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Projectile motion and electric field strength

  • Thread starter mohemoto
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  • #1
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Homework Statement



An electron is launched at a α=38.5° angle and speed of 4.12×10^6 m/s from the positive plate of the parallel plate capacitor shown. If the electron lands d=3.03 cm away, what is the electric field strength inside the capacitor?

Homework Equations



t = d/v
v0y = v*sin0
v0x = v*cos0
x(t) = v0t + 1/2at^2



The Attempt at a Solution



So I first solved for v0y, since the electric field strength is perpendicular and only depends on vertical velocity.

v0y = 2.57 x 10^6

I then found time, after finding the hypotenuse of the distance

d = sqrt(0.0303^2 + 0.0303^2)
d = 0.04

t = d/v
t = 0.04 / 4.12 x 10^6
t = 1.04 x 10^-8

After this, I know I need to find acceleration using
x(t) = v0y*t + 1/2at^2

And this is where I have problems. I'm not exactly sure that the above procedure is correct, but assuming that it is, do I make the position zero? So is it:

v0y = 2.57 x 10^6
t = 1.04 x 10^-8
a = ?
x(t) = 0

I know that after this I just have to substitute into
Eq = ma

And solve for E, but I'm having trouble with solving for acceleration..please help!

Thank you very much!
 

Answers and Replies

  • #2
106
1
t = d/v
v0y = v*sin0
v0x = v*cos0
x(t) = v0t + 1/2at^2
You might have an easier time solving this problem if you clarify these equations. Although I could be mistaken since I don't have a diagram, this seems like a generic projectile motion problem with an E&M twist. Rather than using a=9.8 m/s2 (which is the case in most simple projectile motion problems), you're solving for an unknown acceleration, from which you can calculate the electric field strength. The key concept here, as in most other simple projectile motion problems, is that you can split the problem into two components: a horizontal one, and a vertical one. In the horizontal direction (parallel to the plates), there is no net force. Thus:

[tex] \Delta x(t) = v_{0x} t = v_0 cos( \alpha ) t [/tex]

In the vertical direction, there is a net force due to the electric field, resulting in a net (but constant) acceleration. We can therefore write:

[tex] \Delta y(t) = v_{0y} t + \frac{1}{2} a t^2 = v_0 sin( \alpha ) t + \frac{1}{2} a t^2 [/tex]

Finally, we need to figure out what [tex] \Delta x [/tex] and [tex] \Delta y [/tex] are. I think you're a bit confused on this point:

I then found time, after finding the hypotenuse of the distance

d = sqrt(0.0303^2 + 0.0303^2)
d = 0.04
You might be overthinking this. The electron lands 3.03cm away--is this in the x-direction, the y-direction, or some combination therein? And what does the word "land" tell you about the electron's position along the y-direction?

Once you have [tex] \Delta x [/tex] and [tex] \Delta y [/tex], you should be able to solve the system of equations for the acceleration, then use that to find the electric field strength.
 
Last edited:
  • #3
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Yes, it is a projectile motion question, sorry about that!

Okay, so is the 3.03cm in the x direction then? So do I just do t = d/v and solve for t, with d being 3.03 cm and v being 4.12×10^6 m/s? Or is it the v in the x direction that I use?

I'm still slightly confused..I understand the concept behind all of this, but I don't really understand why you need both Delta_x and Delta_y to solve this..
 
  • #4
106
1
Do you have a diagram of this problem? If so, try drawing a coordinate axis, the initial velocity vector, and perhaps a sketch of the path of the electron, on it (what shape will the latter be?). If that doesn't help, try labeling distances such as that 3.03cm; it might help you visualize exactly what the directions are, and how the equations correspond to them.
 

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