Question

A charged particle of mass m and charge q initially at rest is released in an electric field of magnitude E . It's kinetic energy after time t will be ??

Answer 1

Answer:

Given:

Mass of charged particle = m

Charge = q

Electric Field Intensity = E

To find:

Kinetic energy after time t

Concept:

In presence of the Electric Field, the charge will experience a force. This force will provide acceleration leading to increase in speed of the particle.

Calculation:

∴ Force experienced = charge × Field

=> Force = q × E

=> Force = qE

∴ Acceleration = Force/mass

=> a = (qE)/m

Applying equation of motion :

∴ v = u + at

=> v = 0 + {(qE)/m}t

=> v = (qEt)/m

So Kinetic Energy is given as :

∴ KE = ½mv²

=> KE = ½ m {(qEt/m)}²

=> KE = (qEt)²/2m

So final answer :

$\sf{ \green{ \bold{ \huge{KE = \dfrac{ {(qEt)}^{2} }{2m}}}}}$

Answer 2

$\underline{ \boxed{ \bold{ \mathfrak{ \purple{ \huge{Answer}}}}}}$

Given :

mass of particle = m

charge on particle = q

initial velocity = 0 m/s

Intensity of electric field = E

To Find :

Kinetic energy of charged particle after time t...

Concept :

Here charged particle is released into electric field and we know that every charged particle is experienced force by electric field. So, the net force on the particle is electric force.

$\rm \: mathematically \implies \underline{ \boxed{ \rm{ \bold{ \pink{ F{ \tiny{net}} = ma = qE}}}}}$

Calculation :

$\implies \rm \: ma = qE \\ \\ \therefore \rm \: \blue{a = \frac{qE}{m} } \\ \\ \dag\rm \: as \: per \: kinematic \: equation \\ \\ \implies \rm \: v = u + at \\ \\ \dag \rm \: here \: u = 0 \\ \\ \therefore \rm \: \green{v = \frac{qEt}{m} } \\ \\ \implies \rm \: KE = \frac{1}{2} m {v}^{2} \\ \\ \implies \rm \: KE= \frac{1}{2} m ({ \frac{qEt}{m} )}^{2} \\ \\ \therefore \: \underline{ \boxed{ \bold{\rm{ \orange{KE = \frac{ {q}^{2} {E}^{2} {t}^{2} }{2m} }}}}} \: \red{ \star}$