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Ainat [17]
2 years ago
12

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?
Mathematics
1 answer:
Naya [18.7K]2 years ago
3 0

<span>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 </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

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

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

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