Question

A particle of mass $m$ and charge $-q$ enters the region between the two charged plates initially moving along x-axis with speed $v_x$ as shown in figure. The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. The vertical deflection of the particle at the far edge of the plate is

• 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

The correct option is A $\frac{qE{L}^2}{2m{v}_x^2}$

Here, in the vertical direction,

Initial velocity, $v=0$

acceleration, $a=\frac{F}{m}=\frac{qE}{m}(\because F=qE)...\left(i\right)$

Time taken to cross the field, $t=\frac{distance}{velocity}=\frac{L}{v_x}....\left(ii\right)$

($\because$ velocity the horizontal direction is constant)

$\because s=vt+\frac{1}{2}a{t}^2$

$\therefore$ deflection

$y=0+\frac{1}{2}\left(\frac{qE}{m}\right){(\frac{L}{v_x}.)}^2$

[Using (i) and (ii)]

$\therefore y=\frac{qE{L}^2}{2m{v}_x^2}...\left(iii\right)$