Chapter 17, Problem 65GP

To determine

The angle with which the electron shall travel after it passes between the plates.

Answer to Problem 65GP

  $15.86°$

Explanation of Solution

Given:

The plates are $\text{6}\text{.5}\text{cm}$ long and $\text{1}\text{.3}\text{cm}$ apart with a potential difference of $\text{250}\text{V}$ and the potential at the horizontal plate is $\text{5500 V}$ .

Formula used:

The potential energy is given by

  $\text{PE = qV}$

Where,

  $\text{q}$ is the charge of an electron.

  $\text{V}$ is the voltage applied.

The kinetic energy is given by $\text{K}\text{.E=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Where,

  $\text{m}$ is the mass of an electron.

  $\text{v}$ is the velocity of the electron.

The time taken by the electron to travel the horizontal distance is $\text{t =}\frac{\text{x}}{\text{v}}$

Where,

  $\text{x}$ is the distance travelled.

The electric field acting on the charged particle is $\text{F = qE}$

Where,

  $\text{E}$ is the electric field applied.

According to Newton’s second law,

  $\text{F = ma}$

Where,

  $\text{a}$ is the acceleration of the particle.

The angle of the direction of the path of the electron is given by $\text{tanθ =}\frac{{\text{v}}_{\text{y}}}{{\text{v}}_{\text{x}}}$

  $\text{tan θ =}\frac{\text{E×x}}{\text{2V}}$

Calculation:

The horizontal component of the velocity will not change as the electron passed through the plates and the potential energy is equal to the kinetic energy. Now, let ${\text{v}}_{\text{y}}$ be the final velocity and ${\text{u}}_{\text{x}}$ be the initial velocity in the time period $\text{t}$ .So the acceleration is $\frac{{\text{v}}_{\text{y}}{\text{- u}}_{\text{x}}}{\text{t}}$ . Apply these values to find the velocity in y-direction with ${\text{u}}_{\text{x}}=0$ . Now apply the values to find $\text{tan}\text{θ}$ and then take the inverse to find $\text{θ}$ which is the required angle.

The electric field in the y direction is $\frac{\text{250}}{\text{0}\text{.013}}$

  $\text{=}\text{19230}\text{.77}\text{V/m}$

The angle $\text{θ}$ is ${\text{θ = tan}}^{\text{-1}}\left(\frac{\left(19230.77\times {10}^{-2}\right)\left(6.5\right)}{2\left(2200\right)}\right)$

  ${\text{θ =tan}}^{\text{-1}}\left(0.284\right)$

  $\text{θ=15}{\text{.86}}^0$

Conclusion:

The angle with which the electron shall travel after it passes between the plates.