## Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between t

Question

Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.

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4 weeks 2021-08-19T13:07:48+00:00 1 Answers 3 views 0

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