Question

A uniform electric field ${}^{\prime }{E}^{\prime }$ exist between two parallel plates having opposite charge on each plate. An electron enters the field between the plates with speed $v_0$ as shown in the figure. If length of each plates is$1$ then angle with which electrons deviates from its initial path is (m is the mass of electrons)

Answer 1

Hint: The presence of an electric charge causes an alteration in space known as the electric field. The electric force between a source charge and a test charge is mediated by the electric field. The field is a vector, so it points away from positive charges and toward negative charges by definition.

Complete step-by-step solution:
A continuous charge delivery system is one in which the charge is distributed evenly through the conductor. In a continuous charge device, an infinite number of charges are packed tightly together with just a small gap between them.
Now, coming to the given question;
Uniform Electric field = $E$
Speed of electron entering = $v_0$
Length of each plate =$l$
Mass of electron =$m$

The acceleration of electron is = ${\displaystyle \frac{eE}{m}}$
The horizontal velocity remains $v_0$ as there is no acceleration in this direction.
Thus, the time taken in crossing field $t={\displaystyle \frac{l}{v_0}}$
The upward component of velocity of electron as it emerges from the field region is
$\Rightarrow V_y=\alpha t={\displaystyle \frac{eEl}{m{v}_0}}$
The horizontal component of velocity remains,
$v_y=v_0$
The angle =$0$, made by the resultant velocity with the original direction is given by,
$\tan \theta ={\displaystyle \frac{v_y}{v_x}}={\displaystyle \frac{eEl}{m{v}_{0}x{v}_0}}$
Thus, the electron deviates by an angle $\theta ={\tan }^{-1}\left({\displaystyle \frac{eEl}{m{v_0}^2}}\right)$

 Note:Positive or negative charges pass through the electric field lines. Surface applications such as wound healing, corneal repair, and even brain and spinal stimulation with closely spaced, implanted electrodes are well suited for such sources.