- #1
PFStudent
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Homework Statement
38. Particles 2 and 4, of charge [itex]-e[/itex], are fixed in place on the y axis, at [itex]{y}_{2}[/itex] = -10.0 cm and [itex]{y}_{4}[/itex] = 5.00 cm. Particles 1 and 3, of charge [itex]-e[/itex], are placed on the x-axis and can be moved along the x axis. Particle 5, of charge [itex]+e[/itex], is fixed at the origin. Initially particle 1 is at [itex]{x}_{1}[/itex] = -10.0 cm and particle 3 at [itex]{x}_{3}[/itex] = 10.0 cm.
(a) To what x value must particle 1 be moved to rotate the direction of the net electrostatic force [itex]\vec{F}_{net}[/itex] on particle 5 by 30 degrees counterclockwise?
(b) With particle 1 fixed at its new position, to what x value must you move particle 3 to rotate [itex]\vec{F}_{net}[/itex] back to its original direction?
Homework Equations
Coulomb's Law
Vector Form:
[tex]
\vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}
[/tex]
Scalar Form:
[tex]
{F}_{12} = \frac{k_{e}{q_{1}}{q_{2}}}{{r_{12}}^2}
[/tex]
Magnitude Form:
[tex]
|\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{12}}^2}
[/tex]
Superposition of Forces
[tex]
\sum{\vec{F}_{m}} = {\sum_{i=1}^{n}}{\vec{F}_{mn}}
[/tex]
The Attempt at a Solution
So, there are five particles: [itex]{q}_{1}[/itex] and [itex]{q}_{3}[/itex] placed on the x-axis and [itex]{q}_{2}[/itex] and [itex]{q}_{4}[/itex] placed on the y-axis and [itex]{q}_{5}[/itex] placed at the origin.
I note that there are two unknowns: the net force on particle 5, [itex]\sum\vec{F}_{5}[/itex] and the distance, [itex]{r}_{51}[/itex].
Let,
[tex]
{r}_{51} = x
[/tex]
What I did was setup the [itex]x[/itex] and [itex]y[/itex] components for the net force and then solve each of the them for the (magnitude of the) net force and set them equal to each other and then solved for x.
[tex]
\sum\vec{F}_{5_{x}} = {{F}_{{51}_{x}}}{\hat{i}} + {{F}_{{53}_{x}}}{\hat{i}}
[/tex]
[tex]
{\sum}{\vec{F}_{5_{x}}} = {{F}_{51}}{cos{\theta_{51}}}{\hat{i}} + {{F}_{53}}{cos{\theta_{53}}}{\hat{i}}
[/tex]
[tex]
{\sum}{\vec{F}_{5_{x}}} = {\frac{{k_{e}}{q_{5}}{q_{1}}}{{{r}_{51}}^{2}}}{cos{\theta_{51}}}{\hat{i}} + {\frac{{k_{e}}{q_{5}}{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}{\hat{i}}
[/tex]
[tex]
{\sum}{{F}_{5}}{cos{\theta}}{\hat{i}} = {\frac{{k_{e}}{q_{5}}{q_{1}}}{{{r}_{51}}^{2}}}{cos{\theta_{51}}}{\hat{i}} + {\frac{{k_{e}}{q_{5}}{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}{\hat{i}}
[/tex]
[itex]{\theta}[/itex] = 30 degrees
[tex]
{\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{{r}_{51}}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]
[/tex]
[tex]
{\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{\left(x\right)}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]
[/tex]
[tex]
{\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{x}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]
[/tex]
Doing the same for the [itex]y[/itex] components leads to,
[tex]
\sum\vec{F}_{5_{y}} = {{F}_{{52}_{y}}}{\hat{j}} + {{F}_{{53}_{4}}}{\hat{j}}
[/tex]
[tex]
{\sum}{{F}_{5}}{sin{\theta}}{\hat{j}} = {\frac{{k_{e}}{q_{5}}{q_{2}}}{{{r}_{52}}^{2}}}{sin{\theta_{52}}}{\hat{j}} + {\frac{{k_{e}}{q_{5}}{q_{4}}}{{{r}_{54}}^{2}}}{sin{\theta_{54}}}{\hat{j}}
[/tex]
[itex]{\theta}[/itex] = 30 degrees
[tex]
{\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{sin}{\theta}}}\left[{\frac{{q_{2}}}{{{r}_{52}}^{2}}}{sin{\theta_{52}}} + {\frac{{q_{4}}}{{{r}_{54}}^{2}}}{sin{\theta_{54}}}\right]
[/tex]
Letting,
[tex]
{\sum}{{F}_{5}} = {\sum}{{F}_{5}}
[/tex]
[tex]
\left({\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{x}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]\right) = \left({\frac{{{k}_{e}}{{q}_{5}}}{{sin}{\theta}}}\left[{\frac{{q_{2}}}{{{r}_{52}}^{2}}}{sin{\theta_{52}}} + {\frac{{q_{4}}}{{{r}_{54}}^{2}}}{sin{\theta_{54}}}\right]\right)
[/tex]
Solving for x and simplifying leads to,
[tex]
{x} = {\sqrt{\frac{{{q}_{1}}}{{\frac{{cot}{\theta}}{{cos}{\theta}_{51}}\left[\frac{{q}_{2}}{{{r}_{52}}^{2}}{{sin}{\theta}_{52}} + \frac{{q}_{4}}{{{r}_{54}}^{2}}{{sin}{\theta}_{54}}\right] - {\frac{{q}_{3}}{{{r}_{53}}^{2}}{{cos}{\theta}_{53}}}}}}}
[/tex]
sig. fig. [itex]\equiv[/itex] 3
Where,
[itex]{\theta}_{}[/itex] = 30 degrees
[itex]{\theta}_{51}[/itex] = 180 degrees
[itex]{\theta}_{52}[/itex] = 270 degrees
[itex]{\theta}_{53}[/itex] = 0 degrees
[itex]{\theta}_{54}[/itex] = 90 degrees
Using the above equation I get,
x = [itex]\sqrt{-0.0016139...}[/itex]
Which is ofcourse wrong because it is imaginary.
So, how do I solve this because I thought for sure my approach was right and I don't think I made any mistakes.
Any help is appreciated.
Thanks,
-PFStudent
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