Solution to Question 38
General Theory for Constant Current Discharge
We desire the discharge current $I$ to be constant. Since current is the rate of flow of charge, $I = -\frac{dq}{dt}$. If $I$ is constant, the charge on the capacitor must decrease linearly with time: $$q(t) = q_0 – I t$$ Applying Kirchhoff’s Voltage Law to the loop at any instant $t$: $$V_C – V_R = 0 \implies \frac{q(t)}{C(t)} – I R(t) = 0$$ Therefore, the fundamental condition for all three schemes is: $$\frac{q_0 – I t}{C(t)} = I R(t)$$
Part (a): Variable Resistor $R(t)$
Here, the capacitance is fixed at $C_0$, and the resistance $R(t)$ is varied. Using the general condition: $$\frac{q_0 – I t}{C_0} = I R(t)$$ $$R(t) = \frac{q_0}{I C_0} – \frac{t}{C_0}$$ We are given that the range of resistance is from $R_0$ to $0$. At $t=0$, the charge is maximum ($q_0$), so the potential difference across the capacitor is maximum. To maintain a specific finite current $I$, the resistance must be at its maximum value $R_0$. At $t=0$: $$R(0) = \frac{q_0}{I C_0} = R_0 \implies I = \frac{q_0}{R_0 C_0}$$ Substituting this value of $I$ back into the expression for $R(t)$: $$R(t) = \frac{q_0}{( \frac{q_0}{R_0 C_0} ) C_0} – \frac{t}{C_0}$$ $$R(t) = R_0 – \frac{t}{C_0}$$ To express this purely in terms of given constants, we can factor out $R_0$: $$R(t) = R_0 \left( 1 – \frac{t}{R_0 C_0} \right)$$
Part (b): Variable Rotating Capacitor
Here, the resistance is fixed at $R_0$, and the capacitance $C(t)$ is varied.
From the loop equation:
$$\frac{q(t)}{C(t)} = I R_0$$
Since the RHS ($I R_0$) is constant, the potential difference across the capacitor must be constant.
$$\frac{q_0 – I t}{C(t)} = \frac{q_0}{C_0} \quad (\text{Value at } t=0)$$
$$C(t) = C_0 \left( \frac{q_0 – I t}{q_0} \right) = C_0 \left( 1 – \frac{I}{q_0} t \right)$$
Using the initial condition from part (a) logic, $I = \frac{q_0}{R_0 C_0}$, we get:
$$C(t) = C_0 \left( 1 – \frac{t}{R_0 C_0} \right)$$
Angular Velocity $\omega(t)$:
For a rotary capacitor with overlapping plates, the capacitance is proportional to the overlap angle $\theta$.
$$C(\theta) = k \theta$$
Differentiating with respect to time:
$$\frac{dC}{dt} = k \frac{d\theta}{dt} = k \omega$$
Since $C_0$ corresponds to the maximum overlap (say $\pi$ for semicircular plates), $k = \frac{C_0}{\pi}$. Thus, $\frac{dC}{dt} = \frac{C_0}{\pi} \omega$.
From our derived equation for $C(t)$, the rate of change is:
$$\frac{dC}{dt} = \frac{d}{dt} \left[ C_0 \left( 1 – \frac{t}{R_0 C_0} \right) \right] = -\frac{1}{R_0}$$
Equating the magnitudes (since $\omega$ is a speed):
$$\frac{C_0}{\pi} \omega = \frac{1}{R_0} \implies \omega = \frac{\pi}{R_0 C_0}$$
Part (c): Variable Linear Capacitor
The law for capacitance variation $C(t)$ remains the same as in part (b) because $R_0$ is fixed:
$$C(t) = C_0 \left( 1 – \frac{t}{R_0 C_0} \right)$$
Distance $x(t)$:
For a parallel plate capacitor, capacitance is inversely proportional to the separation distance $x$:
$$C(x) = \frac{\epsilon_0 A}{x}$$
Let initial distance be $x_0$, so $C_0 = \frac{\epsilon_0 A}{x_0}$.
$$C(t) = \frac{\epsilon_0 A}{x(t)}$$
Substituting these into the $C(t)$ equation:
$$\frac{\epsilon_0 A}{x(t)} = \frac{\epsilon_0 A}{x_0} \left( 1 – \frac{t}{R_0 C_0} \right)$$
$$\frac{1}{x(t)} = \frac{1}{x_0} \left( \frac{R_0 C_0 – t}{R_0 C_0} \right)$$
$$x(t) = x_0 \left( \frac{R_0 C_0}{R_0 C_0 – t} \right)$$
(b) $C(t) = C_0 \left( 1 – \frac{t}{R_0 C_0} \right)$ and $\omega(t) = \frac{\pi}{R_0 C_0}$
(c) $C(t) = C_0 \left( 1 – \frac{t}{R_0 C_0} \right)$ and $x(t) = \frac{x_0 R_0 C_0}{R_0 C_0 – t}$