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having trouble with the following

There are three point charges along the x-axis:

q1 is +3µC, located at the origin

q2 is -5µC, located at x = +0.200m

q3 is -8µC

Where is q3 located if the net force on it is 7.00 N in the -x direction?


The way I see to go about this is as follows:

It is not located at x>0.2, as q2 repulses it and is closer and larger than q1

At 0 < x < 0.2 the force from both q1 and q2 goes in -x direction

At x < 0 It is repulsed by q2 but attracted by the closer q1. Given q2's larger size it may still be in this range


If q3 is at x < 0 (which I know to be correct), then F1 is +x direction and F2 is -x direction, so the problem is expressed as F1 - F2 = -7 N where F1 is the force from q1 on q3 and F2 is the force from q2 on q3.

By Coloumb's Law we have that:

F1 = k(q1q3)/sq(r1) and F2 = k(q2q3)/sq(r2)

r1 equals -x where x is the location of q3 on the x-axis (negative), and r2 equals -x+0.2

F1 - F2 = -7

k(q1q3)/sq(-x) - k(q1q3)/sq(-x+0.2) = -7

k(q1q3) and k(q2q3) are known numbers, and gives:

(2.16*10e-1)/sq(-x) - (3.60*10e-1)/sq(-x+2) = -7


Now in earlier problems in the book steps were simple from here, as the right side was 0; so you could just multiply the equation with sq(-x+2)*sq(-x) and solve the polynomial for x = 0. In this case though that leaves a fourth power polynomial on the right side since it isn't empty, which I have no idea how to practically solve.

Also, plotting the left-side equation I get here into a graph program and identifying F=-7 gives me an x value slightly different from the book (at -0.161, while the book says the solution is -0.144); larger than any rounding errors I can think of would imply.

Basically I've been scratching my head over this for a few hours and can't seem to figure out what I am misunderstanding, if it's the mathematics or some aspect of the physics of it that I'm not getting.

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  • $\begingroup$ A bit hard to read your equations - but be careful with your signs. If you explicitly state that some charges are positive and others are negative, and then again use the sign of the charge in your expression, you may be double counting. I recommend that you use a convention of "positive force points to the right", and all forces are added together. Then the signs ought to work out correctly. $\endgroup$
    – Floris
    Aug 25 '15 at 22:18
  • $\begingroup$ $\frac{0.216}{x^2} - \frac{0.36}{(-x+0.2)^2}=-7$ is your last equation. Try and solve it and see what you get for $x$. Seek the common denominator of the first two terms, then rework to a fourth degree polynomial (!). $\endgroup$
    – Gert
    Aug 25 '15 at 22:37
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    $\begingroup$ Welcome to Physics! Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. $\endgroup$
    – Kyle Kanos
    Aug 26 '15 at 2:54
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Lets to this step by step and take care of the signs!

Let $q_1=3$µC, $q_2=5$µC and $q_3=-8$µC.

The formula for the force, acting on particle one due two the presence of particle two, is given by $$\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{{| \vec{r}_{21}|}^2} {\hat{r}}_{21},$$ where $\hat{r}_{21}$ is the unit vector pointing from charge two to charge one. The overall force on one particle is the sum of all (Coulomb-) forces.

The distance $r_{13}$ is the unknown we are looking for. The distance $r_{23}$ can be written as $r_{23}=0.2-r_{13}$, since we know from your reasoning, that particle 3 is to the left of particle 1, i.e. it has a negative $x$-coordinate. As we are in 1D, we do not have to care about the full vector expression. The total force is then given as $$F_{tot}=F_{13}+F_{23} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 \cdot q_3}{|r_{13}|^2} + \frac{q_2 \cdot q_3}{|r_{23}|^2}\right) \\ =8.988\times 10^9\ \mathrm{N\cdot m^2\cdot C}^{-2} \left( - \frac{24\times 10^{-12} \mathrm{C^2}}{|r_{13}|^2} + \frac{40\times 10^{-12} \mathrm{C^2}}{|0.2\mathrm{m}-r_{13}|^2}\right)\\ =\left(-\frac{0.215712}{|r_{13}|^2} + \frac{0.35952}{|0.2\mathrm{m}-r_{13}|^2} \right) \mathrm{N\cdot m^2}\\ \stackrel{!}{=}-7\mathrm{N} \quad \text{(force is in the negative direction)}$$

You should realize that the signs differ from yours! Solving this equation with whatever methods, yields something like $r_{13}=-0.146\,972 \, \mathrm{m}$, which is close to what your book says.

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  • $\begingroup$ Yep - that's what I got... not quite -0.144 but close. $\endgroup$
    – Floris
    Aug 26 '15 at 0:22

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