Question

A charged particle of mass m and charge q is released from
rest in an electric field of constant magnitude E. The kinetic
energy of the particle after time t is
​

Answer 1

Answer:

Force experienced by the charged particle = Eq

Therefore acceleration of the charged particle = Force/mass =( Eq)/m

Velocity of particle after time "t" be V

V = u +at

=>V = 0 + (Eq/m)*t

=> V = {(E*q*t)/m}

The initial velocity is considered zero

(i.e "u" = 0).

Now kinetic energy = 1/2 mV^2

=> KE = 1/2 *m *[{E*q*t/m}^2]

=> KE = 1/2* [{(E*q*t)^2}/m]

So the final answer is

1/2*[{(E*q*t)^2}/m]

Answer 2

AnsweR

1 / 2 × ( E × Q × T )² / m

ExplanatioN

GiveN:-

1) Charged particle of mass m and charge q is released from rest in an electric field of constant magnitude.

2) Let us find what is the kinetic energy of the particle after time t.

HencE

Eq is the force experienced by the particles.

1. Let A be the acceleration.
2. Let the charged partices here be "CP".
3. Let F be the force.
4. Let M be the mass.

HencE, A of CP = F / M = EQ / M

Let velocity of particle of time be "T" which be "V" ( V = Velocity ).

$\sf V = U + at \\ \sf V = 0 + ( \frac{Eq}{m})t \\ \sf V = (e \times q \times T) \: by \: M$

HencE,

zero ( 0 ) = initial Velocity.

Then..

1 / 2 MV²

= KE = 1 / 2 × m × ( E × Q × T ) / M ) ² = KE = 1 / 2 = 1 / 2 × ( E × Q × T² ) / m

= 1 / 2 × ( E × Q × T )² / m

Hope it helps!

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