An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.50×107 m/s . What was the electron's speed as it left the negative plate?

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Two 1.50 cm diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The  electric field between the disks is 4.60x10⁵ V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.50x10⁷ m/s . What was the electron's speed as it left the negative plate?

Given Information:

d = 2.0 mm = 0.002 m

Electric field = E = 4.60x10⁵ V/m

Vf = 2.50x10⁷ m/s

Required Information:

Electron's Initial speed = Vi = ?

Vi = 1.736x10⁷ m/s

Explanation:

First we need to find out the voltage across the capacitor which is given by

V = E*d

Where E is the electric field between plates and d is the spacing between them

V =  (4.60x10⁵)*(0.002)

V = 920 V  

Now we can find out the Electron's Initial speed

The law of conservation of mechanical energy states that the total mechanical energy of the system is conserved.

K denotes Kinetic energy and U denotes potential energy

ΔK + ΔU = 0

ΔK = -ΔU

Where ΔK = 1/2mVf² - 1/2mVi² and -ΔU = qΔV

Where m is the mass of the electron and is equal to 9.11x10⁻³¹ kg and q is the charge of electron and is equal to 1.602x10⁻¹⁹ coulombs

1/2mVf² - 1/2mVi² = qΔV

Rearrange the equation to make Vi² subject of the equation

1/2m(Vf² - Vi²) = qΔV

Vf² - Vi² = 2qΔV/m

Vi² = Vf² - 2qΔV/m

Vi² = (2.50x10⁷)² - 2*(1.602x10⁻¹⁹)*(920)/9.11x10⁻³¹

Vi² = 3.014x10¹⁴ m/s

Take square root

Vi =

Vi = 1.736x10⁷ m/s

Hence, the electron's speed was 1.736x10⁷ m/s as it left the negative plate.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )