Chapter 18, Problem 21 Interactive LearningWare 18.2 provides one approach to so

Question

Chapter 18, Problem 21 Interactive LearningWare 18.2 provides one approach to solving problems such as this one. The - +8.51HC, the other two have identical magnitudes, but opposite signs: 92-4.27HC and - 4.2c (a) mass of 1.50 g and it were free to move, what would be its acceleration? drawing shows three point charges fwed in place. The charge at the coordnate origin hes a value of and a- +4.27C. (a) Determine the net force exerted on a, by the other twos charges. (b) If o, had a at the coordinate origin hes a value of 130 1.30 (a) Number (b) Number Units m mis kg M" p" '" Question Attempts:00' 3 used Copyrighe C, 2000-2015 by ehn waer & Sons, Inc. or retaned companies. All rights reserved. 4.16.1.7 2009-2015 20hn woty & Sons. Inc. A" Rights 2000-2015 2ohn Wiley & Sons, Inc, Al Rights Reserved.A Dvision of 2on Wiley & Sans Ic Prhoo hner 0 8 9 6 3

Explanation / Answer

a)
Q1 = 8.51uC
Q2 = -4.27 uC
Q3 = 4.27 uC
Value of K * Q1*Q2 = 8.99 * 10^9 * 8.51 * 10^-6 * 4.27 * 10^-6 = 0.327
Value of K * Q1*Q3 = 8.99 * 10^9 * 8.51 * 10^-6 * 4.27 * 10^-6 = 0.327

Force on Q1 due to Q2 F21 = K*q1*q2 / (1.3)^2 (Towards Q1)
F21 = 0.327 / (1.3)^2 = 0.193 N

Resolving Forces -
X axis - F21 COS(23) (+ve Xaxis)
Yaxis - F21 SIN(23) (+ve Yaxis)

Force on Q1 due to Q3 F23 = K*Q1*Q3 / (1.3)^2  (Away from Q2)

F31 =  0.327 / (1.3)^2
F31 = 0.193 N

Resolving Forces -
X axis - F31 COS(23) (-ve X axis)
Yaxis - F31 SIN(23) (+ve Y axis)

X axis Component will cancel out each other.
Y axis Component will add up.

Net Force F =  F31 SIN(23) +  F21 SIN(23)
F = 2 * 0.193 * sin(23)
F = 0.1508 N

b)
Force = Mass * Acceleration
0.1508 = 1.5 * 10^-3 * Acceleration
Acceleration a = (0.1508 * 10^3 )/1.5 m/s^2
Acceleration a = 100.5 m/s^2

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