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

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE P23.35? Give your answer (a) in component form and (b) as a magnitude and angle measured cw or ccw (specify which) from the positive x-axis.

Short Answer

Expert verified

a. The electric field at the dot is1.0×105N/Ci^-3.6×104N/Cj^.

b. The angle made by the electric filed is 19.4°counter clockwise direction from the positive X-axis.

Step by step solution

01

Formula for electric field

The electric field is,

E=14πε0qr2

permittivity of free space is ε0.

02

Diagram for electric field position

In the diagram below, two positive charges and one negative charge are located in the three corners of a rectangle.

The electric field generated by the positive charge at the dot is directed away from the positive charge. The electric field generated by the negative charge at the dot is directed towards the negative charge.

03

Explanation (part a)

a.

Calculate the component form of the electric field strength at the place given by the dot in the diagram.

The angle formed by the vertical and the diagonal is,

θ=tan-12.0m4.0m

=26.56°

The component form of the resultant electric field at the dot is as follows:

Er=Exi^+Eyj^

From figure,

Ex=E1-E3cosθ

Ey=E3sinθ-E2

04

Explanation (part a) solution

a.

x-component of the net electric field at the dot.

The magnitude of the electric field due to the charge one is

E1=14πε0q1r12

The magnitude of the electric field due to the charge two is,

E2=14πε0q2r22

The magnitude of the electric field due to the charge three is,

role="math" localid="1651413236226" E3=14πε0q3r32

05

Explanation (part a)

The value,

r3=(2.0cm)2+(4.0cm)2

=4.47cm

Ex=14πε0q1r12-14πε0q3r32cosθ

So,

Ex=9×109C2/N·m2(5.0nC)(2.0cm)2-9×109C2/N·m2(5.0nC)(4.47cm)2cos26.56°

=9×109C2/N·m25.0nC10-9C1nC2.0cm1.0m100cm2-5.0nC10-9C1nC4.47cm1.0m100cm2cos26.56°

=1.0×105N/C

06

Explanation (part a)

For ycomponent,

Ey=14πε0q3r32sinθ-14πε0q2r22

So,

Ey=9×109C2/N·m2(5.0nC)(4.47cm)2(sin26.56)-(10.0nC)(4.0cm)2

=9×109C2/N·m25.0nC10-9C1nC4.47cm1.0m100cm2(sin26.56)-10.0nC10-9C1nC4.0cm1.0m100cm2

=-3.6×104N/C

So electric dot is,

Er=1.0×105N/Ci^+-3.6×104N/Cj^

07

Explanation (part b)

b.

Magnitude of the electric field at the dot is given by the expression,

E=Ex2+Ey2

Substitute all values,

E=1.0×105N/C2+-3.6×104N/C2

=1.1×105N/C

08

Explanation part b

angle of electric field is,

θ'=tan-1EyEx

θ'=tan-1-3.6×104N/C1.0×105N/C

=-19.4°

Here, negative sign indicates counter clockwise direction from the positivex-axis.

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

FIGURE Q23.6shows a hollow soda straw that has been uniformly charged with a positive charge. What is the electric field at the center (inside) of the straw? Explain.

Honeybees acquire a charge while flying due to friction with the air. A 100mgbee with a charge of +23pCexperiences an electric force in the earth’s electric field, which is typically 100N/C, directed downward.

a. What is the ratio of the electric force on the bee to the bee’s weight?

b. What electric field strength and direction would allow the bee to hang suspended in the air?

You’ve been assigned the task of determining the magnitude and direction of the electric field at a point in space. Give a step-by-step procedure of how you will do so. List any objects you will use, any measurements you will make, and any calculations you will need to perform. Make sure that your measurements do not disturb the charges that are creating the field.

You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing a microscopic oscillator to help keep time, and you’ve been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small, positively charged metal ring. His claim is that the negative charge will undergo simple harmonic motion at a frequency determined by the amount of charge on the ring.

a. Consider a negative charge near the center of a positively charged ring centered on the z-axis. Show that there is a restoring force on the charge if it moves along the z-axisbut stays close to the center of the ring. That is, show there’s a force that tries to keep the charge at z=0. b. Show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3,RandQare the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a 2.0μmdiameter ring charged to 1.0×10-13C.

FIGUREP23.41 is a cross section of two infinite lines of charge that extend out of the page. Both have linear charge density λ. Find an expression for the electric field strength E at height y above the midpoint between the lines.

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