Question

106 m/s in a uniform 1.9 x 105 N/C electric field. The field accelerates the Problem 6: An electron has an initial velocity of 5.25 electron in the direction opposite to its initial velocity. Part (a) What is the direction of the electric field? MultipleChoice 1) The field is in the direction of the electron's initial velocity 2) The field is in the direction to the right of the clectron's initial velocity 3) The ficld is in the opposite dircction of the elcctron's initial velocity 4) The field is in another direction not listed here Part (b) How far does the electron travel before coming to rest in m? Numeric : A numeric value is expected and not an expression Part (c) How long does it take the clectron to come to rest in s? Numeric A numeric value is expected and not an expression. Part (d) What is the magnitude of the electron's velocity (in m/s) when it returns to its starting point in the opposite direction of its initial velocity? Numeric : A numeric value is expected and not an expression

Answer 1

(a) 1) The field is in the direction of the electron's initial velocity

The electric field is in a direction opposite to the initial velocity of the electron.

Let's remind that, when an electric charge is immersed in an electric field:

- if the charge is positive, the charge experiences a force in the same direction as the electric field direction

- if the charge is negative, the charge experiences a force in the opposite direction to the electric field direction

In this case, we have an electron: so the electric force exerted on the electron will be in a direction opposite to the direction of the electric field. Since the electron is accelerated in a direction opposite to the electron's initial velocity, this means that the electric force is in a direction opposite to the initial velocity, and so the electric field must be in the same direction as the electron's initial velocity.

(b)
`4.13\cdot 10^(-4) m`

We have:

Electron's initial velocity:
`u=5.25\cdot 10^6 m/s`

Electric field magnitude:
`E=1.9 \cdot 10^5 N/C`

Electron charge:
`q=-1.6\cdot 10^(-19) C`

Mass of the electron:
`m=9.11\cdot 10^(-31)kg`

The electric force exerted on the electron is:

`F=qE=(-1.6\cdot 10^(-19) C)(1.9\cdot 10^5 N/C)=-3.04\cdot 10^(-14)N` (the negative sign means the direction of the force is opposite to its initial velocity)

The electron's acceleration is given by:

`a=(F)/(m)=(3.04\cdot 10^(-14) N)/(9.11\cdot 10^(-31) kg)=-3.34\cdot 10^(16) m/s^2`

Now we can use the SUVAT equation:

`v^2 - u^2 = 2ad`

where

v = 0 is the final speed (the electron comes to rest)

d is the total distance travelled by the electron

Solving for d,

`d=(v^2-u^2)/(2a)=(0-(5.25\cdot 10^6 m/s)^2)/(2(-3.34\cdot 10^(16) m/s^2))=4.13\cdot 10^(-4) m`

(c)
`1.57\cdot 10^(-10)s`

We can use the following equation:

`a=(v-u)/(t)`

where we have

`a=-3.34\cdot 10^(16)m/s^2` is the electron's acceleration

v = 0 is its final speed

`u=5.25\cdot 10^6 m/s` is the initial speed

t is the time it takes for the electron to come at rest

Solving for t,

`t=(v-u)/(a)=(0-(5.25\cdot 10^6 m/s))/(-3.34\cdot 10^(16) m/s^2)=1.57\cdot 10^(-10)s`

(d)
`5.25\cdot 10^6 m/s`

This part of the problem is symmetrical to the previous part. In fact, the force exerted on the electron is the same as before (in magnitude), but in the opposite direction. This also means that the acceleration is the same (in magnitude), but in the opposite direction.

So we have:

u = 0 is the initial speed of the electron

`a=3.34\cdot 10^(16)m/s^2`

`d=4.13\cdot 10^(-4) m` is the distance covered to go back

So we can use the following equation:

`v^2 - u^2 = 2ad`

to find v, the new final speed:

`v=√(u^2 +2ad)=\sqrt{0^2 + 2(3.34\cdot 10^(16) m/s^2)(4.13\cdot 10^(-4) m)}=5.25\cdot 10^6 m/s`