Particle 1 of charge q1 �� ��5.00q and particle 2 of charge q2 �� ��2.00q are fixed to an x axis. (a) as a multiple of distance l, at what coordi- nate on the axis is the net electric field of the particles zero?

Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get 

k q1 / d^2 = - k q2 / (d-1)^2 

Rearranging to calculate for d:

 (d-1)^2/d^2 = -q2/q1 = 0.4 
 d^2-2d+1 = 0.4d^2 
0.6d^2-2d+1 = 0  
d = 2.72075922005613 
d = 0.612574113277207 

We pick the value that is > q2 hence,

d = 2.72075922005613*L

d = 2.72*L

216

Step-by-step explanation:

Use the multiples of 27:

27, 54, 81, 108, 135, 162, 189, 216

↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑

×1 ×2 ×3 ×4 ×5 ×6 ×7 ×8

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