Question

Two parallel plates$1.0cm$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?

Short answer

The negavie plate is$d\approx 1\mathrm{cm}$

Step-by-step solution

Step 1:Introduction

Equation gives the force between the panels.:$ma=qE$

For electron

$a_1=\frac{qE}{m_e}$

For proton

$a_2=\frac{qE}{m_p}$

If distance $d$travel by electron of time $t$when meet the proton:

$d=ut+\frac{1}{2}a{t}^2$

Substitute $u=0$

$d=\frac{1}{2}{a}_1{t}^2$.........$\left(1\right)$

At same time $t$distance travel by proton when meet the electron:

$(1-d)=\frac{1}{2}{a}_2{t}^2$.......$\left(2\right)$

Substitution

From $1$and $2$, we find next

$\to {10}^{-2}\times 1=\frac{1}{2}\left(a_1+a_2\right)t^2$

$\to t^2=\frac{2\times {10}^{-2}}{\left(a_1+a_2\right)}$

Now, value for $t^2$, substitute in the equation for $d$.

$d=\frac{1}{2}{a}_1{t}^2$

$=\frac{1}{2}·\frac{a_1\times 2\times {10}^{-2}}{\left(a_1+a_2\right)}$

$=\frac{1}{2}·\frac{2\times \frac{qE}{m_e}\times {10}^{-2}}{\left[\frac{qE}{m_e}+\frac{qE}{m_p}\right]}$

$=\frac{{10}^{-2}\times m_p}{m_e+m_p}$

$=\frac{1.67\times {10}^{-27}\times {10}^{-2}}{\left[9.11\times {10}^{-31}+1.67\times {10}^{-27}\right]}$

$=0.99\approx 1\mathrm{cm}$

value of electron and proton

The electron and proton $\approx 1cm$pass each other.