Answer:
The distance from the positive plate at which the two pass each other is 0.0023 cm.
Explanation:
We need to find the acceleration of each particle first. Let's use the electric force equation.
[tex]F=Eq[/tex]
[tex]ma=Eq[/tex]
For the proton
[tex]m_{p}a_{p}=Eq_{p}[/tex]
[tex]a_{p}=\frac{Eq_{p}}{m_{p}}[/tex]
[tex]a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}[/tex]
[tex]a_{p}=6.19*10^{10}\: m/s^{2}[/tex]
For the electron
[tex]m_{e}a_{e}=Eq_{e}[/tex]
[tex]a_{e}=\frac{Eq_{e}}{m_{e}}[/tex]
[tex]a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}[/tex]
[tex]a_{e}=1.14*10^{14}\: m/s^{2}[/tex]
Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.
[tex]x_{p}+x_{e}=0.0426[/tex]
Both of them have an initial speed equal to zero. So we have:
[tex]\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426[/tex]
[tex]t^{2}(a_{p}+a_{e})=2*0.0426[/tex]
[tex]t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}[/tex]
[tex]t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}[/tex]
[tex]t=2.73*10^{-8}\: s[/tex]
With this time we can find the distance from the positive plate (x(p)).
[tex]x_{p}=\frac{1}{2}a_{p}t^{2}[/tex]
[tex]x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}[/tex]
[tex]x_{p}=0.0023\: cm[/tex]
Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.
I hope it helps you!
