Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K$$_1$$, K$$_2$$ and K$$_3$$. The first capacitor is filled as shown in figure I, and the second one is filled as shown in figure II.

If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E$$_1$$ refers to capacitor I and E$$_2$$ to capacitor II):

To find the ratio of the energy stored in the two configurations, we first need to determine the equivalent capacitance of both setups.

The energy stored in a capacitor connected to a potential $$V$$ is given by:

$$U = \frac{1}{2} C V^2$$

Since both modified capacitors are charged to the same potential $$V$$, the ratio of their energies is simply the ratio of their capacitances:

$$\frac{E_1}{E_2} = \frac{\frac{1}{2} C_1 V^2}{\frac{1}{2} C_2 V^2} = \frac{C_1}{C_2}$$

Let the standard capacitance without any dielectric be $$C_0 = \frac{\varepsilon_0 A}{d}$$.

1. Analysis of Configuration (I) - Series Combination

In Figure I, the dielectrics split the total plate separation distance $$d$$ into three equal horizontal layers of thickness $$\frac{d}{3}$$. However, each layer retains the full area $$A$$.

This forms three separate capacitors connected in series:

• $$C_a = K_1 \frac{\varepsilon_0 A}{d/3} = 3K_1 C_0$$
• $$C_b = K_2 \frac{\varepsilon_0 A}{d/3} = 3K_2 C_0$$
• $$C_c = K_3 \frac{\varepsilon_0 A}{d/3} = 3K_3 C_0$$
• $$C_x = K_1 \frac{\varepsilon_0 (A/3)}{d} = \frac{K_1}{3} C_0$$
• $$C_y = K_2 \frac{\varepsilon_0 (A/3)}{d} = \frac{K_2}{3} C_0$$
• $$C_z = K_3 \frac{\varepsilon_0 (A/3)}{d} = \frac{K_3}{3} C_0$$
• Correct Option: A

Using the reciprocal rule for a series combination ($$\frac{1}{C_1} = \frac{1}{C_a} + \frac{1}{C_b} + \frac{1}{C_c}$$):

$$\frac{1}{C_1} = \frac{1}{3C_0} \left( \frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3} \right)$$

$$\frac{1}{C_1} = \frac{1}{3C_0} \left( \frac{K_2K_3 + K_3K_1 + K_1K_2}{K_1K_2K_3} \right)$$

$$C_1 = 3C_0 \left[ \frac{K_1K_2K_3}{K_1K_2 + K_2K_3 + K_3K_1} \right]$$

2. Analysis of Configuration (II) - Parallel Combination

In Figure II, the dielectrics split the plate area $$A$$ into three equal vertical sections of area $$\frac{A}{3}$$. However, each section spans the full separation distance $$d$$.

This forms three separate capacitors connected in parallel:

Adding them directly for a parallel combination ($$C_2 = C_x + C_y + C_z$$):

$$C_2 = \frac{C_0}{3} (K_1 + K_2 + K_3)$$

3. Calculating the Energy Ratio ($$\frac{E_1}{E_2}$$)

Now we compute the ratio $$\frac{C_1}{C_2}$$:

$$\frac{E_1}{E_2} = \frac{3C_0 \left[ \frac{K_1K_2K_3}{K_1K_2 + K_2K_3 + K_3K_1} \right]}{\frac{C_0}{3} (K_1 + K_2 + K_3)}$$

Simplifying the constants ($$3 \div \frac{1}{3} = 9$$), the $$C_0$$ terms cancel out:

$$\frac{E_1}{E_2} = \frac{9K_1K_2K_3}{(K_1 + K_2 + K_3)(K_1K_2 + K_2K_3 + K_3K_1)}$$

(Note: The options write the sum term slightly differently as $$(K_1+K_2+K_3)$$, which perfectly matches the algebraic structure of the denominator).