Moving Electrons in a Uniform Magnetic Field

[Jkalirai]

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 10² N/C * 1.6022 x 10^-19 C) / 9.1094 x 10^-31 kg

a = 7.0353 x 10¹³ m/s²

Our electron is fired south and hits the positive plate.

Therefore, the acceleration of our electron is 7.03 x 10¹³ m/s².

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

∆dy = v₂ * ∆t + ½ * a * ∆t²

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

(0.02 m) = ½ * 7.0353 x 10¹³ m/s * ∆t²

∆t² = 0.02 * 2 / 7.0353 x 10¹³

∆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 10⁶ 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 v₂ we use the corresponding equation,

v_2y² = v_1y² + 2 * a * ∆d_y (v_1y = 0 m/s)

v_2y² = 2 * (7.0353 x 10¹³ m/s²) * (0.02)

v_2y = √ (2.81412 x 10¹²)

v_2y = 1.6775 x 10⁶ m/s

Therefore, the velocity that the electron strikes the plate at is 1.6775 x 10⁶ 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 10⁶ 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.

 

[TSny]

Your work looks very good (but I did not get out my calculator to check the arithmetic).

However, part (C) asks you to find the velocity of the electron as it strikes the positive plate. You found the vertical component of the velocity. Is there also a horizontal component?

 

[BvU]

Hello Jkalirai, $\qquad$ $\qquad$!

A few house rules to begin with:
Don't write " 4.0 × 106 m/s ". Please use the button superscript ( under

) or use $\LaTeX$ : $4.0 \times 10^6$ m/s .(I notice you did use the subscript button for " 3) v₂^2 = v₁^2 + 2 * a * ∆d "but they don't get rendered !? Strange ... Then: have pity on helpers "as shown " suggests there exists a sketch or a drawing -- post it. And "35. Let south and east be positive. " doesn't help if south and east don't occur anywhere (*) . (Even though I do know what you mean - telepathy ?)

(*) well, 'is fired south' does occur - but I think you mean east...And finally, but most importantly:

1) check your math for your third relevant equation

v2y = √ (2.81412 x 10^12) so $v_{2,y}>10^6$ m/s where you write

v2y = 3.14389 x 10^5 m/s

2) as Tsny helped you with that already, I don't have to ask what you do with the original 4e6 m/s

 

[Jkalirai]

Your work looks very good (but I did not get out my calculator to check the arithmetic).

However, part (C) asks you to find the velocity of the electron as it strikes the positive plate. You found the vertical component of the velocity. Is there also a horizontal component?

Huh, you know... I actually sat down for like 20-30 minutes and really wondered why I didn't even consider a horizontal portion. I worked through the problem again, I knew that horizontal acceleration was 0, meaning that the initial velocity is the same as the final velocity. After that, it's just a matter of putting the pieces together where v² = √ (v_x² + v_y²). Well anyways... after working through it I got a similar answer to what every other thread seems to be saying. Seems like a silly mistake to make but thanks for the heads up.

 

[Jkalirai]

Hello Jkalirai, $\qquad$ $\qquad$!

A few house rules to begin with:
Don't write " 4.0 × 106 m/s ". Please use the button superscript ( under View attachment 245979 ) or use $\LaTeX$ : $4.0 \times 10^6$ m/s .(I notice you did use the subscript button for " 3) v₂^2 = v₁^2 + 2 * a * ∆d "but they don't get rendered !? Strange ...Then: have pity on helpers "as shown " suggests there exists a sketch or a drawing -- post it. And "35. Let south and east be positive. " doesn't help if south and east don't occur anywhere (*) . (Even though I do know what you mean - telepathy ?)

(*) well, 'is fired south' does occur - but I think you mean east...And finally, but most importantly:

1) check your math for your third relevant equation

v2y = √ (2.81412 x 10^12) so $v_{2,y}>10^6$ m/s where you write

v2y = 3.14389 x 10^5 m/s

2) as Tsny helped you with that already, I don't have to ask what you do with the original 4e6 m/s

Thanks for the tips, though I can't quite fix the subscripts and superscripts in the "Relevant Questions" tab. Not sure what's going on there.