Question

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much time will elapse before it returns to its starting point?

Answer 1

time will elapse before it return to its staring point is 23.6 ns

Step-by-step explanation:

given data

speed u = 2.45 ×
`10^(7)` m/s

uniform electric field E = 1.18 ×
`10^(4)` N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 ×
`10^(-31)` kg ×a = 1.18 ×
`10^(4)` × 1.602 ×
`10^(-19)`

a = 20.75 ×
`10^(14)` m/s²

so acceleration is 20.75 ×
`10^(14)` m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 ×
`10^(7)` = 0 + 20.75 ×
`10^(14)` (t)

t = 11.80 ×
`10^(-9)` s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 ×
`10^(-9)`

time is 23.6 ×
`10^(-9)` s

time will elapse before it return to its staring point is 23.6 ns