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}.**