- #1
PFStudent
- 170
- 0
Homework Statement
13. In. Fig. 21-26, particle 1 of charge +1.0 [itex]\mu[/itex]C and particle 2 of charge -3.0 [itex]\mu[/itex]C are held at separation of L = 10.0 cm on an x axis. If particle 3 of known charge [itex]q_{3}[/itex] is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the
(a) x and
(b) y coordinates?
http://img329.imageshack.us/img329/3658/physicsj1bz9.jpg
Homework Equations
Coulomb's Law
Vector Form:
[tex]
\vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{r^2}\hat{r}_{12}
[/tex]
Scalar Form:
[tex]
|\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{r^2}
[/tex]
The Attempt at a Solution
[itex]q_{1} = +1.0 x 10^{-6} C[/itex]
[itex]q_{2} = -3.0 x 10^{-6} C[/itex]
[itex]L = 0.100 m[/itex]
[itex]x = ?[/itex]
[itex]y = ?[/tex]
There can only be two possible scenarios,
Scenario A
http://img266.imageshack.us/img266/8128/physicsjabc5.jpg
and
Scenario B
http://img266.imageshack.us/img266/1875/physicsjbuh1.jpg
The reason that charge [itex]q_{3}[/itex] can not be between charges [itex]q_{1}[/itex] and [itex]q_{2}[/itex] is because the forces would not balance each other but instead would add vectorally in the same direction, and therefore the net force on charge [itex]q_{3}[/itex] would not be zero.
Beginning with scenario A,
[tex]
|\vec{F_{31}}| = |\vec{F_{32}}|
[/tex]
[itex]r_{31} = L+x[/itex] and [itex]r_{32} = x[/itex]
[tex]
\frac{k_{e}|q_{3}||q_{1}|}{r_{31}^2} = \frac{k_{e}|q_{3}||q_{2}|}{r_{32}^2}
[/tex]
[tex]
\frac{|q_{1}|}{(L+x)^2} = \frac{|q_{2}|}{(x)^2}
[/tex]
Solving for x,
[tex]
x = \frac{-L}{1\mp\sqrt{\frac{|q_{1}|}{|q_{2}|}}}
[/tex]
And evaluating for x with [itex]sig. fig. \equiv 2[/itex],
x = -0.24 m, -0.06 m
Continuing with scenario B,
[tex]
|\vec{F_{31}}| = |\vec{F_{32}}|
[/tex]
[itex]r_{31} = x[/itex] and [itex]r_{32} = L+x[/itex]
[tex]
\frac{k_{e}|q_{3}||q_{1}|}{r_{31}^2} = \frac{k_{e}|q_{3}||q_{2}|}{r_{32}^2}
[/tex]
[tex]
\frac{|q_{1}|}{(x)^2} = \frac{|q_{2}|}{(L+x)^2}
[/tex]
Solving for x,
[tex]
x = \frac{\pm L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{1\mp\sqrt{\frac{|q_{1}|}{|q_{2}|}}}
[/tex]
Now simplifying by factoring out a [itex]\mp[/itex] out of the denominator,
[tex]
x = \frac{\pm L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{\mp\left(\mp1+\sqrt{\frac{|q_{1}|}{|q_{2}|}}\right)}
[/tex]
Now noting that the following is true,
[tex]
\frac{\pm}{\mp} = \frac{+}{-} or \frac{-}{+} = -
[/tex]
Our equation simplifies to
[tex]
x = \frac{-L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{\left(\mp1+\sqrt{\frac{|q_{1}|}{|q_{2}|}}\right)}
[/tex]
[tex]
x = \frac{-L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{\sqrt{\frac{|q_{1}|}{|q_{2}|}}\mp1}
[/tex]
And evaluating for x with [itex]sig. fig. \equiv 2[/itex],
x = +0.14 m, -0.04 m
Therefore, of scenarios A and B there are four possible answers,
A: x = -0.24 m, -0.06 m
B: x = +0.14 m, -0.04 m
The book lists, x = 0.14 m as the answer, but that can’t be the only answer as that only fits with scenario B.
Then, if x= 0.14 m is the only answer what justifies it to be so out of all the other solutions, is it because it is the only positive solution?
What is the other answer for scenario A?
And the answer for scenario B (x = 0.14m) seems to fit more with scenario A, then why did I not get this answer while solving for solutions in scenario A?
Are the other solutions for x real solutions to this problem?
Thanks,
-PFStudent
Last edited by a moderator: