Problem S: Four point charges of equal magnitude Q = 35 nC are placed on the corners of a e of sides D1 = 22 cm and D2 = 9 cm. The charges on the left side of the rectangle are while the charges on the right side of the rectangle are negative. Refer to the figure. +Q -Q rt (e) Calculate the value of the vertical component of the net force, in newtons. D2 +Q DI rt (f) Calculate the magnitude of the net force, in newtons, on the charge located at the lower left corner of the rectangle. t (g) Calculate the angle, in degrees between -180° and +180°, that the net force makes, measured from the positive horizontal direction.

Answered: Problem S: Four point charges of equal magnitude Q = 35 nC are placed on the corners of a e of sides D1 = 22 cm and D2 = 9 cm. The charges on the left side of…

College Physics

11th Edition

ISBN: 9781305952300

Author: Raymond A. Serway, Chris Vuille

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

Problem 8: Four point charges of equal magnitude Q = 35 nC are placed on the corners of a
rectangle of sides D1 = 22 cm and D, = 9 cm. The charges on the left side of the rectangle are
positive while the charges on the right side of the rectangle are negative. Refer to the figure.
+Q
Part (e) Calculate the value of the vertical component of the net force, in newtons.
F, =
D2
+Q
DI
Part (f) Calculate the magnitude of the net force, in newtons, on the charge located at the lower left corner of the rectangle.
F =
Part (g) Calculate the angle, in degrees between -180° and +180°, that the net force makes, measured from the positive horizontal direction.

Expert Solution

Step 1

a)To find the net force, Force between Pairs at the sides and the diagonal pairs must be calculated.

Net force,

$F = F_{1} + F_{2} + F_{3} + F_{4} + F_{5} + F_{6} = - F_{1} J + F_{2} i + F_{3} j - F_{4} i + F_{5} \cos 45^{0} i + F_{5} \sin 45^{0} j - F_{6} \cos 45^{0} i + F_{6} \sin 45^{0} j \frac{}{} = {\frac{- k {(35 \times 10^{- 9})}^{2}}{9^{2}} + \frac{k (35 \times 10^{- 9})}{9^{2}} + 2 (\frac{k {(35 \times 10^{- 9})}^{2}}{{(\sqrt{22^{2^{}} + 9^{2}})}^{2}} \times \sin 45)} j + {\frac{k {(35 \times 10^{- 9})}^{2}}{22^{2}} - \frac{k {(35 \times 10^{- 9})}^{2}}{22^{2}} + \frac{k {(35 \times 10^{- 9})}^{2}}{{(\sqrt{22^{2} + 9^{2}})}^{2}} \cos 45 - \frac{k {(35 \times 10^{- 9})}^{2}}{{(\sqrt{22^{2^{}} + 9^{2}})}^{2}} \cos 45} i$

The horizontal component of net force is

$= 2 \times \frac{k {(35 \times 10^{- 9})}^{29}}{{(\sqrt{22^{2} + 9^{2}})}^{2}} \times \sin 45^{0} i = 27.6 \times 10^{- 9} i N$

Magnitude is $27.6 \times 10^{- 9} N$

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON