/*! 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} Q. 17 The electric field strength2.0 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The electric field strength2.0cmfrom the surface of a10cmdiameter metal ball is50,000N/C. What is the charge (innC)on the ball?

Short Answer

Expert verified

Charge on the ball is27.2nC.

Step by step solution

01

Formula for finding charge

Metal sphere has radius of5cm.

Electric field at from surface is 50,000N/C.

Electric field due to sphere of radiusRand chargeQat distance isr≥R

E=Q4πϵoR2......(1)

02

Calculation for charge

Using equation(1),

Calculate chargeQat metal sphere.

Q=E4πϵoR2

=(50,000N/C)9×109-1(0.02m+0.05m)2

localid="1648628325367" =(50,000N/C)9×109-1(0.07)2

=(50,000N/C)9×109-1(0.0049)

Q=27.2nC

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

What are the strength and direction of the electric field at the position indicated by the dot in FIGUREEX23.3?Specify the direction as an angle above or below horizontal.

A sphere of radius R has charge Q. The electric field strength at distancer>RisEi. What is the ratio Ef/Eiof the final to initial electric field strengths if

(a) Q is halved,

(b) R is halved, and

(c) r is halved (but is still>R)? Each part changes only one quantity; the other quantities have their initial values

A problem of practical interest is to make a beam of electrons turn a 90°corner. This can be done with the parallel-plate capacitor shown in FIGURE. An electron with kinetic energy 3.0×10-17Jenters through a small hole in the bottom plate of the capacitor.

a. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

b. What strength electric field is needed if the electron is to emerge from an exit hole 1.0cmaway from the entrance hole, traveling at right angles to its original direction?

Hint: The difficulty of this problem depends on how you choose your coordinate system.

c. What minimum separation dminmust the capacitor plates have?

Two10cmdiameter charged rings face each other,20cmapart. Both rings are charged to+20nC. What is the electric field strength at (a) the midpoint between the two rings and (b) the center of the left ring?

One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms30μm diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page at a speed of 20m/s. Along the way, the drops pass through two horizontal, parallel electrodes that are 6.0mmlong,4.0mm wide, and spaced 1.0mm apart. The distance from the center of the electrodes to the paper is 2.0cm. To form the tallest letters, which have a height of 6.0mm, the drops need to be deflected upward (or downward) by 3.0mm. What electric field strength is needed between the electrodes to achieve this deflection? Ink, which consists of dye particles suspended in alcohol, has a density of 800kg/m3 .

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.