/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A parallel plate capacitor has a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 68

A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?

Short Answer

Expert verified
Solution: The capacitance of the parallel plate capacitor is approximately \(8.33 \times 10^{-11} \,\mathrm{F}\).

Step by step solution

01

Convert given quantities to standard units

Firstly, we convert given quantities to standard units. The charge given in microcoulombs should be converted to coulombs, and the plate separation given in millimeters should be converted to meters. Charge, \(Q = 0.020 \times 10^{-6} \,\mathrm{C}\). Meanwhile, Separation, \(d = 0.40 \times 10^{-3} \,\mathrm{m}\)
02

Apply the capacitance formula

Now apply the capacitance formula \(C = \frac{Q}{V}\), where \(Q\) is the charge on the plates and \(V\) is the potential difference. Substitute the given quantities and solve for capacitance, \(C\). \(C = \frac{0.020 \times 10^{-6}}{240}\)
03

Calculate the capacitance

Calculate the capacitance by dividing the charge by the potential difference. \(C = \frac{0.020 \times 10^{-6}}{240}\) \(C = 8.333 \times 10^{-11} \,\mathrm{F}\)
04

Express the result

Finally, express the capacitance to an appropriate number of significant figures and report the result in standard units: Capacitance, \(C = 8.33 \times 10^{-11} \,\mathrm{F}\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

An electron is moved from point \(A\), where the electric potential is \(V_{A}=-240 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=-360 \mathrm{V}\). What is the change in the electric potential energy?
Draw some electric field lines and a few equipotential surfaces outside a negatively charged hollow conducting sphere. What shape are the equipotential surfaces?
The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Linear Momentum

Read Explanation

Nuclear Physics

Read Explanation

Dynamics

Read Explanation

Fundamentals of 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.