- #1
msw1
- 5
- 2
- Homework Statement
- Two particles are located on the x axis of a Cartesian coordinate system. Particle 1 carries a charge of +3.0 nC and is at x = -30 mm and particle 2 carries a charge of -3.0 nC and is at x = 30 mm . Particle 3, which carries a charge of +5.0 μC , is located on the positive y axis 90 mm from the origin. What is the magnitude of the vector sum of the electric forces exerted on particle 3?
- Relevant Equations
- F = k(Q1Q2/r^2)
c = sqrt(a^2+b^2)
[tex]r_{13}=r_{23}=\sqrt{(30*10^{-3})^2+(90*10^{-3})^2}=\sqrt{9*10^{-3}}\\
F^E_{13}=F^E_{23}=9E9\cdot\frac{5*10^{-9}\cdot3*10^{-9}}{9*10^{-3}}=1.5*10^{-5}\\
\theta=tan^{-1}(\frac{90*10^{-3}}{30*10^{-3}})=71.565\,degrees\\
\vec{F}^E_{13}=<F^E_{13}cos\theta, F^E_{13}sin\theta> = <4.743*10^{-6}, 1.423*10^{-5}>\\
\vec{F}^E_{23}=<F^E_{23}cos(-\theta), F^E_{23}sin(-\theta)> = <4.743*10^{-6}, -1.423*10^{-5}>\\
[/tex]
Because particle 3 is positively charged, particle 1 is positively charged, and particle 2 is negatively charged, and the magnitude of the force by 1 on 3 is the same as magnitude of the force by 2 on 3, the vertical components of the force should cancel out, while the horizontal components should be additive, so my final answer was
[tex]F=2*(4.743*10^{-6})=9.487*10^{-6}[/tex]
But this is incorrect. Any help would be appreciated.
F^E_{13}=F^E_{23}=9E9\cdot\frac{5*10^{-9}\cdot3*10^{-9}}{9*10^{-3}}=1.5*10^{-5}\\
\theta=tan^{-1}(\frac{90*10^{-3}}{30*10^{-3}})=71.565\,degrees\\
\vec{F}^E_{13}=<F^E_{13}cos\theta, F^E_{13}sin\theta> = <4.743*10^{-6}, 1.423*10^{-5}>\\
\vec{F}^E_{23}=<F^E_{23}cos(-\theta), F^E_{23}sin(-\theta)> = <4.743*10^{-6}, -1.423*10^{-5}>\\
[/tex]
Because particle 3 is positively charged, particle 1 is positively charged, and particle 2 is negatively charged, and the magnitude of the force by 1 on 3 is the same as magnitude of the force by 2 on 3, the vertical components of the force should cancel out, while the horizontal components should be additive, so my final answer was
[tex]F=2*(4.743*10^{-6})=9.487*10^{-6}[/tex]
But this is incorrect. Any help would be appreciated.