Question

The figure shows an arrangement of four charged particles, with angle θ = 37.0 ˚ and distance d = 3.00 cm. Particle 2 has charge q₂ = 6.40 × 10^-19 C; particles 3 and 4 have charges q₃ = q₄ = -1.60 × 10^-19 C. (a) What is the distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero? (b) If particles 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)?

this website has the figure

https://www.chegg.com/homework-help/questions-and-answers/figure-shows-arrangement-four-charged-particles-angle-theta-370-degree-distance-d-300-cm-p-q3603840?trackid=06f98d39478b&strackid=d121f996f8aa

Answer 1

let dist of 3 and 4 from 1 be r then cos37 =d/r => r= d/cos37

Force due to 3 and 4 on 1 will have one horizontal component and one verticalcomponent each but the vertical components cancel out due to symmetry. and horizontal component will be cos 37 times the force=>

F3 = kq1q3/r^2 =kq1q3/d^2 *(cos37)^2 => horizontal component=F3cos37=kq1q3/d^2 *(cos37)^3 this is the same force by q4 also due to symmetry.

Force due to q2: kq1q2/(d+D)^2 since q2 have opposite sign from q3 , their forces will be opposite. so, for net force on q1 to be 0, F2= -2*F3cos37 =>kq1q2/(d+D)^2= - 2kq1q3/d^2 *(cos37)^3 =>-q2/2*q3=(cos37)^3 * (d+D)^2/d^2

cos37~0.8 , q₂ = 6.40 × 10^-19 C and q₃ = q₄ = -1.60 × 10^-19 C and d=0.03m=> 6.4/3.2 =0.512 (0.03+D)^2/0.0009

=> (0.03+D)^2 =2/568.89 => D= 0.059-0.03 = 0.029m

b)If particles 3 and 4 were moved closer to the x axis maintaining symmetry, the horizontal components due to these masses increase so, the opposing force due to q2 needs to increase so, 2 should be closer to 1 ie D will be lees than as of in part a