Question

A block having mass m and charge q is
resting on a frictionless plane at a
distance L from the wall as shown. A
uniform electric field E is switched on as
shown. If the collision of the block with
wall is perfectly elastic, the time taken
by the block to return to the initial
position is​

Answer 1

Answer:

Time taken by block to return to initial position is 2√2mL∈o/aσ

Explanation:

due to Electric filed the block will accelerates uniformaliy towards the wall :

acceleration (a)=F/m=qE/m  (where q=σ/2∈o)

Initially block was at the rest so according to Second Equation of Motion the time taken by the block to reach the Wall will be :

L=1/2at^2  (i-e t=√2L/a)

putting values :

=√2mL/aE

L=√4mL∈o/aσ

After the collision with the wall the block rebound with same speed with opposite to acceleration ,coming to rest travelling distance L in the same time "t" .Executing oscillatory motion with the distance "L" and time period (2t) given as :

T=2t=2√4mL∈o/aσ  

calculation is attached in attachment for clarification.

Hope You find it easy to understand  :)