91Ó°ÊÓ

Separation of two copper plates,$L=0.05\text{m}$

We take the positive direction to be to the right in the figure. The acceleration of the proton is${\mathbf{a}}_p=\mathbf{eE}/{\mathbf{m}}_p$and the acceleration of the electron is,${\mathbf{a}}_e=–\mathbf{eE}/{\mathbf{m}}_e$where,E is the magnitude of the electric field,${\mathbf{m}}_p$is the mass of the proton, and${\mathbf{m}}_e$is the mass of the electron. We take the origin to be at the initial position of the proton. Then, the coordinate of the proton at time tisand $\mathbf{x}=(\mathbf{1}/\mathbf{2}){\mathbf{a}}_p{\mathbf{t}}^2$the coordinate of the electron is.$\mathbf{x}=\mathbf{L}+(\mathbf{1}/\mathbf{2}){\mathbf{a}}_e{\mathbf{t}}^2$They pass each other when their coordinates are the same, or

$\frac{1}{2}{a}_p{t}^2=L+\frac{1}{2}{a}_e{t}^2$

This means${\mathbf{t}}^2=\mathbf{2}\mathbf{L}/({\mathbf{a}}_p–{\mathbf{a}}_e)$and

$\begin{array}{c}x=\frac{a_p}{a_p−a_e}L\\ =\frac{eE/m_p}{(eE/m_p)+(eE/m_e)}L\\ =\left(\frac{m_e}{m_e+m_p}\right)L\end{array}$

Formula:

The distance of proton and electron from the positive copper plate,

$x=\left(\frac{m_e}{m_e+m_p}\right)L$ (i)

Using the given data in equation (i), we can get the value of the required distance as follows:

$\begin{array}{c}x=\left(\frac{\text{9.11}×{\text{10}}^{\text{-31}}\text{kg}}{\text{9.11}×{\text{10}}^{\text{-31}}\text{kg}+\text{1.67}×{\text{10}}^{\text{-27}}\text{kg}}\right)\left(\text{0.050m}\right)\\ =2.7×{10}^{−5}\text{m}\end{array}$

Hence, the value of the distance is.$2.7×{10}^{−5}\text{m}$