Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 28

Find the capacitance of a capacitor that stores \(350 \mu \mathrm{J}\) when the potential difference across its plates is \(100 \mathrm{V}\).

Short Answer

Expert verified
The capacitance of the capacitor is found to be \(7 \times 10^{-12} F\), or \(7 pF\).

Step by step solution

01

Identify given quantities

The problem provides the following data: Energy (U) stored in the capacitor is \(350 \mu J\) and the potential difference (V) across the capacitor is \(100 V\).
02

Convert unit of energy

The given energy is in microjoules \(\mu J\). To proceed with the calculation in SI units, it is crucial to convert it to standard units. The energy in Joules (J) is \(350 \mu J = 350 \times 10^{-6} J\).
03

Use the energy stored in a capacitor formula

The energy (U) stored in a capacitor is given by the formula \(U = \frac{1}{2}C V^2\). We need to rearrange this equation to solve for C. The formula for capacitance then becomes \(C = \frac{2U}{V^2}\).
04

Substitute the given values

Substitute \(U = 350 \times 10^{-6} J\) and \(V = 100 V\) into the capacitance formula: \(C = \frac{2 \times (350 \times 10^{-6})}{100^2}\).
05

Compute the value

Solve the equation to find the capacitance (C). Round to a suitable number of significant figures.

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

An uncharged capacitor has parallel plates \(5.0 \mathrm{cm}\) on a side, spaced \(1.2 \mathrm{mm}\) apart. (a) How much work is required to transfer \(7.2 \mu \mathrm{C}\) from one plate to the other? (b) How much work is required to transfer an additional \(7.2 \mu \mathrm{C} ?\)

Two capacitors are connected in series and the combination is charged to \(100 \mathrm{V}\). If the voltage across each capacitor is \(50 \mathrm{V}\), how do their capacitances compare?

Two positive point charges are infinitely far apart. Is it possible, using a finite amount of work, to move them until they're a small distance \(d\) apart?

Your company is still stuck with those 2 - \(\mu\) F capacitors from Problem 50. They turn out to be so cheap that their capacitances are all too low, ranging from \(1.7 \mu \mathrm{F}\) to \(1.9 \mu \mathrm{F}\). A colleague suggests you put variable "trimmer" capacitors in parallel with the cheap capacitors and adjust the combination to precisely \(2.00 \mu \mathrm{F} .\) The available trimmers have variable capacitance from \(25 \mathrm{nF}\) to \(350 \mathrm{nF}\). Will they work?

Which can store more energy: a \(1.0-\mu \mathrm{F}\) capacitor rated at \(250 \mathrm{V}\) or a 470 -pF capacitor rated at \(3 \mathrm{kV} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Fields in Physics

Read Explanation

Electricity

Read Explanation

Modern Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electricity and Magnetism

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