Chapter 17, Problem 53P

To determine

To Estimate: The value of electric field.

Answer to Problem 53P

  $E=2.89\times {10}^5\text{V/m}$

Explanation of Solution

Given:

The electrons are accelerated by $\text{7 kV}$ horizontally and then they pass from a uniform electric filed for a distance of $\text{2}\text{.8 cm}$ and then deflect upwards to $\text{22 cm}$ top screen and $\text{11 cm}$ above the centre.

Formula used:

The potential energy is given by

  $PE=eV$

Where,

e is the charge on electron.

V is the voltage applied.

The kinetic energy is given by

  $KE=\frac{1}{2}m{v}^2$

Where,

m is the mass of the electron.

v is the velocity of the electron.

The force is given as

  $F=eE$

  $F=ma$

Where,

a is the acceleration of the electron.

E is the electric field applied.

The time taken to travel the distance is

  $t=\frac{s}{v}$

Where,

s is the horizontal distance covered by the electron.

v is the speed of electron.

The vertical displacement of the electron is

  $d=v_{i}t+\frac{1}{2}a{t}^2$

Where,

v_i is the initial velocity of the electron

t is the time taken.

a is the acceleration.

d is the displacement of the electron.

Calculation:

The final velocity can be given as:

  $\begin{array}{l}{v}_f=\left(\frac{0.11\text{m}}{0.22\text{ m}}\right)v_i\\ v_f=\frac{1}{2}{v}_i\mathrm{...................................}(1)\end{array}$

Using the law of conservation of energy,

  $\begin{array}{l}mgh=\frac{1}{2}m{v}_i^2\\ 7\times {10}^3\text{V}=\frac{1}{2}m{v}_i^2\mathrm{..........................}(2)\end{array}$

Also,

  $\begin{array}{l}{p}_f=m{v}_f\\ Ft\cos {30}^0=m{v}_f\mathrm{............................}(3)\end{array}$

But,

  $F=\frac{dU}{dx}=E$

Hence,

  $\begin{array}{l}E\left(\frac{d}{v_i}\right)\cos {30}^0=m{v}_f\\ E\left(\frac{d}{v_i}\right)\cos {30}^0=m\left(\frac{1}{2}{v}_i\right)\\ E=\frac{\frac{1}{2}m{v}_i^2}{d\cos {30}^0}\end{array}$

Substituting all the values,

  $\begin{array}{l}E=\frac{7\times {10}^3\text{V}}{\left(0.028\text{ m}\right)\cos {30}^0}\\ E=2.89\times {10}^5\text{V/m}\end{array}$

Conclusion:

The required values is $E=2.89\times {10}^5\text{V/m}$