Question

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage according to the law $\epsilon_r = \alpha V$, where $\alpha = 1$ per volt. A similar (but containing no dielectric) capacitor charged to a voltage $V = 240$ volt is connected in parallel to the first "non-linear" uncharged capacitor. Determine the final voltage $V_f$ across the capacitors (in volts).

• A. 15

Answer 1

Answer: (A)

15 V

Answer 2

Here's how to determine the final voltage $V_f$ across the capacitors:

1. Nonlinear Capacitor: The dielectric constant is given by $\epsilon_r = \alpha V$, where $\alpha = 1 \, \text{V}^{-1}$. The capacitance becomes $C(V) = \epsilon_0 \epsilon_r \frac{A}{d} = \epsilon_0 \frac{A}{d} V$. The charge on this capacitor at voltage $V$ is $Q_1 = C(V)V = \epsilon_0 \frac{A}{d} V^2$.

2. Linear Capacitor: The second capacitor (without dielectric) has capacitance $C_2 = \epsilon_0 \frac{A}{d}$. Initially, it holds a charge $Q_{2, \text{initial}} = C_2 \cdot 240 = \epsilon_0 \frac{A}{d} \cdot 240$.

3. Total Initial Charge: The nonlinear capacitor is initially uncharged, so the total charge in the system is $Q_{\text{total}} = \epsilon_0 \frac{A}{d} \cdot 240$.

4. After Parallel Connection: Both capacitors share a common final voltage $V_f$. Their charges are:

   • Nonlinear capacitor: $Q_1 = \epsilon_0 \frac{A}{d} V_f^2$
   • Linear capacitor: $Q_2 = \epsilon_0 \frac{A}{d} V_f$

5. Charge Conservation: Using conservation of charge:

   $$\epsilon_0 \frac{A}{d} \cdot 240 = \epsilon_0 \frac{A}{d} (V_f^2 + V_f)$$

   Canceling the common factor $\epsilon_0 \frac{A}{d}$, we get $V_f^2 + V_f - 240 = 0$.

6. Solving the Quadratic Equation:

   $$V_f = \frac{-1 \pm \sqrt{1 + 4 \cdot 240}}{2} = \frac{-1 \pm \sqrt{961}}{2} = \frac{-1 \pm 31}{2}$$

   Taking the positive root, $V_f = \frac{30}{2} = 15$ volts.

Therefore, the final voltage $V_f = 15$ V.