac{\varepsilon_0 l}{d} \left[ (l-x) + \frac{2 \varepsilon_r x}{\varepsilon_r + 1} \right] $$Simplify the term in brackets:
$$ C(x) = \frac{\varepsilon_0 l}{d} \left[ l + x \left( \frac{2 \varepsilon_r}{\varepsilon_r + 1} – 1 \right) \right] = \frac{\varepsilon_0 l}{d} \left[ l + x \left( \frac{\varepsilon_r – 1}{\varepsilon_r + 1} \right) \right] $$2. Current Flow
Since the sheet is pulled “slowly”, we assume the voltage across the capacitor remains approximately $V$ (quasi-static approximation). The current is generated due to the change in capacitance.
$$ I = \frac{dQ}{dt} = V \frac{dC}{dt} = V \frac{dC}{dx} \frac{dx}{dt} $$The sheet is pulled out with speed $v$, so the length inside $x$ decreases: $\frac{dx}{dt} = -v$.
$$ \frac{dC}{dx} = \frac{\varepsilon_0 l}{d} \left( \frac{\varepsilon_r – 1}{\varepsilon_r + 1} \right) $$Substituting this into the current equation:
$$ I = V \left[ \frac{\varepsilon_0 l}{d} \left( \frac{\varepsilon_r – 1}{\varepsilon_r + 1} \right) \right] (-v) $$Magnitude of current (constant):
$$ I = \frac{\varepsilon_0 l v V}{d} \left( \frac{\varepsilon_r – 1}{\varepsilon_r + 1} \right) $$3. Heat Developed
The total heat developed in resistance $R$ is $H = \int I^2 R dt$. Since $I$ is constant, $H = I^2 R \Delta t$.
The time taken to pull the sheet of length $l$ completely out is $\Delta t = l/v$.
$$ H = \left[ \frac{\varepsilon_0 l v V}{d} \left( \frac{\varepsilon_r – 1}{\varepsilon_r + 1} \right) \right]^2 R \left( \frac{l}{v} \right) $$ $$ H = \frac{\varepsilon_0^2 l^2 v^2 V^2}{d^2} \left( \frac{\varepsilon_r – 1}{\varepsilon_r + 1} \right)^2 R \frac{l}{v} $$