(a) What is the energy stored in the 10.0μFcapacitor of a heartdefibrillator charged to 9.00×103V? (b) Find the amount of storedcharge.

Short Answer

Expert verified
  1. The stored energy is 405 J.
  2. The charge stored is 0.09 C

Step by step solution

01

Given Data

The capacitance of the capacitor is:

C=(10.0μF)(1F106μF)=10.0×10-6F

The voltage between the capacitor plates is:ΔV=9.00×103V

02

Calculating for part (a) energy of capacitor

The stored energy for a capacitor of capacitancewith a potential difference between the platesis given by:

UE=12C(ΔV)2

a)

The capacitance C is defined as charge stored per unit potential difference

C=Q(ΔV)

Thus, entering the values of C and ΔV, we obtain:

UE=12(10.0×10-6F)(9.00×103V)=405J

Therefore, stored energy isrole="math" localid="1654921309243" UE=405 J

03

Calculating for part (b), stored charge

b)

The stored charge is given as

Q=C×ΔV

Substitute numerical values:

Q=(10.0×10-6F)(9.00×103V)=0.09C

Therefore, the charge accumulated on capacitor plates islocalid="1654921503322" Q=0.09C

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 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

See all solutions

Recommended explanations on Physics Textbooks

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