Chapter 23: Q. 25 (page 654)

Two parallel plates $1.0 c m$ 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

Expert verified

The negavie plate is $d \approx 1 cm$

Step by step solution

Step 1:Introduction

Equation gives the force between the panels.: $m a = q E$

For electron

$a_{1} = \frac{q E}{m_{e}}$

For proton

$a_{2} = \frac{q E}{m_{p}}$

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

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

Substitute $u = 0$

$d = \frac{1}{2} a_{1} t^{2}$ ......... $(1)$

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

$(1 - d) = \frac{1}{2} a_{2} t^{2}$ ....... $(2)$

Substitution

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

$\to 10^{- 2} \times 1 = \frac{1}{2} (a_{1} + a_{2}) t^{2}$

$\to t^{2} = \frac{2 \times 10^{- 2}}{(a_{1} + a_{2})}$

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

$d = \frac{1}{2} a_{1} t^{2}$

$= \frac{1}{2} \cdot \frac{a_{1} \times 2 \times 10^{- 2}}{(a_{1} + a_{2})}$

$= \frac{1}{2} \cdot \frac{2 \times \frac{q E}{m_{e}} \times 10^{- 2}}{[\frac{q E}{m_{e}} + \frac{q E}{m_{p}}]}$

$= \frac{10^{- 2} \times m_{p}}{m_{e} + m_{p}}$

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

$= 0.99 \approx 1 cm$

value of electron and proton

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

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Short Answer

Step by step solution

Step 1:Introduction

Substitution

value of electron and proton

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