Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 61

Considering the dielectric strength of air, what is the maximum amount of charge that can be stored on the plates of a capacitor that are a distance of \(15 \mathrm{~mm}\) apart and have an area of \(25 \mathrm{~cm}^{2}\) ?

Short Answer

Expert verified
Answer: The maximum charge that can be stored on the plates of the capacitor is \(6.6375 \times 10^{-7} \mathrm{~C}\).

Step by step solution

01

Convert the given dimensions to meters

First, we need to convert the given dimensions (distance and area) to meters for consistency in calculations: - The distance between the plates is given as \(15 \mathrm{~mm}\). Convert this to meters: \(15 \mathrm{~mm} = 0.015 \mathrm{~m}\). - The area of the plates is given as \(25 \mathrm{~cm}^{2}\). Convert this to square meters: \(25 \mathrm{~cm}^{2} = 2.5 \times 10^{-3} \mathrm{~m}^{2}\).
02

Calculate the maximum electric field

The dielectric strength of air is given as \(3 \times 10^{6} \mathrm{~V/m}\). This is the maximum electric field that air can withstand without breaking down. We can calculate the maximum electric field in air using: \(E = \frac{V}{d}\) where \(E\) is the electric field, \(V\) is the voltage, and \(d\) is the distance between the plates. We can use the dielectric strength of air to find the maximum voltage, \(V_{max}\), that the capacitor can handle: \(V_{max} = E_{max} \times d\) Substituting the values into the formula, we get: \(V_{max} = (3 \times 10^{6} \mathrm{~V/m}) \times (0.015 \mathrm{~m}) = 45,000 \mathrm{~V}\)
03

Calculate the capacitance

Using the capacitance formula: \(C = \epsilon_{0} \frac{A}{d}\) where \(\epsilon_{0}\) is the vacuum permittivity constant, \(A\) is the area of the plates, and \(d\) is the distance between the plates. The value of \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m}\). Substitute the values to find the capacitance: \(C = (8.85 \times 10^{-12} \mathrm{~F/m}) \times \frac{2.5 \times 10^{-3} \mathrm{~m}^{2}}{0.015 \mathrm{~m}} = 1.475 \times 10^{-11} \mathrm{~F}\)
04

Calculate the maximum charge stored

Now that we have the capacitance and the maximum voltage that the capacitor can handle, we can find the maximum charge using: \(Q = C \times V\) where \(Q\) is the charge stored, \(C\) is the capacitance, and \(V\) is the voltage across the plates. Substitute the values: \(Q = (1.475 \times 10^{-11} \mathrm{~F}) \times (45,000 \mathrm{~V}) = 6.6375 \times 10^{-7} \mathrm{~C}\) Therefore, the maximum amount of charge that can be stored on the plates of the capacitor is \(6.6375 \times 10^{-7} \mathrm{~C}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

An isolated solid spherical conductor of radius \(5.00 \mathrm{~cm}\) is surrounded by dry air. It is given a charge and acquires potential \(V\), with the potential at infinity assumed to be zero. a) Calculate the maximum magnitude \(V\) can have. b) Explain clearly and concisely why there is a maximum.

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?

The distance between the plates of a parallel plate capacitor is reduced by half and the area of the plates is doubled. What happens to the capacitance? a) It remains unchanged. b) It doubles. c) It quadruples. d) It is reduced by half.

Fifty parallel plate capacitors are connected in series. The distance between the plates is \(d\) for the first capacitor, \(2 d\) for the second capacitor, \(3 d\) for the third capacitor, and so on. The area of the plates is the same for all the capacitors. Express the equivalent capacitance of the whole set in terms of \(C_{1}\) (the capacitance of the first capacitor).
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Electrostatics

Read Explanation

Work Energy and Power

Read Explanation

Radiation

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free