Understanding the Problem: Parallel Plate Capacitor with Dielectric

The question asks for the ratio of the capacitance of a parallel plate capacitor when a dielectric slab is inserted between its plates compared to its capacitance without the dielectric. The dielectric slab has a specific thickness and covers the same area as the plates.

Let's break down the problem:

• We have a parallel plate capacitor with plate area \( A \) and plate separation \( d \).
• A dielectric slab with dielectric constant \( k \) and thickness \( t = \frac{3d}{4} \) is placed between the plates. The slab has the same area \( A \).
• The remaining space between the plates is \( d - t = d - \frac{3d}{4} = \frac{d}{4} \), which is filled with air (or vacuum), having a dielectric constant of approximately 1.
• This setup with the dielectric slab can be thought of as two capacitors connected in series: one capacitor filled with the dielectric and the other filled with air.

Calculating Capacitance without Dielectric (C₀)

The capacitance of a parallel plate capacitor with vacuum or air between the plates is given by the formula:

\( C_0 = \frac{\epsilon_0 A}{d} \)

where:

• \( \epsilon_0 \) is the permittivity of free space.
• \( A \) is the area of the plates.
• \( d \) is the distance between the plates.

Calculating Capacitance with Dielectric Slab (C)

When the dielectric slab of thickness \( t = \frac{3d}{4} \) is inserted, it divides the region between the plates into two parts:

1. A region of thickness \( t_1 = \frac{3d}{4} \) filled with the dielectric (dielectric constant \( k \)).
2. A region of thickness \( t_2 = d - t_1 = d - \frac{3d}{4} = \frac{d}{4} \) filled with air (dielectric constant \( k_{air} \approx 1 \)).

These two regions can be considered as two capacitors connected in series. Let the capacitance of the dielectric-filled part be \( C_1 \) and the air-filled part be \( C_2 \).

The formula for capacitance with a dielectric is \( C = \frac{k \epsilon_0 A}{t} \), where \( t \) is the thickness filled by the dielectric.

For the dielectric-filled part:

\( C_1 = \frac{k \epsilon_0 A}{t_1} = \frac{k \epsilon_0 A}{3d/4} = \frac{4k \epsilon_0 A}{3d} \)

For the air-filled part:

\( C_2 = \frac{k_{air} \epsilon_0 A}{t_2} \)

Since \( k_{air} \approx 1 \):

\( C_2 = \frac{1 \cdot \epsilon_0 A}{d/4} = \frac{4 \epsilon_0 A}{d} \)

When capacitors are in series, the reciprocal of the total capacitance is the sum of the reciprocals of individual capacitances:

\( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \)

Substitute the expressions for \( C_1 \) and \( C_2 \):

\( \frac{1}{C} = \frac{1}{\frac{4k \epsilon_0 A}{3d}} + \frac{1}{\frac{4 \epsilon_0 A}{d}} \)

\( \frac{1}{C} = \frac{3d}{4k \epsilon_0 A} + \frac{d}{4 \epsilon_0 A} \)

To add these fractions, find a common denominator, which is \( 4k \epsilon_0 A \):

\( \frac{1}{C} = \frac{3d}{4k \epsilon_0 A} + \frac{d \cdot k}{4 \epsilon_0 A \cdot k} \)

\( \frac{1}{C} = \frac{3d + kd}{4k \epsilon_0 A} \)

\( \frac{1}{C} = \frac{d(3 + k)}{4k \epsilon_0 A} \)

Now, invert to find \( C \):

\( C = \frac{4k \epsilon_0 A}{d(k + 3)} \)

Finding the Ratio of Capacitances (C/C₀)

We need to find the ratio of the capacitance with the dielectric \( C \) to the capacitance without the dielectric \( C_0 \):

\( \text{Ratio} = \frac{C}{C_0} \)

Substitute the expressions for \( C \) and \( C_0 \):

\( \text{Ratio} = \frac{\frac{4k \epsilon_0 A}{d(k + 3)}}{\frac{\epsilon_0 A}{d}} \)

To simplify, multiply the numerator by the reciprocal of the denominator:

\( \text{Ratio} = \frac{4k \epsilon_0 A}{d(k + 3)} \times \frac{d}{\epsilon_0 A} \)

Cancel out the common terms \( \epsilon_0 A \) and \( d \):

\( \text{Ratio} = \frac{4k}{k + 3} \)

The ratio of the capacitance with the dielectric inside to its capacitance without the dielectric is \( \frac{4k}{k + 3} \).

Summary of Calculation Steps

1. Calculated initial capacitance \( C_0 \) without the dielectric.
2. Identified the system with the dielectric as two capacitors in series: dielectric-filled and air-filled.
3. Calculated the capacitance of the dielectric-filled part (\( C_1 \)).
4. Calculated the capacitance of the air-filled part (\( C_2 \)).
5. Used the series capacitance formula (\( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \)) to find the total capacitance \( C \) with the dielectric.
6. Calculated the ratio \( \frac{C}{C_0} \).

Revision Table: Parallel Plate Capacitor Concepts

Additional Information: Electric Field and Dielectrics

When a dielectric material is placed in an electric field, its molecules polarize. This polarization creates an internal electric field within the dielectric that opposes the external field. The net electric field inside the dielectric is thus reduced compared to the field in vacuum.

The dielectric constant \( k \) quantifies this reduction: \( k = \frac{E_0}{E} \), where \( E_0 \) is the electric field in vacuum and \( E \) is the net electric field in the dielectric. Since \( E < E_0 \), the dielectric constant \( k \) is always greater than 1 for any material dielectric.

For a parallel plate capacitor, the voltage difference \( V \) is related to the electric field \( E \) and plate separation \( d \) by \( V = Ed \). Since the dielectric reduces the electric field \( E \) for the same charge \( Q \) on the plates, the voltage \( V \) is also reduced. Because capacitance is defined as \( C = Q/V \), a smaller voltage \( V \) for the same charge \( Q \) means a higher capacitance \( C \). This is why inserting a dielectric increases the capacitance of a capacitor.

In the case of a partially filled capacitor like this problem, the electric field is different in the dielectric region and the air region. The voltage across the capacitor is the sum of the voltage drops across the dielectric part and the air part. This naturally leads to treating the setup as two capacitors in series, as the voltage divides across the different regions, and the charge on each 'section' is the same as the charge on the capacitor plates.