Coulomb's Law, Charges and Acceleration

In summary, the electron was fired at 4.0 e+6 m/s horizontally between the parallel plates, and the electric field between the plates was 4.0 e+2 N/C. The separation of the plates was 2.0 cm. The electron traveled a distance of 9.6 e+2 m and had a velocity of 1.7 × 10⁶ m/s when it hit the plate.
  • #1

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 = xi + yi ∙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 = xi + yi ∙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 = yi + a ∙ t

= 0 + 7.0 × 10¹³ m/s² ∙ 2.4 × 10ˉ⁸ s

= 1.7 × 10⁶ m/s

Thank you for your help :)
 
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  • #2
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
 
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  • #3
BvU said:
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
 
  • #4
You already used the 2 to divide the 7 and got 3.5. Nice way to confuse yourself... :)
 
  • #5
BvU said:
You already used the 2 to divide the 7 and got 3.5. Nice way to confuse yourself... :)
ya :/ .. ha

BvU said:
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.

BvU said:
In part c I wonder what happened to the original 4 106 m/s
Glancing over this ...
 
Last edited:
  • #6
Code:
[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[/FONT]
 
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  • #7
BvU said:
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.?
 
  • #8
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)
 
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  • #9
BvU said:
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 :)
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of physics that describes the electrostatic interaction between charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is Coulomb's Law related to charges?

Coulomb's Law is directly related to charges, as it quantifies the force between two charged particles. The magnitude of the force is determined by the charges of the particles and the distance between them.

3. Can Coulomb's Law be used to calculate acceleration?

No, Coulomb's Law alone cannot be used to calculate acceleration. It only describes the force between two charged particles and does not take into account the masses of the particles. To calculate acceleration, we need to use Newton's second law, which relates force, mass, and acceleration.

4. How does distance between charges affect the force according to Coulomb's Law?

According to Coulomb's Law, the force between two charged particles is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force between them decreases.

5. What is the unit of charge used in Coulomb's Law?

The unit of charge used in Coulomb's Law is the Coulomb (C). It is a derived unit in the SI system, and it represents the amount of electric charge transferred in one second by a current of one ampere.

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