Question

Asked Jan 28, 2020

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*q*_{1} = -5.61*q* and particle 2 of charge *q*_{2} = +3.31*q* are fixed to an *x* axis. As a multiple of distance *L*, at what coordinate on the axis is the net electric field of the particles zero?

Step 1

Let the electric field due to charge *q*_{1 }be denoted as *E*_{1}, whereas the electric field due to charge *q*_{2 }be denoted as *E*_{2}.

Let *x *denote the distance to the right of the charge *q*_{2} ,along the positive *x *axis, at which the net electric field of due to the two particles is zero.

Step 2

From the figure , it can be inferred that the charge

The net electric field *E*_{net} at a distance *x* to the right, from the charge *q*_{2 }along the *x *axis due to both charges is,

Step 3

Since the net electric field at a distance *x* to the right of the charge *q*2 is zero, substitute 0 for *E*net , -5.61*q *for *q*1 , 3.31*q* for *q...*

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