Physics 2 Chapter 21 Problem 22 Chapter 21, Problem 022 Your answer is partially

Question

Physics 2 Chapter 21 Problem 22 Chapter 21, Problem 022 Your answer is partially correct. Try agan. The figure shows an arrangement of four charged partides, with angle 0-35.0 . and distance d . 3.00 cm. Part de 2 has charge q 9.60 x 101, C; particles 3 and 4 have charges Oz-G4--1.60 × 10 " c (a) what is the distance D between te ang n and part de 2 if the net electrostatic force on particle 1 due to the other parties is zero? (b) If parte es 3 and 4 were, moved closer to the x axis but maintained their symmetry about that axis, would the required value of D be greater than, less than, or the same as in part (a) (a) Number o.029 Unitt the tolerance as +/-2%

Explanation / Answer

given

theta = 35 deg

d = 3 cm = 0.03 m

q1 = ?

q2 = 9.6*10^-19 C

q3 = q4 = -1.6*10^-19 C

a) D = ?

Net force on 1 is 0

so, from symmetry, the y component of the force due to particle 3 and 4 cancel out at the location of particle 1, and only the horizontal components act

so, from coloumbs law

2kq3q1cos(theta)/(d/cos(theta))^2 + kq2q1/(D + d)^2 = 0

2*1.6*10^-19*cos^3(35)/0.03^2 = 9.6*10^-19/(D + 0.03)^2

3.2*cos^3(35)/0.03^2 = 9.6/(D + 0.03)^2

D = 0.04 m = 4.008 cm

b) as the particels 3 and 4 are brought closer to the origin, the cos(theta) value increases

hence the Value of D decreases

from this equation

2kq3q1cos(theta)/(d/cos(theta))^2 + kq2q1/(D + d)^2 = 0

2q3cos^3(theta)/(d)^2 + q2/(D + d)^2 = 0

so when cos(theta) increases ( when theta is decreased), value od D decreases

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