Two identical parallel plate capacitors of capacitance C each, have plates of area A, separated by a

Q: Why are the dielectrics in Fig I considered to be in parallel?

Because the top and bottom plates are continuous across all three dielectrics. This means the potential difference $V$ across each dielectric slab is exactly the same, which is the defining characteristic of a parallel combination.

Q: Why does the area become $A/3$ in Fig I, but the distance remains $d$?

The dielectrics are placed side-by-side, so they share the total area $A$ equally, giving each an area of $A/3$. However, they all span the entire gap between the top and bottom plates, so their thickness remains $d$.

Q: In Fig II, why is it a series combination?

The dielectrics are stacked vertically. When a voltage is applied across the outer plates, the same amount of charge is induced on the boundary of each dielectric layer. Equal charge across components is the hallmark of a series circuit.

Q: Why does the energy ratio simplify to just the ratio of capacitances?

The formula for the energy stored in a capacitor is $E = \frac{1}{2}CV^2$. The problem states that both modified capacitors are charged to the *same potential* $V$. Therefore, the $V^2$ terms cancel out when you take the ratio, leaving only $C_I / C_{II}$.

Q: What happens if the dielectrics had different thicknesses in Fig II?

If the thicknesses were different (say $d_1, d_2, d_3$), you would use those specific values in the capacitance formula $C = \frac{K \varepsilon_0 A}{d_i}$ for each layer. The series combination formula $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$ would still apply, but the final algebraic expression would be more complex.

Question

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 Fig. I, and the second one is filled as shown in Fig. 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)) :

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the First Setup

Let's start by looking at the first capacitor configuration.

Notice how the three dielectric slabs are placed side-by-side between the plates.

Because they are connected across the same two plates, the potential difference across each section is identical.

**This means we can treat this setup as three separate capacitors connected in parallel.**

### Capacitance of the First Capacitor

Now, let's calculate the equivalent capacitance for this parallel combination.

Each individual capacitor has a plate area of $A/3$, and a plate separation of $d$.

We can write down the capacitance for each section as:

$$C_1 = \frac{K_1 \varepsilon_0 (A/3)}{d}$$

$$C_2 = \frac{K_2 \varepsilon_0 (A/3)}{d}$$

$$C_3 = \frac{K_3 \varepsilon_0 (A/3)}{d}$$

Since they are in parallel, we simply add them up to find the equivalent capacitance $C_I$.

Factoring out the common terms, we get the total capacitance:

$$C_I = \frac{\varepsilon_0 A}{3d} (K_1 + K_2 + K_3)$$

### Analyzing the Second Setup

Moving on to the second capacitor, the dielectrics are stacked vertically, one on top of the other.

In this configuration, the charge induced across each dielectric layer is the same.

**Therefore, this setup behaves exactly like three separate capacitors connected in series.**

### Capacitance of the Second Capacitor

Let's find the equivalent capacitance for this series combination.

Here, each capacitor has the full plate area $A$, but the separation distance is only $d/3$.

We write down the individual capacitances:

$$C'_1 = \frac{K_1 \varepsilon_0 A}{d/3}$$

$$C'_2 = \frac{K_2 \varepsilon_0 A}{d/3}$$

$$C'_3 = \frac{K_3 \varepsilon_0 A}{d/3}$$

For a series combination, we add the reciprocals of the capacitances.

Substituting the values, we get an expression for $\frac{1}{C_{II}}$:

$$\frac{1}{C_{II}} = \frac{d}{3\varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3} \right)$$

### Simplifying the Master Equation

Let's simplify this expression to make it easier to work with.

We take the common denominator for the terms inside the bracket, which gives us $K_1K_2K_3$.

$$\frac{1}{C_{II}} = \frac{d}{3\varepsilon_0 A} \left( \frac{K_2K_3 + K_1K_3 + K_1K_2}{K_1K_2K_3} \right)$$

Then, we simply invert the entire equation to find the equivalent capacitance $C_{II}$.

Notice how the $3$ moves to the numerator here:

$$C_{II} = \frac{3\varepsilon_0 A}{d} \left( \frac{K_1K_2K_3}{K_1K_2 + K_2K_3 + K_3K_1} \right)$$

### The Energy Ratio

We are asked to find the ratio of the energies stored in the two capacitors.

The energy stored in a capacitor is given by the formula:

$$E = \frac{1}{2}CV^2$$

Since both capacitors are charged to the exact same potential $V$, the energy stored is directly proportional to their capacitance.

So, the ratio $E_1 / E_2$ is simply the ratio of $C_I$ to $C_{II}$.

### Final Calculation

Now for the final step. Let's substitute the expressions for $C_I$ and $C_{II}$ that we just derived.

$$\frac{E_1}{E_2} = \frac{\frac{\varepsilon_0 A}{3d} (K_1 + K_2 + K_3)}{\frac{3\varepsilon_0 A}{d} \left( \frac{K_1K_2K_3}{K_1K_2 + K_2K_3 + K_3K_1} \right)}$$

The $\frac{\varepsilon_0 A}{d}$ terms cancel out beautifully.

The $3$ in the denominator of $C_I$ multiplies with the $3$ in the numerator of $C_{II}$, giving us a $9$ in the denominator.

And the fraction in the denominator flips up, giving us our final elegant result:

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

This matches option (d) perfectly.

Select Answer:

Visualized Solution

Analyzing the First Capacitor

Capacitance of the First Capacitor

Analyzing the Second Capacitor

Capacitance of the Second Capacitor

Simplifying the Second Capacitance

Ratio of Stored Energies

Final Calculation

The Sigma Insight: Capacitance and Capacitors

Analyzing the First Setup

Capacitance of the First Capacitor

Analyzing the Second Setup

Capacitance of the Second Capacitor

Simplifying the Master Equation

The Energy Ratio

Final Calculation

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