A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation bet

Q: Why are the capacitors considered to be in series?

Because the dielectrics are stacked one after the other along the distance between the plates. The electric field lines pass through both dielectrics sequentially, which is the hallmark of a series connection.

Q: What if the dielectrics were placed side-by-side, dividing the area of the plates?

In that case, the electric field lines would pass through either one dielectric or the other, not both. This would make it a parallel combination, and the equivalent capacitance would be $C_1 + C_2$.

Q: Why did we write $C_1$ and $C_2$ in terms of $C_0$?

It simplifies the calculation immensely! Instead of dealing with $\epsilon_0$, $A$, and $d$ individually, we can just use the given value of $C_0 = 9\text{ pF}$ at the very end.

Q: Does the order of the dielectrics matter?

No! In a series combination, the equivalent capacitance is independent of the order of the capacitors. $\frac{C_1 C_2}{C_1 + C_2}$ is the same as $\frac{C_2 C_1}{C_2 + C_1}$.

Q: What happens to the voltage across each dielectric?

In a series combination, the total voltage is divided between the capacitors in inverse proportion to their capacitances. The capacitor with the smaller capacitance will have a larger voltage drop across it.

Question

A parallel plate capacitor with air between the plates has a capacitance of $9	ext{ pF}$. The separation between its plates is $d$. The space between the plate is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1 = 3$ and thickness $d/3$ while the other one has dielectric constant $K_2 = 6$ and thickness $2d/3$. Capacitance of the capacitor is now

Options:

(A) $1.8	ext{ pF}$

(B) $45	ext{ pF}$

(C) $40.5	ext{ pF}$

(D) $20.25	ext{ pF}$

Sigma Percentile · Accepted Answer

### Visualizing the Capacitor

Imagine you are looking at a simple parallel plate capacitor. Between its plates, there is nothing but air. The plates are separated by a distance $d$, and this basic setup gives us an initial capacitance of $C_0 = 9	ext{ pF}$.

Now, let's spice things up. We fill the space between the plates with two different dielectric materials. The first material has a dielectric constant $K_1 = 3$ and takes up one-third of the distance, so its thickness is $d/3$. The second material has a dielectric constant $K_2 = 6$ and fills the remaining two-thirds of the space, giving it a thickness of $2d/3$.

### The Series Connection Insight

Here is the crucial conceptual leap: how do these two dielectrics interact? Because they are stacked one after the other along the distance between the plates, the electric field lines must pass through both of them sequentially.

This physical arrangement is exactly equivalent to having two separate capacitors connected in **series**. Therefore, we can find the total equivalent capacitance using the standard series formula:
$$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$$

### Calculating Individual Capacitances

Let's break the problem down and calculate the capacitance of each section individually. The general formula for a capacitor with a dielectric is $C = \frac{K \epsilon_0 A}{t}$, where $t$ is the thickness.

For the first section:
$$C_1 = \frac{K_1 \epsilon_0 A}{d/3} = \frac{3 K_1 \epsilon_0 A}{d}$$

For the second section:
$$C_2 = \frac{K_2 \epsilon_0 A}{2d/3} = \frac{3 K_2 \epsilon_0 A}{2d}$$

Now, instead of plugging in messy numbers, let's use a brilliant shortcut. We know that the original air capacitance is $C_0 = \frac{\epsilon_0 A}{d}$. Let's express $C_1$ and $C_2$ in terms of $C_0$.

Substituting $K_1 = 3$:
$$C_1 = 3(3) \left(\frac{\epsilon_0 A}{d}ight) = 9 C_0$$

Substituting $K_2 = 6$:
$$C_2 = \frac{3(6)}{2} \left(\frac{\epsilon_0 A}{d}ight) = 9 C_0$$

Wow! Both sections happen to have the exact same capacitance of $9 C_0$.

### The Final Equivalent Capacitance

Now that we have $C_1$ and $C_2$, finding the equivalent capacitance is a breeze. Since they are equal and in series, the equivalent capacitance is simply half of their individual value. Let's plug them into our series formula to be sure:
$$C_{eq} = \frac{(9 C_0)(9 C_0)}{9 C_0 + 9 C_0} = \frac{81 C_0^2}{18 C_0} = 4.5 C_0$$

Finally, we bring back our original value of $C_0 = 9	ext{ pF}$. 
$$C_{eq} = 4.5 	imes 9	ext{ pF} = 40.5	ext{ pF}$$

And there we have it! The new capacitance of the system is **$40.5	ext{ pF}$**. Always remember, when dielectrics divide the distance, they are in series. If they divide the area, they are in parallel!

Select Answer:

Visualized Solution

Initial Air Capacitor

Inserting Dielectrics

Equivalent Circuit

Individual Capacitances

Substituting Values

Calculating Equivalent Capacitance

Final Answer

The Way Forward

The Sigma Insight: Combination of Capacitors

Visualizing the Capacitor

The Series Connection Insight

Calculating Individual Capacitances

The Final Equivalent Capacitance

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