- #1

Jkalirai

- 4

- 1

- Homework Statement
- An electron is fired at 4.0 × 10^6 m/s horizontally between the parallel plates, as shown, 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 × 10^2 N/C. The separation of the plates is 2.0 cm.

a) Find the acceleration of the electron between the plates.

b) Find the horizontal distance traveled by the electron when it hits the plate.

c) Find the velocity of the electron as it strikes the plate.

- Relevant Equations
- 1) m * a = ε * q

2) ∆d = v * ∆t + ½ * a * ∆t^2

3) v[SUB]2[/SUB][SUP]2[/SUP] = v[SUB]1[/SUB][SUP]2[/SUP] + 2 * a * ∆d

a) We can solve for acceleration by looking at F

_{NETy}

F

_{NETy}= F

_{E}(G is negligible)

F

_{NETy}= m * a

The mass (m) of an electron is 9.1093836 x 10

^{-31}kg.

The elementary charge (q) of an electron is -1.60217662 x 10

^{-19}C

a = ε * q / m

a = (4.0 x 10

^{2}N/C * 1.6022 x 10

^{-19}C) / 9.1094 x 10

^{-31}kg

a = 7.0353 x 10

^{13}m/s

^{2}

Our electron is fired south and hits the positive plate.

Therefore, the acceleration of our electron is 7.03 x 10

^{13}m/s

^{2}

b) To find the horizontal distance, we first solve for ∆t

∆dy = v

_{2}* ∆t + ½ * a * ∆t

^{2}

We know that our initial vertical velocity is 0 m/s.

(0.02 m) = ½ * 7.0353 x 10

^{13}m/s * ∆t

^{2}

∆t

^{2}= 0.02 * 2 / 7.0353 x 10

^{13}

∆t = √ 5.6856 x 10

^{-16}

∆t = 2.3845 x 10

^{-8}s

Now we solve for ∆dx

Our horizontal acceleration is equal to 0 therefore,

∆d

_{x}= v

_{1x}* ∆t

∆d

_{x}= (4.0 x 10

^{6}m/s * 2.3845 x 10

^{-8}s)

∆d

_{x}= 0.09538 m

Therefore, our horizontal distance is equal to 0.096 m [E].

c) To find v

_{2}we use the corresponding equation,

v

_{2y}

^{2}= v

_{1y}

^{2}+ 2 * a * ∆d

_{y}(v

_{1y}= 0 m/s)

v

_{2y}

^{2}= 2 * (7.0353 x 10

^{13}m/s

^{2}) * (0.02)

v

_{2y}= √ (2.81412 x 10

^{12})

v

_{2y}= 1.6775 x 10

^{6}m/s

Therefore, the velocity that the electron strikes the plate at is 1.6775 x 10

^{6}m/s.

I am not sure if my answer for part c) is correct or not. I've seen other people with older threads get answers close to 4.3 x 10

^{6}but it didn't make sense to me because they used the initial velocity of x in place of the initial velocity of y, whereas I stated that my initial velocity of y is equal to 0.

Any advice is appreciated.

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