Answer:
a. 9.52 cm b. 4.34 × 10⁶ m/s
Explanation:
a. The horizontal distance traveled by the electron when it hits the plate.
The electric force F on the electron due to the electric field E of mass, m is
F = -eE = ma
a = -eE/m where a = acceleration of electron
The vertical distance moved by the electron is given by
Δy = ut +1/2at²
u = initial vertical velocity = 0. and take the top plate as y = 0 and bottom plate as y
So,
0 - y = 0 × t + 1/2at²
-y = 1/2at²
substituting a = -eE/m
-y = 1/2(-eE/m)t²
y = eEt²/2m
making t subject of the formula,
t = √(2ym/eE) where t is the time it takes to reach the bottom plate.
Since E = 4.0 × 10² N/C, y = distance between plates = 2.0 cm = 0.02 m, m = 9.109 × 10⁻³¹kg and e = 1.602 × 10⁻¹⁹ C
t = √[(2 × 0.02 m × 9.109 × 10⁻³¹kg)/(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C)]
t = √[(0.36436 × 10⁻³¹kgm)/(6.408 × 10⁻¹⁷ N)]
t = √[(0.0569 × 10⁻¹⁴kgm/N)t
t = 0.238 × 10⁻⁷ s
t = 23.8 × 10⁻⁹ s
t = 23.8 ns
The horizontal distance moved when it hits the plates x = vt where v = initial horizontal velocity = 4.0 × 10⁶ m/s
x = 4.0 × 10⁶ m/s × 23.8 × 10⁻⁹ s
= 0.0952 m
= 9.52 cm
b. The velocity of the electron as it strikes the plate.
To find the velocity of the electron as it strikes the plates, we calculate its final vertical velocity V as it strikes the plate. This is gotten from
v' = u + at since u = 0,
v' = at
= -eEt/m
= -(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C × 0.238 × 10⁻⁷ s)/9.109 × 10⁻³¹kg
= -1.525 × 10⁻²⁴ Ns/9.109 × 10⁻³¹kg
= -0.167 × 10⁷ m/s
= -1.67 × 10⁶ m/s
So, the resultant velocity as it strikes the plate v = √(v'² + v²)
= √((-1.67 × 10⁶ m/s)² + (4 × 10⁶ m/s)²)
= √(2.7889 + 16) × 10⁶ m/s
= √18.7889 × 10⁶ m/s
= 4.335 × 10⁶ m/s
≅ 4.34 × 10⁶ m/s
