Question

A parallel-plate air capacitor has an initial capacitance value of \( 8\,\mu\text{F} \). If the space between the plates is completely filled with a solid dielectric slab having a dielectric constant \( K = 5 \), what will the new capacitance value of the component be?

• A. \( 1.6\,\mu\text{F} \)
• B. \( 40\,\mu\text{F} \)
• C. \( 20\,\mu\text{F} \)
• D. \( 80\,\mu\text{F} \)

Hint:

Inserting a dielectric material always increases the capacitance value (\( C = KC_0 \)). If the capacitor is disconnected from its battery before inserting the slab, the stored charge stays constant while the voltage drops. If it stays connected to the battery, the voltage remains constant while the stored charge increases.

Answer 1

The Correct Option is B

Concept: The capacitance of a parallel-plate structure measures its ability to store electric charge. For plates of area \( A \) separated by an air gap distance \( d \), the initial capacitance value \( C_0 \) is given by: \[ C_0 = \frac{\varepsilon_0 A}{d} \] When a dielectric material fills the gap, the internal electric field polarizes the dielectric atoms, reducing the net potential difference between the plates for a given charge and increasing the capacitance.

Step 1: Set up the capacitance formula for a dielectric filled medium. Introducing a dielectric material replaces the permittivity of free space \( \varepsilon_0 \) with the material's absolute permittivity \( \varepsilon = K\varepsilon_0 \): \[ C_{\text{new}} = \frac{\varepsilon A}{d} = \frac{K\varepsilon_0 A}{d} \] This shows that filling the gap scales the capacitance value directly by the factor of the dielectric constant: \[ C_{\text{new}} = K \cdot C_0 \]

Step 2: Substitute the given numbers to find the final value. Using our given values \( C_0 = 8\,\mu\text{F} \) and \( K = 5 \): \[ C_{\text{new}} = 5 \cdot 8\,\mu\text{F} = 40\,\mu\text{F} \] The component's capacitance increases to exactly \( 40\,\mu\text{F} \).