Chapter 21, Problem 022 The figure shows an arrangement of four charged particle

Question

Chapter 21, Problem 022 The figure shows an arrangement of four charged particles, with angle = 31.0 ° and distance d = 3.00 cm. Particle 2 has charge q2 = 8.00 x 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)? (a) Number units the tolerance is +/-1 in the 3rd significant digit

Explanation / Answer

distance of 1 from 3 or 4.

r = d/ cos(theta)

force on 1 due to either of 3 or 4,

F = k q1 q3 / r^2 = k q1 q3 cos^2(theta) / d^2

net force on 1 due to both of 3 and 4,

F' = 2 F cos(theta) = 2 k q1 q3 cos^3(theta) / d^2

for magnittude of force on 1 due to 2 will be F'.

hence F' = k q1 q2 / (d + D)^2

So, 2 k q1 q3 cos^3(theta) / d^2 = k q1 q2 / (d + D)^2

(d + D)^2 = d^2 q2 / (2 q3 cos^3 (theta))

(A) (3 + D)^2 = 3^2 (8 x 10^-19) / (2 x 1.60 x 10^-19 x (cos31)^3)

(3 + D)^2 = 35.77

D = 3 cm ......Ans

(B) if 3 and 4 comes closer to x axis.

then theta will decrease.

so cos(theta) will increase.

and (3 + D)^2 will decrease.

hence D will be less than we got in (a)

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