Question

Two parallel plate air capacitors have their plate areas 100 and \(500 \mathrm{~cm}^{2}\) respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is \(0.5 \mathrm{~mm}\), what is the distance between the plates of the second capacitor ? (A) \(0.25 \mathrm{~cm}\) (B) \(0.50 \mathrm{~cm}\) (C) \(0.75 \mathrm{~cm}\) (D) \(1 \mathrm{~cm}\)

Short answer

The distance between the plates of the second capacitor is \(0.25\,\mathrm{cm}\).

Step-by-step solution

Write the capacitance formula for both capacitors

We will use the formula for the capacitance of a parallel plate capacitor, which is: \(C = \dfrac{ε_0A}{d}\) For the first capacitor, let's denote the capacitance as \(C_1\), the plate area as \(A_1 = 100\,\mathrm{cm}^2\), and the distance between the plates as \(d_1 = 0.5\,\mathrm{mm}\). For the second capacitor, let's denote the capacitance as \(C_2\), the plate area as \(A_2 = 500\,\mathrm{cm}^2\), and the distance between the plates as \(d_2\), which we need to find.

Set up the equation to find the distance between the plates of the second capacitor

Since both capacitors have the same charge and potential, their capacitances are equal, i.e., \(C_1 = C_2\). Now, using the capacitance formula, we can write the following equation: \(\dfrac{ε_0A_1}{d_1} = \dfrac{ε_0A_2}{d_2}\) We want to find \(d_2\), so let's rearrange the equation to solve for it: \(d_2 = \dfrac{A_2}{A_1} d_1\)

Substitute the given values and solve for the distance

Now, we can substitute the values of \(A_1\), \(A_2\), and \(d_1\) into the equation: \(d_2 = \dfrac{500\,\mathrm{cm}^2}{100\,\mathrm{cm}^2} (0.5\,\mathrm{mm})\) To make the units consistent, let's convert the distance from millimeters to centimeters: \(0.5\,\mathrm{mm} = 0.5\,\mathrm{cm/10} = 0.05\,\mathrm{cm}\) Now, we can plug this value into the equation: \(d_2 = \dfrac{500\,\mathrm{cm}^2}{100\,\mathrm{cm}^2} (0.05\,\mathrm{cm})\)

Calculate the distance between the plates of the second capacitor

Now, let's solve for \(d_2\): \(d_2 = 5 (0.05\,\mathrm{cm}) = 0.25\,\mathrm{cm}\) So, the distance between the plates of the second capacitor is \(0.25\,\mathrm{cm}\). Therefore, the correct option is (A) \(0.25\,\mathrm{cm}\).

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is a device that stores electrical energy within an electric field. It consists of two conducting plates separated by a small distance. When a voltage is applied, opposite charges build up on each plate, creating an electric field. The energy is stored in this field. This simple configuration allows for the study and use of capacitors in various electronic circuits.

Important aspects of a parallel plate capacitor include its plate area and the separation distance between the plates. These factors significantly influence the capacitance of the capacitor, or its ability to store charge. Understanding this relationship is crucial when designing capacitors for specific functions.

Capacitance Formula

The capacitance of a parallel plate capacitor can be calculated using the formula: \[ C = \frac{ε_0 A}{d} \]where:

• \(C\) is the capacitance
• \(ε_0\) is the permittivity of free space
• \(A\) is the area of one of the plates
• \(d\) is the distance between the plates

This formula helps us understand how adjustments in the plate area or separation affect the capacitance.

Increasing the plate area, \(A\), leads to greater capacitance because a larger surface can store more charge. Conversely, increasing the distance between the plates, \(d\), decreases capacitance as the charges are further apart, and the electric field becomes weaker. This is why precise control of these variables is critical in designing capacitors for different electronic applications.

Charge and Potential Relation

The relationship between the charge, voltage (potential difference), and capacitance is given by the equation:\[ Q = C \times V \]where:

• \(Q\) is the charge stored
• \(C\) is the capacitance
• \(V\) is the potential difference across the plates

This equation shows that the stored charge is directly proportional to both the capacitance and the potential difference. Thus, for a given capacitor, increasing the voltage increases the amount of charge it can store. Conversely, for a constant charge, a larger capacitance allows the capacitor to operate at a lower voltage.

This principle is fundamental when designing circuits that require specific voltage levels and charge storage capacities, enabling the efficient functioning of electronic devices.

Plate Area and Distance Relationship

In a parallel plate capacitor, there is a distinct relationship between the plate area, the distance between the plates, and how these affect the capacitor's performance. As described by the capacitance formula, the capacitance is directly proportional to the plate area \(A\) and inversely proportional to the distance \(d\) between the plates.This means:

• An increase in plate area \(A\) enhances the capacitance, allowing for more charge storage.
• An increase in the distance \(d\) reduces the capacitance, as the electric field weakens over greater separation.

These relationships help explain why capacitors with differing geometries have different capabilities. Larger plate areas are preferable for high-capacitance applications. However, the practical limitations often involve maintaining the right distance, balancing the physical size of the capacitor, and its intended environment in an electronic circuit. Understanding these relationships is essential when engineering capacitors for specific tasks, ensuring energy storage devices meet desired operational requirements efficiently.