Question

kindly help me.......​

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:

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

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

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

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

_______________________________________

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

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

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

_______________________________________

Now, In Vertical Direction, we have

Initial Velocity of the particle (u) = 0

Displacement(Vertically) = s

Time taken = t

Therefore, we have

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

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

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

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

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

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