A + 5.7 m C and a - 3.5 m C charge are placed 25 cm apart. Where can a third cha

Question

A + 5.7 mC and a - 3.5 mC charge are placed 25 cm apart.  Where can a third charge be placed so that it experiences no net force?

So we're trying to find r. We want the Forces to be equal, so:

kq1q2/r2 = kq1q2/r2

Q1 is the "third charge," so we can just cancel it out here, as   well as the K. So then we're left with:

(3.5mC)/r2 = (5.7mC)/(r+.25)2

THIS IS THE PART I NEED HELP...don't the R's cancel out? How do they get to to the point where they can do the quadratic equation? I would appreciate it if someone showed me this step by step cause what I do is this.

(3.5)/R^2 = (5.7) / (R^2 + .50R + 0.0625)   and I don't now what to do from here....how do I set it up to get it into the quadratic equation?

Thanks!

A + 5.7 mC and a - 3.5 mC charge are placed 25 cm apart. Where can a third charge be placed so that it experiences no net force? kq1q2/r2 = kq1q2/r2 Q1 is the "third charge," so we can just cancel it out here, as well as the K. So then we're left with: (3.5mC)/r2 = (5.7mC)/(r+.25)2 (3.5)/R^2 = (5.7) / (R^2 + .50R + 0.0625) and I don't know what to do from here how do I set it up to get it into the quadratic equation?

Explanation / Answer

Let's draw a picture:

q1----------------q2-----------------q3
0-----------------x------------------25

First, let's assume q1=+5.7C and q3=-3.5C. Can a charge q2 with no net force (i.e., F=0) be located between q1 and q3? The answer is NO ... because the force of q1 on q2 is to the right and the force of q3 on q2 is also to the right (because q3 is a negative charge). This means a charge with no net force can't be located between q1 and q3. Next, what if we assume q1=+5.7C and q2=-3.5C, can a charge (q3) with no net force be located to the right of q2? The answer is YES ... because the force of q1 on q3 is still to the right and the force of q2 is to the left. So we can label the figure as such:

+5.7C............-3.5C
q1----------------q2-----------------q3
0-----------------25------------------x

Now the force of q1 on q3 = kq1q3/(25+x)^2

and the force of q2 on q3 = kq2q3/x^2

So the total force on q3 = kq1q3/(25+x)^2 + kq2q3/x^2 = 0 (no net force)

Eliminate the common terns (k & q3), so: q1/(25+x)^2 + q2/x^2 = 0

Which we can rewrite as: q1*x^2 + q2*(25+x)^2 = 0

And ... q1*x^2 + q2*(625 + 50x + x^2) = 0

And ... (q1+q2)*x^2 + 50*q2*x + 625*q2 = 0

Substitute the values of q1 and q2: 2.2*x^2 - 175*x - 2187.5 = 0

Solve the quadratic equation for x = -10.98 cm and 90.53 cm

We can toss out x = -10.98 cm by inspection because this would place the charge to the left of q1, so the answer must be x = 90.53 cm.

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