Coulomb's Law, Charges and Acceleration

[julianwitkowski]

Homework Statement

An electron is fired at 4.0 e+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 e+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 as it strikes the plate?

Homework Equations

a = qE/m
F = qE
d = x_i + y_i ∙t + ½ ∙ a ∙ t²

The Attempt at a Solution

I've read the chapter a few times and I think this is right but I want to make sure I haven't made a careless error, Please and thank you for your help!

a) 7.0 e¹³ m/s²

b) This is the one I'm not sure about...

d = x_i + y_i ∙t + ½ ∙ a ∙ t²

d = 3.5 e¹³ m/s² ∙ t² + 0 + 0 = 2e-2 m

t = √ 2e^-2 / 3.5 e¹³ = 2.4 eˉ⁸ s

d = 4.0 e⁶ m/s ∙ 2.4 eˉ⁸ s = 9.6 eˉ² m

c) ...and this...

v = y_i + a ∙ t

= 0 + 7.0 × 10¹³ m/s² ∙ 2.4 × 10ˉ⁸ s

= 1.7 × 10⁶ m/s

Thank you for your help :)

 

[BvU]

Hi. a OK, b method looks good, numbers look good too (except rounding off -- I get 9.5 10^-2 m).
I would use the 10 instead of the e to avoid any possible misunderstanding.

In part c I wonder what happened to the original 4 10⁶ m/s

 

[julianwitkowski]

Hi. a OK, b method looks good, numbers look good too (except rounding off -- I get 9.5 10^-2 m).
I would use the 10 instead of the e to avoid any possible misunderstanding.

In part c I wonder what happened to the original 4 10⁶ m/s

Thank you... I'm checking over that, but one thing I'm curious about is why the time is not = √ (2 ⋅ 2 ⋅10^-2 / 3.5⋅10¹³) = 3.3⋅10ˉ⁸ s

 

[BvU]

You already used the 2 to divide the 7 and got 3.5. Nice way to confuse yourself... :)

 

[julianwitkowski]

You already used the 2 to divide the 7 and got 3.5. Nice way to confuse yourself... :)

ya :/ .. ha

I get 9.5 10^-2

Thanks! I see that I apparently got the time wrong, it's 2.3 not 2.4 ... :) But now I get 9.2E-2 instead of 9.5E-2 ?

I'm going to try a different calculator.

wolfram t = 2.39 ... and I got 0.0956 ... So being as how I hate significant digits I still don't know the best judgement call in situations like this, should I ignore the 6? I think I should I guess because the question gives 2.0 and just to clarify this, the 0 in 2.0 is a significant digit right? I think I heard this does not count as a place holder 0.

In part c I wonder what happened to the original 4 10⁶ m/s

Glancing over this ...

 

[BvU]

[FONT=Courier New]  
9.10900E-31    kg           m
 
 1.60200E-19    C            e
 
 400            N/C          V 
 
 7.0348E+13     m/s2         a = e*V/m
 
 0.02           m            d
 
 5.6860E-16     s2           2*d/a 
 
 2.3845E-08     s            sqrt
 
 4.00E+06       m/s          v
 
 9.538E-02      m            v*t

 

[julianwitkowski]

In part c I wonder what happened to the original 4 10⁶ m/s

Ok about this I'm a bit confused, wouldn't the final velocity be acceleration * time ? Or is your comment in reference to how the electron lost a lot of velocity.?

 

[BvU]

Neither nor.

For part c the exercise asks about the velocity as it strikes the plate. You forgot the horizontal component 4000000 m/s, vertical component a * t indeed.
But you might want to turn that into a magnitude (and perhaps an angle if that's needed -- I normally interpret velocity as a vector, speed as a magnitude)

 

[julianwitkowski]

Neither nor.

For part c the exercise asks about the velocity as it strikes the plate. You forgot the horizontal component 4000000 m/s, vertical component a * t indeed.
But you might want to turn that into a magnitude (and perhaps an angle if that's needed -- I normally interpret velocity as a vector, speed as a magnitude)

Got it, thanks again for the help :)