Question

Question 33:
A particle of mass m and charge (-9) enters the region between the two charged plates
initially moving along x-axis with speed VX (like particle 1 in Fig. 1.33). The length of plate is
L and a uniform electric field E is maintained between the plates. Show that the vertical
deflection of the particle at the far edge of the plate is qEĽ (2m2).
Compare this motion with motion of a projectile in gravitational field discussed in Section
4.10 of Class XI Textbook of Physics.​

Answer 1

Answer:

8.2 hope it is helpful ........I am not sure if it is wrong please explain

Answer 2

QuestioN :-

• A particle of mass 'm' and charge (-q) enters the region between the two charged plates initially moving along with x axis with speed Vx. The lenght of the plate is L and an uniform electric field E is maintained betwn. the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEl^2/(2mVx^2).

GiveN :-

• Mass of particle = 'm'

• Charge of particle = '-q'

• Electric Field = 'E'

• Speed = \sf V_xVx

• Length of the plate = 'L'

As, the Charged particle is moving with Speed(Vx) on an uniform electric field.

Now, Applying Newtons 2nd Law

$\green{ \boxed { \sf \: F = m \: a}}$

$\implies \: \sf \: a \: = \: \frac{F}{m}$

$\implies \: \sf \: a \: = \: \frac{qE}{m} \: \: \: \:[F = qE] \: \: \: \longrightarrow \: (1)$

_________________________________

$\sf \: Now, \: we \: have \: \: \: \boxed{ \purple{ \sf \: Speed = \: \frac{Distance}{time}}}$

$\implies \sf \: \: time = \frac{Distance}{Speed}$

$\therefore \: \sf \: t = \frac{L} { \: \: V_x} \: \: \: \: \: \: \: \longrightarrow \: (2)$

_________________________________

Now, In Vertical Direction, we have

• Initial Velocity of the particle (u) = 0

• Displacement (Vertically) = s

• Time taken = t

Therefore, we have

$\boxed{ \green { \sf \: s = ut + \frac{1}{2} a {t}^{2} }}$

On putting the values of 'a' and 't' from Equation (1) and (2) we get,

$\implies \sf \: s \: = \: (0 \times t)\: + \: \frac{1}{2} \times ( \frac{qE} {m}) \times{(\frac{L} {\: \: V_x})}^{2}$

$\implies \sf \: s \: = \frac{1}{2} \times ( \frac{qE} {m}) \times{(\frac{L} {\: \: V_x})}^{2}$

$\implies \sf \: s \: = \: \frac{qE{L}^{2}}{ \: 2m{V_x}^{2} }$

$\boxed { \sf \: Hence, \: the \: Vertical \: De flection \: is \: \frac{qE{L}^{2}}{2m{V_x}^{2}}}$