Physics Solution – Question 18

Solution to Question 18

1. Initial Steady State (Dielectric Inserted)

Initially, the capacitor is filled with a dielectric of constant $k$. In the steady state, the capacitor is fully charged to the battery potential $V$.

Initial Capacitance: $C_i = kC$
Initial Charge: $Q_i = C_i V = kCV$
Initial Energy stored: $U_i = \frac{1}{2} C_i V^2 = \frac{1}{2} kCV^2$

2. Analysis of the “Quick Pull”

The problem states the plate is “quickly pulled out”. This mechanical process occurs much faster than the electrical time constant of the circuit ($\tau = RC$). Consequently, the charge on the plates remains constant during the pull.

Charge just after pull: $Q’ = Q_i = kCV$ (conserved)
New Capacitance: $C_{new} = C$
Potential just after pull: $V’ = \frac{Q’}{C_{new}} = \frac{kCV}{C} = kV$

The energy stored in the capacitor immediately after the pull is:

$$ U_{after\_pull} = \frac{(Q’)^2}{2C_{new}} = \frac{(kCV)^2}{2C} = \frac{1}{2} k^2 CV^2 $$

3. Final Steady State

After the pull, the capacitor (now at potential $kV$) is connected to the battery (potential $V$). Since $k > 1$, $V’ > V$, so the capacitor will discharge into the battery until its potential equals $V$.

Final Capacitance: $C_f = C$
Final Charge: $Q_f = CV$
Final Energy stored: $U_f = \frac{1}{2} CV^2$

4. Calculation of Heat Generated

We apply the law of conservation of energy for the period after the pull until the new steady state is reached.

Energy Balance Equation: $$ U_{after\_pull} = U_{final} + W_{absorbed\_by\_battery} + H $$ where $H$ is the heat dissipated in the resistance.

Calculation of Charge Flow:
The charge on the capacitor changes from $Q’ = kCV$ to $Q_f = CV$. The charge flowing out of the capacitor and into the positive terminal of the battery is: $$ \Delta q = Q’ – Q_f = kCV – CV = (k-1)CV $$

Work done on the Battery:
Since charge flows into the positive terminal of the battery (against the EMF), the battery absorbs energy: $$ W_{batt} = \Delta q \cdot V = [(k-1)CV] \cdot V = (k-1)CV^2 $$

Solving for Heat ($H$):

$$ \frac{1}{2} k^2 CV^2 = \frac{1}{2} CV^2 + (k-1)CV^2 + H $$ $$ H = \frac{1}{2} k^2 CV^2 – \frac{1}{2} CV^2 – (k-1)CV^2 $$ $$ H = \frac{1}{2} CV^2 \left[ k^2 – 1 – 2(k-1) \right] $$ $$ H = \frac{1}{2} CV^2 \left[ k^2 – 1 – 2k + 2 \right] $$ $$ H = \frac{1}{2} CV^2 (k^2 – 2k + 1) $$ $$ H = \frac{1}{2} CV^2 (k-1)^2 $$

The correct expression for the heat generated is: $$ \frac{1}{2} CV^2 (k-1)^2 $$ (Option d)