GATE EE 2014 Set 2 — Question 18

To find the ratio of the permittivities $\epsilon_1 : \epsilon_2$, we can model the system as three capacitors connected in series.

1. System Geometry and Parameters

Let the area of the plates be $A$. The total distance between the plates is $d$.
The middle dielectric slab of permittivity $\epsilon_2$ has a thickness of $d/2$ and is placed symmetrically. Thus, the remaining space of thickness $d - d/2 = d/2$ is shared equally by the two outer regions of permittivity $\epsilon_1$.This gives the thicknesses of the three regions as:

• First region (permittivity $\epsilon_1$): $d_1 = \frac{d}{4}$
• Middle region (permittivity $\epsilon_2$): $d_2 = \frac{d}{2}$
• Third region (permittivity $\epsilon_1$): $d_3 = \frac{d}{4}$

2. Capacitance of Each Region

The capacitance of each of the three regions can be written as:$$C_1 = \frac{\epsilon_1 A}{d_1} = \frac{\epsilon_1 A}{d/4} = \frac{4 \epsilon_1 A}{d}$$$$C_2 = \frac{\epsilon_2 A}{d_2} = \frac{\epsilon_2 A}{d/2} = \frac{2 \epsilon_2 A}{d}$$$$C_3 = \frac{\epsilon_1 A}{d_3} = \frac{\epsilon_1 A}{d/4} = \frac{4 \epsilon_1 A}{d}$$Since the regions are in series, the same electric displacement field $D$ exists across all regions, meaning the charge $Q$ on each equivalent series capacitor is the same:$$Q = C_1 V_1 = C_2 V_2 = C_3 V_3$$where $V_1$, $V_2$, and $V_3$ are the potential drops across the respective regions.

3. Using the Potential Distribution

The total potential difference across the capacitor is:$$V_{\text{total}} = 10\text{ V} - 0\text{ V} = 10\text{ V}$$The potential drop across the first region (between one of the plates and the nearest dielectric interface) is given as:$$V_1 = 2\text{ V}$$By symmetry, the potential drop across the third region is also:$$V_3 = V_1 = 2\text{ V}$$The potential drop across the middle region $V_2$ is:$$V_2 = V_{\text{total}} - (V_1 + V_3) = 10 - (2 + 2) = 6\text{ V}$$

4. Finding the Ratio $\epsilon_1 : \epsilon_2$

Using the charge equality relation $C_1 V_1 = C_2 V_2$:$$\left(\frac{4 \epsilon_1 A}{d}\right) V_1 = \left(\frac{2 \epsilon_2 A}{d}\right) V_2$$Substitute the values of $V_1 = 2\text{ V}$ and $V_2 = 6\text{ V}$:$$4 \epsilon_1 (2) = 2 \epsilon_2 (6)$$$$8 \epsilon_1 = 12 \epsilon_2$$Solving for the ratio $\epsilon_1 : \epsilon_2$:$$\frac{\epsilon_1}{\epsilon_2} = \frac{12}{8} = \frac{3}{2}$$Thus, the ratio $\epsilon_1 : \epsilon_2$ is $3:2$.Correct Option: C