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Chapter 31: Problem 31

A charge of \(20 \mu \mathrm{C}\) is placed on the positive plate of an isolated parallel-plate capacitor of capacitance \(10 \mu \mathrm{F}\). Calculate the potential difference developed between the plates.

Short Answer

Expert verified
The potential difference is 2 V.

Step by step solution

01

Identify the Given Values

First, let's identify the given values in the problem. We have a charge \( Q = 20 \mu \text{C} \) and a capacitance \( C = 10 \mu \text{F} \). Keep in mind that \( 1 \mu \text{C} = 1 \times 10^{-6} \text{C} \) and \( 1 \mu \text{F} = 1 \times 10^{-6} \text{F} \). So, the given values are \( Q = 20 \times 10^{-6} \text{C} \) and \( C = 10 \times 10^{-6} \text{F} \).
02

Recall the Formula for Potential Difference

The potential difference \( V \) across the plates of a capacitor is given by the formula \( V = \frac{Q}{C} \), where \( Q \) is the charge and \( C \) is the capacitance.
03

Substitute the Known Values into the Formula

Substitute the values of \( Q \) and \( C \) into the formula: \[ V = \frac{20 \times 10^{-6} \text{C}}{10 \times 10^{-6} \text{F}} \].
04

Calculate the Potential Difference

Perform the division to find the potential difference:\[ V = \frac{20}{10} = 2 \text{V} \]. The potential difference developed between the plates is \( 2 \text{V} \).
05

Review the Solution

Check the calculations to ensure they are correct and include units in all steps. The calculations confirm that the potential difference is 2 Volts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Difference
In the world of capacitors, the potential difference, often symbolized as \( V \), plays a crucial role. It is the voltage developed between two points, in this case, the plates of a capacitor. Imagine potential difference as electronic pressure across a capacitor, pushing charges through it. The formula \( V = \frac{Q}{C} \) helps us calculate this voltage. Here, \( Q \) is the charge stored on the plates, and \( C \) is the capacitance, or the capacitor's ability to store charge. To further understand it:
  • Potential difference is proportional to the amount of charge \( Q \).
  • As capacitance \( C \) increases, for the same charge, \( V \) decreases.
These insights help us in analyzing and solving problems, like those involving parallel-plate capacitors.
Charge and Capacitance Relationship
A fundamental aspect of capacitors is the relationship between charge (\( Q \)) and capacitance (\( C \)). This relationship is defined by the equation \( Q = C \times V \), where \( V \) is the potential difference. Breaking it down:
  • Charge \( Q \): It is the amount of electrical charge stored on the capacitor's plates.
  • Capacitance \( C \): This is the capacitor's capacity to store charge per unit potential difference. It determines how much charge will be stored for each volt of potential difference.
  • The relationship is proportional, meaning if you increase \( V \), \( Q \) increases, keeping \( C \) constant.
This simple relationship is key to designing circuits and understanding capacitor behavior under different conditions.
Parallel-Plate Capacitor
A parallel-plate capacitor is one of the most basic types of capacitors and consists of two plates separated by a dielectric material. It is crucial for various electronic applications because of its simplicity and efficiency in storing charge. Let鈥檚 examine its features:
  • Structure: It comprises two flat, parallel conductive plates. The space between the plates is filled with an insulating material known as the dielectric, which increases the capacitor's ability to store charge.
  • Capacitance: The capacitance \( C \) is affected by the area of the plates (\( A \)) and the distance (\( d \)) between them, typically expressed with the formula \( C = \varepsilon \frac{A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric material.
  • Efficiency: By altering the plate size, distance, and dielectric material, we can manipulate how much charge the capacitor can store.
Parallel-plate capacitors are simple yet highly practical components for everyday electronics.

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Most popular questions from this chapter

A cylindrical capacitor is constructed using two coaxial cylinders of the same length \(10 \mathrm{~cm}\) and of radii \(2 \mathrm{~mm}\) and \(4 \mathrm{~mm}\). (a) Calculate the capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii \(4 \mathrm{~mm}\) and \(8 \mathrm{~mm}\). Calculate the capacitance.
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