The figure shows an arrangement of four charged particles, with angle = 35.0 and

Question

The figure shows an arrangement of four charged particles, with angle = 35.0 and distance d = 1.50 cm. Particle 2 has charge q2 = 8.00 × 10-19 C; particles 3 and 4 have charges q3 = q4 = -4.80 × 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)?

Explanation / Answer

a)

distance point 1 to 3 or 4, r = d/cos(theta)

= 1.5/cos(35)

= 1.83 cm

= 0.0183 m

magnitude of force exerted by 3 and 4 on 1,

|F3| = |F4| = k*q1*q3/r^2

= 9*10^9*q1*4.8*10^-19/0.0183^2

= 1.29*10^-5*q1

net force due to F3 and F4, F34 = 2*1.29*10^-7*q1*cos(35)

= 2.113*10^-5*q1 N

now use,

F2 = F34

k*q1*q2/(d + D)^2 = 2.113*10^-7*q1

9*10^9*8*10^-19/(0.015 + D)^2 = 2.113*10^-5

==> D = 0.00346 m

= 0.346 cm

b) less than

because the magnitude of F3 and F4 on the charge increases.

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