Question

The electric field strength is 20,000 N/C inside a parallel-platecapacitor with a 1.0 mm spacing. An electron is released fromrest at the negative plate. What is the electron’s speed when itreaches the positive plate?

Answer 1

2.65 x 10⁶ m/s

Step-by-step explanation:

E = magnitude of electric field = 20,000 N/C

d = spacing between the plates of the capacitor = 1.0 mm = 0.001 m

ΔV = Potential difference between the plates = ?

Potential difference between the plates is given as

ΔV = E d

ΔV = (20000) (0.001)

ΔV = 20 Volts

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of the electron as it reach the positive plate

using the conservation of energy

Kinetic energy gained = electric potential energy lost

(0.5) m v² = q ΔV

(0.5) (9.1 x 10⁻³¹) v² = (1.6 x 10⁻¹⁹) (20)

v = 2.65 x 10⁶ m/s

Answer 2

Step-by-step explanation:

It is given that,

Electric field strength, E = 20,000 N/C

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

`F=qE`

Also,
`ma=qE`

`a=(qE)/(m)`

`a=(1.6* 10^(-19)* 20,000)/(9.1* 10^(-31))`

`a=3.51* 10^(15)\ m/s^2`

Using third equation of motion as :

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

`v=√(2ad)`

`v=\sqrt{2* 3.51* 10^(15)* 0.001}`

v = 2649528.2599 m/s

or

`v=2.64* 10^6\ m/s`

So, the velocity of the electron when it reaches the positive plate is
`2.64* 10^6\ m/s`. Hence, this is the required solution.