Question

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

Short answer

The ratio of the capacitance in the two cases is (C) \(2:1\).

Step-by-step solution

Understand the capacitance formula for parallel plate capacitor

Capacitance (C) between two plates is given by the formula: \(C = \frac{\epsilon A}{d}\) Where ε is the permittivity of the dielectric material between the plates, A is the area of the plates, and d is the distance between them.

Capacitance in the first case (C1)

The first case, the distance between the plates is d. Using the capacitance formula for a parallel plate capacitor: \(C_1 = \frac{\epsilon A}{d}\)

Capacitance in the second case (C2)

In the second case, a metal sheet is introduced between the plates and it has a thickness of d/2. The metal sheet divides the capacitor into two capacitors connected in series, each with a thickness of d/2. The new capacitance (C2) is equal to the capacitance of one of these new capacitors multiplied by 2 in series: \(C'_2 = \frac{\epsilon A}{d/2}\) As the two capacitors are in series, their total capacitance (C2) can be found using the formula for capacitors in series: \(\frac{1}{C_2} = \frac{1}{C'_2} + \frac{1}{C'_2}\) Which can be simplified to: \(\frac{1}{C_2} = 2 \cdot \frac{1}{C'_2}\)

Determine the ratio of capacitance

We now want the ratio \(\frac{C_1}{C_2}\). Let's plug in the values we found for C1 and C2: \(\frac{C_1}{C_2} = \frac{\frac{\epsilon A}{d}}{\frac{1}{2 \cdot \frac{1}{\frac{\epsilon A}{d/2}}}}\) Simplifying: \(\frac{C_1}{C_2} = \frac{2 \epsilon A d}{d \epsilon A}\) Which yields: \(\frac{C_1}{C_2} = \frac{2}{1}\) Hence, the capacitance ratio in the two cases is 2:1. The correct answer is (C) \(2:1\).

Key concepts

Capacitance Formula

When learning about parallel plate capacitors, it's crucial to start with the capacitance formula. Capacitance, denoted by \(C\), represents the ability of a capacitor to store charge. For a simple parallel plate capacitor, the capacitance is determined by the equation:\[C = \frac{\epsilon A}{d}\]Let's break down this formula:

• \(\epsilon\) is the permittivity of the material between the plates. It indicates how easily an electric field can form in the material.
• \(A\) is the area of one of the plates. The larger the area, the more charge the capacitor can hold.
• \(d\) is the distance between the plates. A smaller distance increases capacitance, as the electric field is stronger over a smaller gap.

Understanding this relationship helps explain how modifying any of these factors affects the capacitance of a capacitor.

Capacitance Ratio

In scenarios involving changes to a capacitor's structure, it's often useful to determine the capacitance ratio. This helps us understand the effect brought about by these changes. Consider a situation where a metal sheet is placed between the plates of a parallel plate capacitor.First, calculate the initial capacitance \(C_1\) using the original setup:\[C_1 = \frac{\epsilon A}{d}\]Next, when a metal sheet of thickness \(d/2\) is introduced, the capacitor effectively becomes two capacitors in series, each with a distance of \(d/2\). The capacitance in this new configuration for each section, \(C'\), is:\[C' = \frac{\epsilon A}{d/2} = \frac{2\epsilon A}{d}\]The total capacitance \(C_2\) for the series setup is found using:\[\frac{1}{C_2} = \frac{1}{C'} + \frac{1}{C'} = \frac{2}{C'}\], which simplifies to \[C_2 = \frac{C'}{2}\]Finally, the ratio of the capacitance in the original setup to the new setup is:\[\frac{C_1}{C_2} = 2:1\]This ratio clarifies the decrease in capacitance due to the metal sheet's presence.

Capacitors in Series

When capacitors are connected in series, their combined effects can be quite different compared to a single unit. In such setups, the total capacitance \(C_{total}\) is less than any individual capacitance in the series.The general formula for combining capacitors in series is:\[\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\]This series configuration reduces the overall ability to store charge, as the voltage distributed over each capacitor results in a larger total voltage drop. To visualize this in the context of a parallel plate capacitor with a metal sheet inserted, think of the capacitor being split into halves. Each section acts as a separate capacitor:

• Each section has a capacitance \(C'\) calculated as \(\frac{2\epsilon A}{d}\).
• Connecting these in series according to the formula provides the reduced total capacitance.

Through this example, the principle of capacitors in series is highlighted, showing how overall capacitance is influenced by structural arrangements.