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

Two2.0cmdiameter insulating spheres have a6.0cmspace between them. One sphere is charged to+10nC, the other to-15nC. What is the electric field strength at the midpoint between the two spheres?

Short Answer

Expert verified

The electric field strength at the midpoint| between the two spheres is5.56×105N/C.

Step by step solution

01

Formula for electric field intensity

The intensity of the electric field at a point dunits away from a point charge qcan be calculated using the following formula:

(E)=q/4πεd2NC-1

The vector sum of the intensities produced by the individual charges equals the intensity of the electric field at any point due to a number of charges.

02

Calculation for electric field at the midpoint| between the two spheres 

Distance between the outer edges is 6.7cm.

Diameter of sphere is2.0cm.

Hence the distance between center is(6.7+2.0)=8.7cm.

Mid point is ,

=(8.7/2)cm

localid="1648629844607" =4.35cm=0.0435m.

Charge of spherelocalid="1648629881080" 1islocalid="1648391289846" 41nC.

Charge of spherelocalid="1648629893533" 2is-76nC.

The fields will now all point in the same direction.

As a result, whole field is:

localid="1648630029365" Enet=E1+E2

localid="1648630472057" E1due tolocalid="1648630003851" 42nCwill be:

localid="1648630478720" E1=9×109×41×10-9(0.0435)2

localid="1648630485528" E1=1.95×105N/C

localid="1648630491962" E2due to localid="1648630497375" -76nCwill be:

localid="1648630504401" E2=9×109×76×10-9(0.0435)2

localid="1648630510652" E2=3.61×105N/C

03

Calculation for electric field strength at the midpoint| between the two spheres 

SubstituteE1andE2values we get,

Enet=E1+E2

Enet=1.95×105N/C+3.61×105N/C

Enet=5.56×105N/C

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

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.

A small glass bead charged to +6.0nCis in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0cmfrom the rod’s center. The bead is repelled from the rod with a force of840mN. What is the total charge on the rod?

What are the strength and direction of the electric field at the position indicated by the dot in FIGUREP23.36? Give your answer (a)in component form and (b)as a magnitude and angle measured cwor ccw(specify which) from the positive x-axis.

A 12-cmlong thin rod has the nonuniform charge density λ(x)=(2.0nC/cm)e-|x|/(6.0cm), where localid="1648623923714" xis measured from the center of the rod. What is the total charge on the rod?

Hint: This exercise requires an integration. Think about how to handle the absolute value sign.

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?

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