Question

shows an arrangement of four charged particles, with angle θ = 30.0° and 

distance d = 2.00 cm. Particle 2 has charge q2 = +8.00 × 10-19 C; particles 3 and 4 have charges q3

= q4 = -1.60 × 10-19 C. (a) What is 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)?

Answer 1

d=2.00cm

$\theta=30°\\q_2=8\times 10^{-19}c\\q_3=q_4=-1.6\times 10^{-19}c$

$r=\frac{2d}{\sqrt{3}}$

Net force on q₁ due to to the to charge q₃ q₄ on the y-axis has magnitude of force

$F=2\frac{q_1q_3}{4\pi \epsilon_0 r^2}cos30°$

$q_2=8.00\times10^{-19}c=5.00|q_3|$ then net force will be zero

$\frac{3\sqrt {3}q_1q_3}{16\pi\epsilon_0 d^2}=\frac{q_1q_2}{4\pi \epsilon(D+d)^2}$

$D=d(2\sqrt{\frac{ 5}{3\sqrt{3}}}-1)$

$D=0.9245d$

d=2.00cm

$D=1.8490cm$

(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

$\theta$ Varying then D increase

We can say D will be greater then part (a)

$D>1.8490cm$