Two long parallel plates A and B are separated by a distance of 4   cm with an electric field of 45 . 5 Vm - 1 between the plates directed normally from plate A to plate B , as shown in the figure. An electron is projected from plate A with velocity v at an angle of 30 ° with the surface…

Question

Two long parallel plates A and B are separated by a distance of 4 &#160; cm with an electric field of 45 . 5 Vm - 1 between the plates directed normally from plate A to plate B , as shown in the figure. An electron is projected from plate A with velocity v at an angle of 30 &#176; with the surface of plate A . The maximum value of v so that the electron does not hit plate B is ( Assume gravity free space, charge of electron = 1 . 6 &#215; 10 - 19 C and mass of electron = 9 . 1 &#215; 10 - 31 &#160; kg

Aicharya · Accepted Answer

The figure below represents the electron between two parallel
plates $A$ and $B$ ....

The expression for the maximum height is,

$h_{\max} = \frac{v^{2} \sin^{2} 30 °}{2 a}$

$\frac{4}{100} = \frac{v^{2} \sin^{2} 30 °}{2 a} . . . (1)$

The acceleration of the electron is,

$F = q_{e} E$

$m a = q_{e} E$

$a = \frac{q_{e} E}{m}$

Substitute the values in equation $(1)$ .

$\frac{4}{100} = \frac{v^{2} \sin^{2} 30 °}{2 (\frac{q_{e} E}{m})}$

$\frac{4}{100} = \frac{v^{2} (\frac{1}{4})}{2 (\frac{q_{e} E}{m})}$

Rewrite the above equation.

$v = \sqrt{\frac{4 \times 8}{100} (\frac{q_{e} E}{m_{e}})}$

$v = \sqrt{\frac{4 \times 8}{100} \frac{1.6 \times 10^{- 19} \times 45.5}{9.1 \times 10^{- 31}}}$

$v = 1600 km s^{- 1}$

Two long parallel plates A and B are separated by a distance of 4   cm with an electric field of 45 . 5…

Solution

Two long parallel plates A and B are separated by a distance of 4 &#160; cm with an electric field of 45 . 5…

Solution

Two long parallel plates A and B are separated by a distance of 4 cm with an electric field of 45 . 5…