Question

Two parallel charged plates are placed parallel to $x$ -axis. A particle of mass $m$ and charge $-q$ enters the region between the plates along the axis with a speed $v_x$ as depicted in the adjoining figure. If the length of the plates is $L$ and a uniform electric field $E$ is maintained between the plates, then find the vertical deflection of the particle at the far edge of the plate.

• A. $\frac{qE{L}^2}{2m{v}_x^2}$
• B. $\frac{qE{L}_{}^2}{2m{v}_x}$
• C. $\frac{2m{v}_x^2}{qE{L}^2}$
• D. $\frac{2m{v}_x}{q{E}^{2}L}$

Answer 1

Answer: (A)

Here, in the vertical direction, initial velocity, $v=0$
acceleration, $a=\frac{F}{m}=\frac{q^E}{m}\left(\because F={\left(qE\right)}^{}\right)$
Time taken to cross the field, $t=\frac{distance}{velocity}=\frac{L}{v_x}$
( $\because$ Velocity along the horizontal direction is constant)
$\because s=vt+\frac{1}{2}a{t}^2$
$\because$ Deflection,
$y=0+\frac{1}{2}\left(\frac{q^E}{m}\right){\left(\frac{L}{v_x}\right)}^2$
[Using (i) and (ii)]
$\therefore y=\frac{qE{L}^2}{2m{v}_x^2}$