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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 parallel-plate capacitor having plate area \(25 \mathrm{~cm}^{2}\) and separation \(1.00 \mathrm{~mm}\) is connected to a battery of \(6.0 \mathrm{~V}\) Calculate the charge flown through the battery. How much work has been done by the battery during the process ?

A capacitor with stored energy \(4 \cdot 0 \mathrm{~J}\) is connected with an identical capacitor with no electric field in between. Find the total energy stored in the two capacitors.

A parallel-plate capacitor having plate area \(400 \mathrm{~cm}^{2}\) and separation between the plates \(1.0 \mathrm{~mm}\) is connected to a power supply of \(100 \mathrm{~V}\). A dielectric slab of thickness \(1 \cdot 0 \mathrm{~mm}\) and dielectric constant \(5 \cdot 0\) is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?

Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch \(S\). Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3 . Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

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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