In this example, we explore the effect of dielectrics on the capacitance of capacitors in two distinct configurations. Understanding how dielectrics influence capacitance is crucial for applications in electronics and circuit design.

For part (a), we consider a capacitor where the dielectric fills the top half. This configuration can be treated as two capacitors in series: one with a dielectric and one without. The capacitance of the first capacitor, denoted as \( C_1 \), is given by the formula:

\( C_1 = \frac{2 \kappa \epsilon_0 A}{d} \)

Here, \( \kappa \) represents the dielectric constant, \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates. The second capacitor, \( C_2 \), which does not contain a dielectric, is expressed as:

\( C_2 = \frac{2 \epsilon_0 A}{d} \)

To find the equivalent capacitance \( C \) of the entire system, we use the formula for capacitors in series:

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

Substituting the expressions for \( C_1 \) and \( C_2 \), we have:

\( \frac{1}{C} = \frac{d}{2 \kappa \epsilon_0 A} + \frac{d}{2 \epsilon_0 A} \)

Finding a common denominator leads to:

\( C = \frac{2 \kappa}{1 + \kappa} \cdot \frac{\epsilon_0 A}{d} \)

This result indicates that the effective capacitance resembles that of a capacitor with an effective dielectric constant, reflecting the influence of the dielectric material.

In part (b), we analyze a different configuration where the dielectric occupies only the left half of the capacitor. In this case, the capacitors are in parallel, sharing the same voltage across their plates. The capacitance for the first capacitor with the dielectric, \( C_1 \), is calculated as:

\( C_1 = \frac{\kappa \epsilon_0 A}{2d} \)

For the second capacitor without a dielectric, \( C_2 \), the capacitance is:

\( C_2 = \frac{\epsilon_0 A}{2d} \)

Since these capacitors are in parallel, the total capacitance \( C \) is simply the sum of \( C_1 \) and \( C_2 \):

\( C = C_1 + C_2 = \frac{\kappa + 1}{2} \cdot \frac{\epsilon_0 A}{d} \)

This expression also suggests an effective dielectric constant, demonstrating how the arrangement of dielectrics can modify the overall capacitance of a capacitor system.