For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectri

Q: Why are the two parts considered to be in series and not parallel?

Because the dielectric and air gap are stacked along the distance between the plates. The electric field lines pass through them sequentially, and the potential drops across them one after the other. If the dielectric covered half the area instead, they would be in parallel.

Q: Why does the thickness go in the denominator of the capacitance formula?

Capacitance is inversely proportional to the distance between the plates ($C = \frac{\varepsilon_0 A}{d}$). A smaller distance means the plates are closer, which increases the electric field strength for a given voltage, allowing it to store more charge.

Q: Can we use the direct formula for partially filled capacitors?

Absolutely! The formula $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$ is derived using the exact same series combination logic. Plugging in $t = \frac{3d}{4}$ will give you the same answer.

Q: What happens if we set $K=1$ in the final formula?

If $K=1$, it means the slab is just air. Plugging $K=1$ into $\frac{4K}{K+3}C_0$ gives $\frac{4(1)}{1+3}C_0 = C_0$. This perfectly matches our physical intuition!

Q: Does the position of the dielectric slab matter?

No, it doesn't. Whether the slab is touching the left plate, the right plate, or floating in the middle, the total series capacitance remains exactly the same because addition is commutative.

Question

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4}d$, where $d$ is the separation between the plates of parallel plate capacitor. The new capacitance ($C'$) in terms of original capacitance ($C_0$) is given by the following relation

Options:

(A) $C' = \frac{3+K}{4K}C_0$

(B) $C' = \frac{4+K}{3}C_0$

(C) $C' = \frac{4K}{K+3}C_0$

(D) $C' = \frac{4}{3+K}C_0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a standard parallel plate capacitor with an initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$. Now, we introduce a dielectric slab of constant $K$, but it doesn't fill the entire space between the plates. It only occupies a thickness of $\frac{3d}{4}$. The remaining $\frac{d}{4}$ is just empty space (air).

Because the dielectric slab and the air gap are stacked one after the other along the distance $d$, the electric field lines pass through them sequentially. This physical arrangement perfectly mirrors an electrical circuit where two components are connected in series. Therefore, we can model this partially filled capacitor as two distinct capacitors, $C_1$ and $C_2$, connected in series.

### The Master Equation
Let's determine the individual capacitances. For the dielectric portion ($C_1$), the area remains $A$, but the thickness is $\frac{3d}{4}$.

$$C_1 = \frac{K \varepsilon_0 A}{\frac{3d}{4}} = \frac{4K \varepsilon_0 A}{3d}$$

For the air gap ($C_2$), the area is also $A$, but the thickness is $\frac{d}{4}$, and the dielectric constant is simply $1$.

$$C_2 = \frac{\varepsilon_0 A}{\frac{d}{4}} = \frac{4 \varepsilon_0 A}{d}$$

Now, we apply the formula for capacitors in series. The reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

### Final Calculation
Let's substitute our expressions into the series formula:

$$\frac{1}{C_{eq}} = \frac{3d}{4K \varepsilon_0 A} + \frac{d}{4 \varepsilon_0 A}$$

To simplify, we can factor out the common term $\frac{d}{4 \varepsilon_0 A}$:

$$\frac{1}{C_{eq}} = \frac{d}{4 \varepsilon_0 A} \left( \frac{3}{K} + 1 \right)$$

Finding a common denominator inside the parenthesis gives:

$$\frac{1}{C_{eq}} = \frac{d}{4 \varepsilon_0 A} \left( \frac{3+K}{K} \right)$$

Finally, we invert the entire equation to solve for $C_{eq}$:

$$C_{eq} = \frac{4K \varepsilon_0 A}{(3+K)d}$$

We know that the original capacitance is $C_0 = \frac{\varepsilon_0 A}{d}$. By substituting $C_0$ into our result, we arrive at the elegant final relation:

**$$C' = \frac{4K}{K+3}C_0$$**

*Pro Tip:* You can also solve this instantly using the general formula for a partially filled capacitor: $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$. Substituting $t = \frac{3d}{4}$ will yield the exact same result!

Select Answer:

Visualized Solution

Visualizing the Setup

Series Combination Logic

Individual Capacitances

Applying Series Formula

Algebraic Simplification

Final Capacitance

Alternative Perspective

The Sigma Insight: Combination of Capacitors

Analyzing the Setup Imagine a standard parallel plate capacitor with an initial capacitance $C_{0} = \frac{ε _{0} A}{d}$

The Master Equation Let's determine the individual capacitances

Final Calculation

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