Question

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a potential difference of 200 V. The electron enters through a small hole in the negative plate and moves toward the positive plate. At the time the electron is near the negative plate, its speed is 4.0×105m/s.4.0×105m/s. Assume the electric field between the plates to be uniform, and find the speed of electron at (a) 0.10 cm, (b) 0.50 cm, (c) 1.0 cm, and (d) 1.5 cm from the negative plate, and (e) immediately before it hits the positive plate

Answer 1

a)
`v=1.77* 10^6\ m/s`

b)
`v=3.872* 10^6\ m/s`

c)
`v=5.5* 10^6\ m/s`

d)
`v=6.7* 10^6\ m/s`

e)
`v=7.7* 10^6\ m/s`

Step-by-step explanation:

Given that

d = 2 cm

V = 200 V

`u=4* 10^5\ m/s`

We know that

F = E q

F = m a

E = V/d

So

m a = q .V/d b

`a=(q.V)/(m.d)` ---------1

The mass of electron

`m=9.1* 10^(-31)\ kg`

The charge on electron

`q=1.6* 10^(-19)\ C`

Now by putting the all values in equation 1

`a=(1.6* 10^(-19)* 200)/(9.1* 10^(-31)* 0.02)\ m/s^2`

`a=1.5* 10^(15)\ m/s^2`

We know that

`v^2=u^2+2as`

a)

s = 0.1 cm

`v^2=(4* 10^5)^2+2* 1.5* 10^(15)* 0.1* 10^(-2)`

`v=\sqrt{3.16* 10^(12)}\ m/s`

`v=1.77* 10^6\ m/s`

b)

s = 0.5 cm

`v^2=(4* 10^5)^2+2* 1.5* 10^(15)* 0.5* 10^(-2)`

`v=\sqrt{1.5* 10^(13)}\ m/s`

`v=3.872* 10^6\ m/s`

c)

s = 1 cm

`v^2=(4* 10^5)^2+2* 1.5* 10^(15)* 1* 10^(-2)`

`v=\sqrt{3.06* 10^(13)}\ m/s`

`v=5.5* 10^6\ m/s`

d)

s = 1.5 cm

`v^2=(4* 10^5)^2+2* 1.5* 10^(15)* 1.5* 10^(-2)`

`v=\sqrt{4.5* 10^(13)}\ m/s`

`v=6.7* 10^6\ m/s`

e)

s = 2 cm

`v^2=(4* 10^5)^2+2* 1.5* 10^(15)* 2* 10^(-2)`

`v=\sqrt{6.06* 10^(13)}\ m/s`

`v=7.7* 10^6\ m/s`

Answer 2

a) d=0.1 cm=0.001 m

`v_(f)=1.9*10^(6)m/s`

b) d=0.5 cm=0.005 m

`v_(f)=4.2*10^(6)m/s`

c) d=1.0 cm=0.01 m

`v_(f)=5.9*10^(6)m/s`

d) d=1.5 cm=0.015 m

`v_(f)=7.3*10^(6)m/s`

e) d=2 cm=0.02 m

`v_(f)=8.4*10^(6)m/s`

Step-by-step explanation:

Let's start with the Coulomb force equation:

`F=qE`

• F is the force
• q is the electron charge
• E is the electric field

The force is equal to the mass times the acceleration, so we will have:

`ma=qE`

`a=(qE)/(m)` (1)

Now, the potential difference between the two plates is given by:

`\Delta V=ED`

`E=(\Delta V)/(D)` (2)

• D is the distance between parallel plates (D = 2.0 cm = 0.02 m).

Let's put (2) in (1)

`a=(q\Delta V)/(Dm)` (3)

By the other side, let's recall the relation between velocities and acceleration of a particle, a kinematic equation:

`v_(f)^(2)=v_(i)^(2)+2ad` (4)

• vf is the final velocity
• vi is the initial velocity
• a is the acceleration
• d is the distance

If we take square root in both sides, we will get the final velocity equation that depends on the distance.

`v_(f)=\sqrt{v_(i)^(2)+2ad}` (5)

Let's put (3) in (5)

`v_(f)=\sqrt{v_(i)^(2)+(2dq\Delta V)/(Dm)}`

a) For d=0.1 cm=0.001 m

`v_(f)=\sqrt{v_(i)^(2)+(2dq\Delta V)/(Dm)}`

`v_(f)=\sqrt{(4*10^(5))^(2)+(2*0.001*1.6*10^(-19)*200)/(0.02*9.11*10^(-31))}`

`v_(f)=1.9*10^(6)m/s`

b) For d=0.5 cm=0.005 m

`v_(f)=\sqrt{(4*10^(5))^(2)+(2*0.005*1.6*10^(-19)*200)/(0.02*9.11*10^(-31))}`

`v_(f)=4.2*10^(6)m/s`

c) For d=1.0 cm=0.01 m

`v_(f)=\sqrt{(4*10^(5))^(2)+(2*0.01*1.6*10^(-19)*200)/(0.02*9.11*10^(-31))}`

`v_(f)=5.9*10^(6)m/s`

d) For d=1.5 cm=0.015 m

`v_(f)=\sqrt{(4*10^(5))^(2)+(2*0.015*1.6*10^(-19)*200)/(0.02*9.11*10^(-31))}`

`v_(f)=7.3*10^(6)m/s`

e) For d=2 cm=0.02 m

`v_(f)=\sqrt{(4*10^(5))^(2)+(2*0.02*1.6*10^(-19)*200)/(0.02*9.11*10^(-31))}`

`v_(f)=8.4*10^(6)m/s`

I hope it helps you! :)