Three point charges are arranged along the x axis. Charge q1 = -4.10 nC is locat
ID: 251793 • Letter: T
Question
Three point charges are arranged along the x axis. Charge q1 = -4.10 nC is located at x= 0.230 m and charge q2 = 2.50 nC is at x= -0.330 m . A positive point charge q3 is located at the origin.Three point charges are arranged along the x axis. Charge q1 = -4.10 nC is located at x= 0.230 m and charge q2 = 2.50 nC is at x= -0.330 m . A positive point charge q3 is located at the origin.
Part A:
What must the value of q3 be for the net force on this point charge to have magnitude 4.90 N ?
Part B:
Where along the x axis can q3 be placed and the net force on it be zero, other than the trivial answers of x=+ and x=?
Please provide clear work and if able, clear explanation.
Explanation / Answer
First Charge ,q1 = -4.10 nC
Distance,x1= 0.230 m
Second charge,q2 = 2.50 nC
Distance,x2= -0.330 m
1)So let us find the the field due to the first two charges at the origin, where both fields point to the right.
So, electric field,E = -8.99*109N·m2/C2 * (4.1*10-9C / (0.23m)2 + 2.5*10-9C / (0.33m)2
E = -903 N/C
Then for F = 4.9*10-6 N =903 * q
q=5.42nC it can be + or -
2) It is obvious that the point can't be between the two charges, since the field always points to the right in this range. We know that the point cant be outside the larger charge, or the larger charge would always dominate (by proximity). So we can conclude that it is to the left of the positive charge, where the field due the positive charge is to the left, and the field from the negative charge is to the right. Since their distance is 0.56 m, the distance d from the left charge must be such that
k * q3 * 2.5nC / d2 = k * q3 * 4.1nC / (0.56m + d)2
2.5(0.56 + d)2 = 4.1d2
1.6d2-2.8d-0.784=0
d = -0.618 m, 1.62 m
Since this was set up with d measured to the left of the left charge, our answer is d = 1.62 m
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