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## Homework Statement

An electron has an initial velocity of 5.00 x 10^6 m/s in a uniform 2.00 x 10^5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.

(a) What is the direction of the electric field?

(b) How far does the electron travel before coming to rest?

(c) What is the electron's velocity when it returns to it's starting point?

## Homework Equations

F= Eq

F=ma

Eq=ma

## The Attempt at a Solution

F= Eq

= (2.00 x 10^5)(-1..60x10^-19)

= -3.2 x 10^-14 N

F= ma

-3.2x10^-14= (9.11x10^-31)a

a= -3.51x10^16 m/s^2

Taking downward direction as positive,

v^2= u^2 + 2as

0= 5.00 x 10^6 + 2(-3.51x10^16)s

s= 3.56 x 10^-4 m

Electron is accelerated in the direction towards the positive plate. It is stated that the direction of acceleration is opposite to that of its initial velocity. How is it possible?

Can anyone helps me to figure out what has gone wrong in my equations? Thanks a lot.