2. Suppose that the charges in Example 1.3 are situated as shownin the figure. Find the resultant electric force on the charge at A. R- what is the electric EXAMPLE 1.3Force of Adjacent Charges - Four Charges 1.> Four point charges (two with q = 2.50 × 10-6 C and two with q =-2.50 x 10-6 C) are situated at the corners of a square of side 1.00m as shown. Find the resultant force that the charge at A willexperience due to the charges at the other corners of the square.Given:Find:qa = 9B = +2.50 × 10-6 C9c = 9p = -2.50 × 10-6 Cr = 1.0 mFB on A = ?Fc on A = ?FD on A = ?Fresultant = ?Solution:For FBon Aa.|(+2.50 × 10-6 C)(+2.50 × 10-6 C)|FB on A = k(r)2(9 × 10º N · m²/C²)(1.0 m)²|+6.25 × 10-12 C|1 m²FB on A is repulsive. Since the charges are both positive, the charge at B will nchthe charge away from it. Therefore, FB on A is directed to the left.FB on A = (9 × 10° N · m²/C²)= 0.06 N4/11b.For Fc on A|(+2.50 × 10-6 C)(-2.50 × 10-6 C)|(1.0 m)²Fc on A = k(r)2= (9 × 10° N · m²/C²)|-6.25 × 10-12 C|1.0 m²Fc on A = (9 × 10° N · m² /C²)= 0.06 N2GENERAL PHYSICS 2LEARNING MATERIAL 2Fc on A is attractive since the charges are opposite. The charge at C will attract thecharge at A toward it. Therefore, Fc on A is directed down.c.For Fp on Ar|(+2.50 × 10-6 C)(-2.50 × 10-6 C)|FD on A = k= (9 x 10° N · m²/C²)·(r)²(1.414 m)²Note: We used 1.414 m for distance between D and A by applying the PythagoreanTheorem. 1.00 m is only for the sides.|-6.25 × 10-12 C|Fp on A = (9 × 10° N · m² /C²)= 0.03 N2.00 m²FD on A is attractive since the charges are opposite. Thecharge at D will attract the charge at A toward it.Therefore, FD on A is directed down at 45° with +x-axis.The diagram for the forces is shown in the figure.Now, to find for the resultant force Fresultant, let's use thecomponent method again.ForceHorizontal ComponentVertical ComponentFB on A = 0.06 N0.06 N(cos 180°) = –0.06 N0.06 N(sin 180°) = 0 NFc on A = 0.06 N0.06 N(cos 270°) = 0 N0.06 N(sin 270°) = –0.06 NFp on A = 0.03 N0.03 N(cos 315°) = +0.02 N0.03 N(sin 315°) = –0.02 NFresultantF = -0.04> F, = -0.08 NSolving for the magnitude of the resultant force,Fp on AFresultant =+Fresultant = -(-0.04 N)² + (-0.08 N)² = 0.08 NFD on ASolving for the direction,-0.08||-0.04|FresultaarFc on A0 = tan-1= tan-1= 64.5°rThus, the resultant force is 0.08 N, 64.5° from the negative x-axis.

The given data are listed below with appropriate variables. Here, q denotes the amount of charge…

Answered: Suppose that the charges in Example 1.3…

Suppose that the charges in Example 1.3 are situated as shown in the figure. Find the resultant electric force on the charge at A.

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter21: Electric Fields

Section: Chapter Questions

Problem 51A

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