Calculating Electron Trajectory in a Parallel Plate Conductor

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An electron is fired between two parallel conducting plates with a given speed and angle, and a voltage difference is applied. The attempt at calculating the electron's trajectory involves using equations for force, electric field, and acceleration. The calculated vertical displacement was found to be 0.54 mm, but this result is over 10% off from the expected value. The user has not yet attempted the second part of the problem due to potential errors in the acceleration calculation. Additional clarification on intermediate steps is requested to identify the source of the discrepancy.
laurajean997
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Homework Statement



An electron is fired at a speed vo = 5.6 ✕ 10^6 m/s and at an angle θo = -40° between two parallel conducting plates that are D = 2.5 mm apart, as in Figure P16.66. The voltage difference between the plates is ΔV = 103 V.

p16-66.gif


Homework Equations



F = Eq

E = V/d

a = F/m

The Attempt at a Solution



I tried finding acceleration with (ΔV/D)(q/m) or 103/.0025 * 1.6e-19/9.11e-31

I found Vy with 5.6e6sin(40)

Then i used vyf^2 = vyo^2 + 2aΔy to solve for Δy

then i subtracted my answer from .00125 m and got .54 mm, but the website says I'm off by more than 10%.

I haven't tried part B yet because i could have acceleration wrong and that will just mess up both answers.
 
Last edited:
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Don't forget gravity, or at least check how it compares with the acceleration of the electric field.
Write down equations, instead of narrating your calculations.
 
The acceleration was something x 10^15 so i figured gravity was negligible.
 
laurajean997 said:

Homework Statement



An electron is fired at a speed vo = 5.6 ✕ 10^6 m/s and at an angle θo = -40° between two parallel conducting plates that are D = 2.5 mm apart, as in Figure P16.66. The voltage difference between the plates is ΔV = 103 V.

p16-66.gif


Homework Equations



F = Eq

E = V/d

a = F/m

The Attempt at a Solution



I tried finding acceleration with (ΔV/D)(q/m) or 103/.0025 * 1.6e-19/9.11e-31

I found Vy with 5.6e6sin(40)

Then i used vyf^2 = vyo^2 + 2aΔy to solve for Δy

then i subtracted my answer from .00125 m and got .54 mm, but the website says I'm off by more than 10%.

I haven't tried part B yet because i could have acceleration wrong and that will just mess up both answers.

Hi laurajean997. Welcome to Physics Forums.

Can you show your attempt in a bit more detail? In particular, show the values that you obtained for the intermediate steps.
 
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