Two parallel plates 0.500 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.
How far from the negative plate is the point at which the electron and proton pass each other?
Answer
Expression for acceleration of charged particle is,
a=\frac{q E}{m}
The expression for acceleration of electron is,
a_{e}=\frac{q E}{m_{e}}
Here, a_{e} is the acceleration of electron and m_{e} is the mass of electron.
The expression for acceleration of proton is,
a_{p}=\frac{q E}{m_{p}}
Here, a_{p} is the acceleration of proton and m_{p} is the mass of proton.
Assume that d be the distance of electron from the negative plate.
Expression for the distance is,
s=u t+\frac{1}{2} a t^{2}
Sulostitute d for s, 0 for u, q E / m_{e} for a.
d=\left(\frac{1}{2}\right)\left(\frac{q E}{m_{e}}\right) t^{2} (1)
Then the distance of proton from the positive plate is 0.500 \mathrm{~cm}-d.
Expression for the distance is, s=u t+\frac{1}{2} a t^{2}
Substitute 0.500 \mathrm{~cm}-d for s, 0 for u, q E / m_{p} for a
0.500 \mathrm{~cm}-d=\left(\frac{1}{2}\right)\left(\frac{q E}{m_{p}}\right) t^{2} (2)
Divide equation (1) by equation (2)
\frac{d}{0.500 \mathrm{~cm}-d}=\frac{m_{p}}{m_{e}}
Substitute 1836 m_{e} for m_{p}
\frac{d}{0.500 \mathrm{~cm}-d}=\frac{1836 m_{e}}{m_{e}}
\frac{d}{0.500 \mathrm{~cm}-d}=1836
Rearrange above equation for d
d=0.4997 \mathrm{~cm}
\approx 0.500 \mathrm{~cm}
Therefore, the distance from the negative plate at which the electron and proton pass each other is 0.500 \mathrm{~cm}.
