A particle of mass m carrying charge q is released from rest in uniform electric field of intensity E. The kinetic energy acquired by the particle after moving a distance of x is (neglect gravitational force)

Question

A particle of mass m carrying charge q is released from rest in uniform electric field of intensity E. The kinetic energy acquired by the particle after moving a distance of x is (neglect gravitational force) (A) q E x (B)  q E x2  (C)  q E2 x  (D)  q2 E x

Tardigrade Admin · Accepted Answer

Velocity acquired by particle in electric field

v^{2} = u^{2} + 2 a s

or

v^{2} = 0 + 2 a x

v^{2} = 2 a x

Hence, kinetic energy of particle
KE =

\frac{1}{2} m v^{2}

=

\frac{1}{2} m (2 a x)

= amx
but F = q E
or ma = qE
or a =

\frac{qE}{m}

Hence, KE =

\frac{qE}{m} \times m x = q E x

Q. A particle of mass m carrying charge q is released from rest in uniform electric field of intensity E. The kinetic energy acquired by the particle after moving a distance of x is (neglect gravitational force)

$q E x$

$q E x^{2}$

$q E^{2} x$

$q^{2} E x$

$q^{2} E^{2} x$

Velocity acquired by particle in electric field
$v^{2} = u^{2} + 2 a s$
or $v^{2} = 0 + 2 a x$
$v^{2} = 2 a x$
Hence, kinetic energy of particle
KE = $\frac{1}{2} m v^{2}$
= $\frac{1}{2} m (2 a x)$
= amx
but F = q E
or ma = qE
or a = $\frac{qE}{m}$
Hence, KE = $\frac{qE}{m} \times m x = q E x$

Q. A particle of mass m carrying charge q is released from rest in uniform electric field of intensity E. The kinetic energy acquired by the particle after moving a distance of x is (neglect gravitational force)

qEx

qEx2

qE2x

q2Ex

q2E2x