# [SOLVED] Physics problem based on a picture, but hard!?

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Physics problem based on a picture, but hard!?

Two charged particles, with charges q_1=q and q_2=4q, are located at a distance d=2.00cm apart on the x axis. A third charged particle, with charge q_3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.

Find the position of charge 3 when q = 1.00 nC.

The question is: Which of the following sketches represents a possible configuration for this problem?

Based on this picture: by

q3 must be closer to q1 than to q2 for the force magnitudes to be the same, simply because q2 has 4 times the charge as q1. Therefore, B and D are eliminated right off.

The question acts about the MAGNITUDE of the force that q1 and q2 exert on q3; we don't have to worry about the direction of the force. The magnitudes are given by q1q3/(r13)^2 and q2q3/(r23)^2, but we know that q1q3 is 1/4 what q2q3 is. Therefore, r23 must be twice what r13 is, to make these magnitudes the same. In math-speak:

q*q/(r13)^2 = q*4q/(r23)^2, so that 1/(4r13^2) = 4/(4r23^2)

==> r23^2 = 4r13^2

==> r23 = 2*r13.

We have two possibilities: q3 is between q1 and q2, or (reading left-to-right) their positions are q3, q1, q2.

q3 Between:

We know that r12 = 2.00 cm. Also, r13 + r23 = r12, so that r13 + 2*r13 = 2 cm, and so r13 = 0.66 cm. Here, q3 is 1/3 of the way between q1 and q2.

q3 to the left:

We know that r12 = 2.00 cm. Also, r13 + r12 = r23, so that r13 + 2 cm = 2 r13. Thus, r13 = 2cm (the same distance as r12).

From the diagrams given, if they're anything to scale, the only possible answer is (A), as in (C), r13 is obviously quite a bit less than r12. If they're NOT drawn to scale, then both (A) and (C) could be answers.

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+1 vote