A parallel plate capacitor is of area 6 cm^2 and a separation 3 mm. The gap is filled with three die

Q: The question says 'equal thickness', so shouldn't the thickness be $d/3$ for each?

This is a common trap! While the text says 'equal thickness', the phrase '(see figure)' is crucial. The diagram clearly shows the dielectrics placed side-by-side. In this physical arrangement, they all span the entire distance between the plates, so their thickness is $d$. It is their area that is divided into $A/3$.

Q: Why are these considered to be in parallel and not in series?

Capacitors are in parallel when they share the same potential difference. Here, the top surface of all three dielectrics touches the top plate, and the bottom surface touches the bottom plate. Therefore, the voltage across each dielectric section is exactly the same, making it a parallel combination.

Q: What would the arrangement look like if they were in series?

If they were in series, the dielectrics would be stacked one on top of the other, like layers in a sandwich. In that case, each layer would have the full area $A$, but the thickness would be divided into $d/3$ for each layer.

Q: Why didn't we use the values $6 \text{ cm}^2$ and $3 \text{ mm}$ in the calculation?

When you set up the equation equating the equivalent capacitance of the three sections to the capacitance of a single equivalent dielectric, the term $\frac{\varepsilon_0 A}{d}$ appears on both sides and cancels out. This means the equivalent dielectric constant depends only on the arrangement and the individual $K$ values, not on the specific dimensions of the capacitor.

Q: Can we always just take the average of the dielectric constants for parallel combinations?

You can take the simple arithmetic average ONLY if the dielectrics divide the area equally (i.e., each has area $A/n$). If they had different areas, you would need to use a weighted average based on their respective areas.

Question

A parallel plate capacitor is of area $6 	ext{ cm}^2$ and a separation $3 	ext{ mm}$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$ and $K_3 = 14$. The dielectric constant of a material which give same capacitance when fully inserted in above capacitor, would be

Options:

(A) 4

(B) 36

(C) 12

(D) 14

Sigma Percentile · Accepted Answer

### The Beauty of Dielectrics in Capacitors

Capacitors are fascinating devices that store electrical energy, and introducing a dielectric material between their plates is a classic way to enhance their capacitance. But what happens when we don't just use one dielectric, but a combination of several? This problem takes us on a journey to understand exactly that, challenging us to find a single equivalent dielectric constant that can replace a complex arrangement.

### Analyzing the Setup: Decoding the Diagram

The problem presents us with a parallel plate capacitor of area $A = 6 	ext{ cm}^2$ and a separation $d = 3 	ext{ mm}$. The gap is filled with three different dielectric materials with constants $K_1 = 10$, $K_2 = 12$, and $K_3 = 14$.

Now, here is where we must be careful. The text mentions "equal thickness", which might initially make you think they are stacked on top of each other. However, the phrase "(see figure)" is our guiding light. Looking at the diagram, it is crystal clear that the dielectrics are placed **side-by-side**.

What does this physical arrangement mean mathematically? 
1. Because they span from the top plate to the bottom plate, each dielectric has the full thickness $d$.
2. Because they are placed side-by-side and divide the total space equally, each dielectric occupies exactly one-third of the total area, meaning the area for each is $A/3$.

### The Master Equation: Parallel Combination

Since the top surfaces of all three dielectrics touch the top conducting plate, and their bottom surfaces touch the bottom conducting plate, the potential difference $V$ across each of them is identical. In the world of circuits, when components share the same potential difference, they are connected in **parallel**.

We can treat this system as three separate capacitors connected in parallel. Let's write down the capacitance for each individual section using the standard formula $C = \frac{K \varepsilon_0 A}{d}$:

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

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

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

### The Equivalent Capacitance

For capacitors in parallel, the equivalent capacitance is simply the algebraic sum of the individual capacitances:

$$C_{eq} = C_1 + C_2 + C_3$$

Substituting our expressions, we get:

$$C_{eq} = \frac{K_1 \varepsilon_0 A}{3d} + \frac{K_2 \varepsilon_0 A}{3d} + \frac{K_3 \varepsilon_0 A}{3d}$$

Factoring out the common terms, we arrive at a beautiful, symmetric equation:

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

### The Final Calculation

The ultimate goal is to find a single material with an equivalent dielectric constant $K$ that would provide the exact same capacitance if it filled the entire space between the plates. The capacitance of this hypothetical single capacitor would be:

$$C_{eq} = \frac{K \varepsilon_0 A}{d}$$

By equating our two expressions for $C_{eq}$, we set up the final stage of our calculation:

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

Notice how the geometric parameters $\frac{\varepsilon_0 A}{d}$ elegantly cancel out from both sides. This tells us a profound truth: the equivalent dielectric constant in this specific parallel arrangement is independent of the actual area or separation distance! It is simply the arithmetic mean of the individual dielectric constants:

$$K = \frac{K_1 + K_2 + K_3}{3}$$

Plugging in the given values:

$$K = \frac{10 + 12 + 14}{3} = \frac{36}{3} = 12$$

Our equivalent dielectric constant is **12**.

### The Way Forward: A Thought Experiment

Before we wrap up, let's ponder a classic variation of this problem. What if the text was literal, and the dielectrics were actually stacked one on top of the other, like layers in a cake?

In that scenario, each dielectric would have the full area $A$, but only one-third of the thickness ($d/3$). Because the charge would have to pass through them sequentially, they would act as capacitors in **series**. The equivalent dielectric constant would then be calculated using the harmonic mean:

$$K_{eq} = \frac{3}{\frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3}}$$

Always let the diagram be your ultimate guide in determining whether the setup is in series or parallel!

Select Answer:

Visualized Solution

Analyzing the Capacitor Geometry

Identifying the Combination

Capacitance of Individual Sections

Equivalent Capacitance

Equivalent Dielectric Constant

Calculating $K$

What if they were stacked?

The Sigma Insight: Capacitance and Capacitors

The Beauty of Dielectrics in Capacitors

Analyzing the Setup

The Master Equation

The Equivalent Capacitance

The Final Calculation

The Way Forward

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