Chapter 17, Problem 54P

To determine

The range of deflection by the horizontal plate.

Answer to Problem 54P

  $\pm {\text{2 × 10}}^{\text{5}}\text{V/m }$

Explanation of Solution

Given:

The width of the screen is $\text{30}\text{cm}$ and is $\text{34}\text{cm}$ from the $\text{2}\text{.6}\text{cm}$ deflection plate and the electrons are accelerated by $\text{6}\text{.0}\text{kV}$ .

Formula used:

The distance from the center of the screen to the bright spot on the screen is

  $\text{y}\text{=}\frac{\text{d}}{\text{2}}$

Where,

  $\text{d}$ is the width of the screen.

The law of conservation of energy

  ${\text{K}}_{\text{i}}\text{+}{\text{U}}_{\text{i}}\text{=}\text{Constant}$

Where,

  ${\text{K}}_{\text{i}}$ is the kinetic energy.

  ${\text{U}}_{\text{i}}$ is the potential energy.

The velocity of the electron is $\text{v}\text{=}\sqrt{\frac{\text{2eV}}{\text{m}}}$

Where,

  $\text{e}$ is the charge of an electron. i.e. $\text{1}{\text{.6×10}}^{\text{-19}}\text{C}$ .

  $\text{V}$ is the voltage applied.

  $\text{m}$ is the mass of electron i.e. $\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}$ .

Time taken to travel a distance of $\text{L}$ meters by an electron in the electric field is $\text{t}\text{=}\frac{\text{L}}{\text{v}}$

The electric force acting on the electron along the field $\left(\text{E}\right)$ is

  $\text{F}\text{=}\text{eE}$

As per Newton’s second law:

  $\text{F}\text{=}\text{ma}$

The acceleration, $\text{a}\text{=}\frac{\text{eE}}{\text{m}}$

The velocity of the deflected electron is $\text{v}\text{=}\frac{\text{eEt}}{\text{m}}$

Where, $\text{t}$ is the time period

Calculation:

The distance from the center of the screen to the bright spot on screen where the electron hit is

  $\text{y=}\frac{\text{30}}{\text{2}}\text{cm}$

  $\text{y=15}\text{cm}$

  $\text{y=0}\text{.15}\text{m}$

The velocity of the electron is

  $\text{v}=\sqrt{\frac{\left(2\right)\left(1.6\times {10}^{-19}\right)\left(6\times {10}^3\right)}{\left(9.11\times {10}^{-31}\right)}}$

  $=\sqrt{\frac{1.92\times {10}^{-15}}{9.11\times {10}^{-31}}}$

  $=\sqrt{2.1075\times {10}^{15}}$

  $=4.59\times {10}^7$

The time taken to travel a distance of $\text{0}\text{.026 m}$ is

  $=\frac{\text{0}\text{.026}}{\text{4}{\text{.59×10}}^{\text{7}}}$

  $\text{=}\text{5}{\text{.66×10}}^{\text{-10}}\text{s}$

The time taken to travel a distance of $\text{0}\text{.34 m}$ is

  $=\frac{\text{0}\text{.34}}{\text{4}{\text{.59×10}}^{\text{7}}}$

  $\text{=}\text{7}{\text{.40×10}}^{\text{-9}}\text{s}$

The electric field is given by

  $=\frac{\left(\text{9}{\text{.11×10}}^{\text{-31}}\right)\left(\text{0}\text{.15}\right)}{\left(\text{1}{\text{.6×10}}^{\text{-19}}\right)\left(\text{5}{\text{.66×10}}^{\text{-10}}\right)\left(\text{7}{\text{.40×10}}^{\text{-9}}\right)}$

  $=\frac{\text{1}{\text{.3665×10}}^{\text{-31}}}{\text{6}{\text{.7014×10}}^{\text{-37}}}$

  ${\text{= 2×10}}^{\text{5}}\text{V/m}$

The range is $\pm {\text{2×10}}^{\text{5}}\text{V/m}$

Conclusion:

The required value is $\pm {\text{2 × 10}}^{\text{5}}\text{V/m }$ .