A charged particle of mass m and charge q is released from rest in uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after t second is

Question

A charged particle of mass m and charge q is released from rest in uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after t second is (A) (E q2 M/2 t2) (B) (2 E2 t2/m q) (C) (E2 q2 t 2/2 m) (D)  ( E q m/ t )

Tardigrade Admin · Accepted Answer

Force on particle = F = qE Hence, acceleration of particel = a =( F / m )=( qE / m ) Initial speed = u =0 Hence, final velocity = v = u + at =( qEt / m ) Kinetic energy = K =(1/2) mv 2=(1/2) m (( qEt / m ))2 Longrightarrow K=(E2 q2 t2/2 m)

Q. A charged particle of mass $m$ and charge $q$ is released from rest in uniform electric field $E$ . Neglecting the effect of gravity, the kinetic energy of the charged particle after t second is

$\frac{E q ^{2} M}{2 t ^{2}}$

$\frac{2 E ^{2} t ^{2}}{m q}$

$\frac{E ^{2} q ^{2} t ^{2}}{2 m}$

$\frac{E q m}{t}$

Force on particle $= F = qE$
Hence, acceleration of particel $= a = \frac{F}{m} = \frac{qE}{m}$
Initial speed $= u = 0$
Hence, final velocity $= v = u + a t = \frac{qEt}{m}$
Kinetic energy $= K = \frac{1}{2} m v^{2} = \frac{1}{2} m (\frac{qEt}{m})^{2}$
$⟹ K = \frac{E ^{2} q ^{2} t ^{2}}{2 m}$

Q. A charged particle of mass m and charge q is released from rest in uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after t second is

2t2Eq2M​

mq2E2t2​

2mE2q2t2​

tEqm​