Question

A charged particle of charge Q released in electric field
$e = e_{0}(1 - {ax}^{2} )$
from origin. find position when its kinetic energy again become zero​

Answer 1

Answer:

A force is generated because of the elective field

F=qE⇒a=mqE.

As the particle starts from rest and acceleration is constant applying equation of motion.

V=u+at

V=0+mqEt

V=mqET

Then, KE21mv2=21m(mqEt)2=21mm2q2E2t2=2mq2E2t2

Answer 2

A charged particle of charge Q released in electric field E = E₀(1 - ax²) from origin. initially the particle is at rest at x = 0.

We have to find the position when its kinetic energy again becomes zero.

solution : from work energy theorem,

change in kinetic energy = work done by the external agent

⇒∆K = 0 [ as work done by the external agent is zero. ]

F = qE

and ∆K = ∫ F dx

⇒∫qE₀(1 - ax²)dx = 0

⇒qE₀∫(1 - ax²) dx = 0

⇒x - ax³/3 = 0

⇒1 = ax²/3

⇒x² = 3/a

⇒x = √(3/a)

Therefore the position when its kinetic energy again becomes zero is √(3/a)