How Do You Solve for Charges Using Coulomb's Law Equation?

[exitwound]

Homework Statement

Homework Equations

F=kQq/d^2

The Attempt at a Solution

I've tried working on this for two days and can't figure it out.

In (a), the cumulative force on A is the sum of the force from B and the force from C. or:

$$F=\frac{k Q_a Q_b}{d^2} + \frac{k Q_a Q_c}{d^2}$$

In (b), the same applies.

however, I can't figure out what to do with these equations in order to isolate Qb or Qc. If I use a negative d (-d) as a distance from A-->B in (b), then I get two equations that are identical, but shouldn't be. If I move the origin, it doesn't seem to matter either.

I don't know how to start this problem.

 

[faradayslaw]

Qc/Qb=1.328 (sorry)

Ok, so in the first scenario, both charges B and C exert forces to the left, on charge A, since all charges are positive. However, in figure b, charge b exerts a force to the right, while charge C exerts a force to the left, on charge a. Hence, in fig. a,
-2.03x10^-23==-kQAQB/r^2-kQAQC/r^2
while in fig b
-2.86x10^-24==-kQAQC/r^2+kQAQB/r^2

If you factor out k, QA, and r^2, keeping in mind that RB=RC (distance from a to b, and a to c are the same in both figures), and divide the two equations, you can get QC/QB which is 1.328

Hope this helps,

 

[Dadface]

Because of the direction of the force you know that Qc is bigger than Qb so write an equation like you did for part a but getting the directions right.Now divide one equation by the other and tidy it up.

 

[faradayslaw]

Because of the direction of the force you know that Qc is bigger than Qb so write an equation like you did for part a but getting the directions right.Now divide one equation by the other and tidy it up.

You are correct, QC should be larger than Qb, but when I worked it out, I got QC/QB==1.328, I don't know why...

corrected, see above, I had a problem with the signs in my initial equation...

 

[exitwound]

I'm still absolutely lost on this.

Faraday, I understand what you did taking into account the negative Force due to B in the second example. However, I don't know what to do with the equations at this point.

I end up with:

(a) F=(kQa/d^2)(Qb+Qc)

(b) F=(kQa/d^2)(Qc-Qb)

I don't understand where to go from here.

 

[faradayslaw]

I'm still absolutely lost on this.

Faraday, I understand what you did taking into account the negative Force due to B in the second example. However, I don't know what to do with the equations at this point.

I end up with:

(a) F=(kQa/d^2)(Qb+Qc)

(b) F=(kQa/d^2)(Qc-Qb)

I don't understand where to go from here.

You can now divide the two equations above, so F1/F2==(QB+QC)/(QC-QB), hence, qc/qb is 1.328. (I had a problem with my signs in the initial solution.)

 

[exitwound]

Why should I do that? I am not following the logic, I guess.

 

[ideasrule]

Why should I do that? I am not following the logic, I guess.

Both equations have "d", which you don't know but can eliminate by dividing the equations. After dividing, you have the ratio F1/F2, which you can calculate, as well as Qb and Qc. You'll have to rearrange to get an expression for the ratio Qb/Qc.

 

[exitwound]

Okay. 1.328 is right, and I did the simplification on paper here as well. Ends up looking like:

$$\frac{Q_c}{Q_b} = \frac {F_1+F_2}{F_1-F_2}$$

I don't know if I ever would have figured out to divide one by the other though.