Q53P - Fundamentals Of Physics

Question

Two large parallel copper plates are $\mathbf{5}\mathbf{.0}\mathbf{ }\mathbf{\text{cm}}$ apart and have a uniform electric field between them as depicted in Fig. 22-60. An electronis released from the negative plate at the same time that a proton is released from the positive plate. Neglect the force of the particles on each other and find their distance from the positive plate when they pass each other. (Does it surprise you that you need not know the electric field to solve this problem?)

 

Step-by-Step Solution

Answer

Their distance from the positive plate is.$2.7 \times {10}^{-5}\text{m}$

Step 1: The given data

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

Step 2: Understanding the concept of electrostatic force

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 t is and $\mathbf{x}=(\mathbf{1}/\mathbf{2}){\mathbf{a}}_p{\mathbf{t}}^2$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)

Step 3: Calculation of the distance of electron and proton from the positive plate

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

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

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