Question

A parallel plate capacitor has two layers of dielectric as shown in figure. If this capacitor is connected across a battery, then the ratio of potential difference across the dielectric layer is

• A. $\frac{4}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

3/2

Answer 2

The given setup shows a parallel plate capacitor with two dielectric layers placed side-by-side, effectively forming two capacitors in series.

1. Identify the components:

   • Layer 1: Dielectric constant $K_1 = 2$, thickness $d_1 = d$.
   • Layer 2: Dielectric constant $K_2 = 6$, thickness $d_2 = 2d$. Let the area of the capacitor plates be $A$.

2. Capacitance of each layer: The capacitance of a parallel plate capacitor with a dielectric of constant $K$ and thickness $t$ is given by $C = \frac{K \epsilon_0 A}{t}$.

   • Capacitance of the first layer ($C_1$): $C_1 = \frac{K_1 \epsilon_0 A}{d_1} = \frac{2 \epsilon_0 A}{d}$
   • Capacitance of the second layer ($C_2$): $C_2 = \frac{K_2 \epsilon_0 A}{d_2} = \frac{6 \epsilon_0 A}{2d} = \frac{3 \epsilon_0 A}{d}$

3. Potential difference across layers: Since the two dielectric layers are in series, the charge ($Q$) stored on each equivalent capacitor is the same. The potential difference across a capacitor is given by $V = \frac{Q}{C}$.

   • Potential difference across the first layer ($V_1$): $V_1 = \frac{Q}{C_1} = \frac{Q}{\frac{2 \epsilon_0 A}{d}} = \frac{Qd}{2 \epsilon_0 A}$
   • Potential difference across the second layer ($V_2$): $V_2 = \frac{Q}{C_2} = \frac{Q}{\frac{3 \epsilon_0 A}{d}} = \frac{Qd}{3 \epsilon_0 A}$

4. Ratio of potential differences: The ratio of potential difference across the dielectric layers is $\frac{V_1}{V_2}$: $\frac{V_1}{V_2} = \frac{\frac{Qd}{2 \epsilon_0 A}}{\frac{Qd}{3 \epsilon_0 A}}$ $\frac{V_1}{V_2} = \frac{Qd}{2 \epsilon_0 A} \times \frac{3 \epsilon_0 A}{Qd}$ $\frac{V_1}{V_2} = \frac{3}{2}$

The ratio of potential difference across the dielectric layers is $\frac{3}{2}$.