/*! 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} Q58P The charge center of a thundercl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The charge center of a thundercloud, drifting 3.0km above the earth’s surface, contains 20C of negative charge. Assuming the charge center has a radius of 1.0km, and modeling the charge center and the earth’s surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

Short Answer

Expert verified

(a) C=9.27nF

(b)VCG=2.16GV

(c)E=7.2×105V/m

(d)U=21.6GJ.

Step by step solution

01

Parameters.

d=3.0km=3.0×103mQ=20Cr=1km=1×103m

02

(a) Calculating the capacitance of the system.

The capacitance is given by

C=∈0Ad (1)

All is known except A so consider the cloud to be a circular disk and the area is given by

A=πr2=3.14×106m2

Substituting in (1) yields

C=8.85x10-12×3.14×10630×103=9.27×10-9F.

03

(b) Calculating the potential difference between the charge center and ground.

The voltage difference between the cloud and the ground is given by

VCG=QC=209.27×10-9=2.16×109V.

04

(c) Calculating the average strength of the electric field.

The electric field flows through the air is given by

EA=Q∈0⇒E=QA∈0 (3)

Substituting in (3) yields

E=208.85×10-12×3.14×106=7.2×105V/m

05

(d) Calculating the electrical energy stored in the system.

The energy stored in system is given buy

U=12QV=12×20×2.16×109=2.16×1010J.

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 91Ó°ÊÓ!

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 particle with mass1.81×10-3kgand a charge of has1.22×10-8C, at a given instant, a velocityV⇶Ä=(3.00×104m/s).What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic fieldB=(1.63T)i+(0.980T)j^?

(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

Ordinary household electric lines in North America usually operate at 120 V . Why is this a desirable voltage, rather than a value considerably larger or smaller? On the other hand, automobiles usually have 12 V electrical systems. Why is this a desirable voltage?

In the circuit in Fig. E25.47, find (a) the rate of conversion of internal (chemical) energy to electrical energy within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor.

(See Discussion Question Q25.14.) An ideal ammeter A is placed in a circuit with a battery and a light bulb as shown in Fig. Q25.15a, and the ammeter reading is noted. The circuit is then reconnected as in Fig. Q25.15b, so that the positions of the ammeter and light bulb are reversed. (a) How does the ammeter reading in the situation shown in Fig. Q25.15a compare to the reading in the situation shown in Fig. Q25.15b? Explain your reasoning. (b) In which situation does the light bulb glow more brightly? Explain your reasoning.

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.