Question

# In this problem, consider electrostatic forces only. In the figure, there is a particle of charge...

In this problem, consider electrostatic forces only. In the figure, there is a particle of charge +Q at x = 0, and there is a particle of charge +16Q at x = +4a. You are going to bring in a third particle, with a charge of −4Q, and place it at an appropriate spot on the x-axis. Use Q = 60.0 ✕ 10−6C, and a = 40.0 cm. Also, use k = 9.00 ✕ 109 N · m2/C2. (a)First, place the particle of charge −4Q at the correct location so that the +Q charge experiences no net force because of the other two charged particles. In that situation, calculate the magnitude of the force that just one of the other charged particles exerts on the +Q charge. (b)Instead, at what location on the x-axis would you place the particle of charge −4Q so that the +16Q charge experiences no net force because of the other two charged particles? Note that we're looking for an answer in units of centimeters. (c)Finally, place the particle of charge −4Q at the correct location so that the −4Q charge experiences no net force because of the other two charged particles. In that situation, calculate the magnitude of the force that just one of the other charged particles exerts on the −4Q charge.

Ans:-

Given data q1 = Q = 60*10^-6C, x = 0

q2 = 16Q = 60*16*10^-6 =960*10^-6C , x= 4a = 4*40 = 160cm = 1.6m

q3 = -4Q =- 64*10^-6C

a] F1 = K*q1*q3 /r1^2

= 9*10^9*60*10^-6*-64*10^-6 /r^2 = -34.56/r1^2

F2 = K*q2*q1 /r2^2

= 9*10^9* 960*10^-6*60*10^-6/(1.6)^2 = 202.5N

F1 + F2 = 0

-34.56/r1^2 + 202.5 = 0

r1^2 = 34.56/202.5 = 0.17

r1 = 0.4m

b}F1 = Kq2*q3 /r1^2 = 9*10^9 * 960*10^-6*-64*10^-6 / r1^2 =-552.96/r1^2

F2 = Kq2 q1 /r2^2 = 202.5N

F1 +F2 = 0

-552.96/r1^2 +202.5 =0

r1 = 1.7m

c]F1 Kq1*q3/r1^2 = -34.56/r^2

F2 = Kq2*q3/r2^2 = -552.96/(1.6-r)r^2

34.56/r^2 = 552.96/(1.6-r)^2

(1.6-r)^2*34.56 = 552.96*r^2

2.56-3.2r+r^2 = 16r^2

-15r^2-3.2r+2.56= 0

15r^2 + 3.2r -2.56= 0

By solving we get

r = 0.32m

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