Question

Figure 21-30shows an arrangement of four charged particles, with angle θ = 30.0 ˚ and distance d = 3.00 cm. Particle 2 has charge q2 = 9.60 × 10-19 C; particles 3 and 4 have charges q3 = q4 = -4.80 × 10-19 C. 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?

Answer 1

Lets i be the component of force in the direction of the x-axis and j be the component force in direction of the y-axis.
Force of q3 on q1 = kq3q1/(4πεd^2) * (cos(30)i + sin(30)j)

Force of q4 on q1 = kq4q1/(4πεd^2) * (cos(30)i - sin(30)j)

Force of q2 on q1 = kq2q1/(4πε(d+D)^2)(-i)

Since electrostatic force on q1 is zero

kq3q1/(4πεd^2) * (cos(30)i + sin(30)j) + kq4q1/(4πεd^2) * (cos(30)i - sin(30)j) = kq2q1/(4πε(d+D)^2)i

and q3 = q4

2kq3q1/(4πεd^2) *(cos30) = kq2q1/(4πε(d+D)^2)

(√3)q3/d^2 = q2/(d+D)^2

(d+D)^2 = q2* d^2 /(√3q3)

(d+D) = 3.22

D= 3.22 -3

D = 0.22 cm