Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 1.0-mm-diameter oil droplet (density role="math" localid="1649088632675" 900kg/m3 ) is negatively charged with the addition of 25 extra electrons. It is released from rest 2.0mm from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of 3.5m/s. What is the surface charge density of the plane?

Short Answer

Expert verified

The surface charge density of the plane is6.4x10-6C/m²

Step by step solution

01

Given information and formula used

Given :

Diameter of oil droplet : 1.0-mm

Density : 900kg/m3

Charge : Negatively charged with 25extra electrons.

It is released from rest : 2.0mm

Collides with a speed of : 3.5m/s.

Theory used :

Conservation of Energy says :

12mv2=qEd

02

Calculating the surface charge density of the plane 

Mass of the droplet is :

ρ×43πr3=900×43π×(0.5×10-6)3=4.7x10-16kg

Mass charge on the droplet :

ne=25×1.6×10-19=40×10-19C

From energy conservation :

E=12×mv²qd=12×4.7×10-16×(3.5)240×10-19×2×10-3=3.6x105N/C

We know that for electric field owing to infinite plane sheet:

E=η2ε0η=E2ε0=2×8.85×10-12×3.6×105=6.4x10-6C/m²

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

Two10cmdiameter charged rings face each other,20cmapart. The left ring is charged to-20nCand the right ring is charged to+20nc.

a. What is the electric fieldE, both magnitude and direction, at the midpoint between the two rings?

b. What is the force on a proton at the midpoint?

FIGURE shows a thin rod of length Lwith total charge Q. Find an expression for the electric fieldE at point P. Give your answer in component form.

An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charges in opposite directions. The dipole moment of an induced dipole is directly proportional to the electric field. That is, p=αE, where αis called the polarizability of the molecule. A bigger field stretches the molecule farther and causes a larger dipole moment.

a. What are the units of α?

b. An ion with charge qis distancerfrom a molecule with polarizability α. Find an expression for the force Fionondipole.

The surface charge density on an infinite charged plane is -2.0×10-6C/m2 . A proton is shot straight away from the plane at 2.0×106m/s. How far does the proton travel before reaching its turning point?

A small segment of wire in FIGURE Q23.4contains 10nCof charge.

a. The segment is shrunk to one-third of its original length. What is the ratio of λf/λi, where λiandλf are the initial and final linear charge densities?

b. A proton is very far from the wire. What is the ratio Ff /Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

c. Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to keep the linear charge density unchanged?

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