Physics 2 Chapter 21 Problem 22 Chapter 21, Problem 022 The figure shows an arra
ID: 1636488 • Letter: P
Question
Physics 2 Chapter 21 Problem 22 Chapter 21, Problem 022 The figure shows an arrangement of four charged partidles, with angle e-35.0 and diatance d- 3.00 cm. Particde 2 has charge q-9.60 x 101 C: particles 3 and 4 have charges -q4--1.60 x 10 " c. (a) what is the dstance D betwee. the ongin and partde 2 the net electrostatic force on partide due to the other part des " zero? (b) parbdes 3 and 4·ere movied dloser to the x axis but maintained their symmetry about thet axis, would the required value of D be greater than, less than, or the same as in part (a)? (a) Number (b) D the tolerance is +/-2% Click if you would like to Show Work for this question: gen show worlk UnitsExplanation / Answer
If the charge on particle 1 is q then the force on it from particle 2 is 8kq/(d+D)^2, where k=1/40.
The y-components of the forces on particle 1 from charges 3 and 4 are equal and opposite, so they cancel; the x-components are each -1.6kqcos/r^2 where r is the distance between 1 and 4.
We have rcos=d and cos=½3, so (cos)^2=3/4 and cos/r^2=cos(cos)^2/d^2=½33/4d^2.
So for the sum of x-components of forces on 1 from charges 2, 3 and 4 to be zero we require
8kq/(d+D)^2 - 2*1.6kq*½33/4d^2 = 0.
Insert d=3cm and rearrange. We get
(3+D)^2 = 180/33
3+D = 5.9
Therefore D = 2.9 cm.