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).​

Answer 1

$\blue {\sf{ \underline{ \underline{ANSWER}}}}$

Given :

• Mass of particle = 'm'
• Charge of particle = '-q'
• Electric Field = 'E'
• Speed = $\sf V_x$
• Length of the plate = 'L'

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

Now, Applying Newtons 2nd Law

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

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

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

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

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

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

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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,

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

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

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

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

Answer 2

Answer:

Mass of particle = 'm'

Charge of particle = '-q'

• Electric Field = 'E'

Speed = VX

Length of the plate = 'L'

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

Now, Applying Newtons 2nd Law

m a

a

F m

qE

a =

[F =qE]

(1)

Now, we have

time

.. t= Vx

Speed

Distance Speed

→ (2)

Distance

time

Now, In Vertical Direction, we have

Initial Velocity of the particle (u)

• Displacement(Vertically) = s

Time taken = t

Therefore, we have

1 s=ut + Eat? 2

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

s = (0×t) + i x () x (뉴)" 1

S 2 Ix (%) x (뉴)?

S = 2 qEL? 2mV ?

Hence, the Vertical Deflection is

= 0

qEL2

2mV, 2