Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2

Question

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.312m to the right of Q1. Q3 is located 0.158 m to the right of Q2.

Part 1: In the above, Q1 = (1.56 * 10^-6 C), Q2 = (-3.03*10^-6 C) and Q3 = (3.33 * 10^-6 C). Calculate the total force on Q2. Give with the plus sign for a force directed to the right.

Part 2: Now the charges Q1 = (1.56 * 10^-6 C) and Q2 = (-3.03 * 10^-6 C) are fixed at their positions, distance 0.257 m apart, and the charge Q3 = (3.33 * 10^-6 C) is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

Explanation / Answer

For part 1)

I used .257 for the distance between Q1 and Q2 since that was given for part B. Using the .312 distance gives an answer to part A of 3.20 N to the right, however, you said that was wrong in two different answers. Perhaps it was a typo, so I will try the .257 you typed in for part B. Using that you get...

Q1 pulls Q2 to the left. The force is by F = kqq/r^2

F = (9 X 10^9)(1.56 X 10^-6)(3.03 X 10^-6)/(.257)^2 = .437 N to the left

Q3 pulls Q2 to the right

F = (9 X 10^9)(3.33 X 10^-6)(3.03 X 10^-6)/(.158)^2 = 3.64 N to the right

The net force is 3.64 - .644 = 2.99 N to the right

Part 2)

The only location for this to happen will be to the left of Q1

We can say

kqq/r^2 = kqq/r^2

That is

(9 X 10^9)(1.56 X 10^-6)(3.33 X 10^-6)/x^2 = (9 X 10^9)(3.03 X 10^-6)(3.33 X 10^-6)/(.257 + x)^2

Simplify

.04675/x^2 = .090809/(.066 + .514x + x^2)

Cross multiply

3.0855 X 10^-3 + .02403x + .04675x^2 = .090809x^2

Then in standard form, .044059x^2 -.02403x - 3.0855 X 10^-3 = 0

Put that into the quadratic formula and get x = .652 or -.107

Since the value of x has to be positive, we can eliminate the negative root

The positive root of .652 means that the distance is .652 m to the left of Q1

For your answer key, they will want -.652 m

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