Respuesta :
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 = ?
Answer:
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 = [tex]\sqrt{3.014*10^{14} }[/tex]
Vi = 1.736x10⁷ m/s
Hence, the electron's speed was 1.736x10⁷ m/s as it left the negative plate.
