Where Did I Go Wrong in Calculating the Charge's Movement?

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SUMMARY

The discussion focuses on the miscalculations involved in determining the movement of a charge in an electric field. The electric field (E) was calculated as 2x105 V/m using a voltage (V) of 4x103 volts and a distance (d) of 0.02 meters. The force (F) on the charge, calculated using F=qE, resulted in 0.3204 N, leading to an incorrect acceleration (a) of 3.517x1029 m/s2 and an implausible time (T) of 2.84x10-32 seconds. The primary errors identified include the incorrect value of the elementary charge and the misuse of projectile motion equations.

PREREQUISITES
  • Understanding of electric fields and their calculations
  • Familiarity with Newton's second law (F=ma)
  • Knowledge of basic kinematics and projectile motion
  • Concept of elementary charge and its correct value (1.602x10-19 C)
NEXT STEPS
  • Review the principles of electric fields and their effects on charged particles
  • Study the correct application of Newton's second law in electric fields
  • Learn about kinematic equations for projectile motion, specifically $$y=\frac{1}{2}at^2$$
  • Examine the significance of the elementary charge in physics calculations
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Students and educators in physics, particularly those focusing on electromagnetism and kinematics, as well as anyone involved in troubleshooting calculations related to electric fields and charged particle motion.

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Homework Statement
Given a charge -e moving through a set of charged plates at a velocity of 0.8 x10^7 m/s. The plates have a voltage of 4000v, how long can the plates be without the charge -e impacting one of them?
-e = -1.602 x 10^-6 C
Me = 9.11 x 10^-31 kg
Relevant Equations
E=v/d
F=qE=ma
Electric field between the plates
V=4x10^3
d=0.02
E=4x10^3/0.02
E=2x10^5 V/m

Calculate the force on charge -e
-e = -1.602x10^-6
Me = 9.11x10^-31
F=qE
F=(1.602x10^-6)(2x10^5)
F=0.3204 N

Using F=ma
a=F/Me
a=0.3204/9.11x10^-31
a=3.517x10^29 m/s

Time = distance / speed
T=.01 (half distance of plates with charge in the middle) / 3.517x10^29
T =2.84x10^-32 seconds

Calculate the distance the charge would move at 0.8x10^7 m/s in 2.84x10^-32 seconds

Distance = speed x time
Distance = 0.8x10^7 x 2.84x10^-32
Distance = 2.272x10-25 mtrs
I think I have royally screwed up my calculations somewhere here as a) the charge is moving substantially faster than the speed of light. And b) the length or distance seems really really small?
 
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A few points to consider...

If 'e' is meant to be the elementary charge, then e = 1.602x10⁻¹⁹C, not 1.602x10⁻⁶C.

The particle (it may not be an electron) behaves like a simple projectile, with an initial horizonal velocity, in a gravitational field. Make sure you know how to do simple projectile calculations!

Watch out for careless mistakes. E.g the unit for acceleration is not m/s!
 
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Except the mistake regarding the value of acceleration due to the electric field, which mistake is due to wrong charge of electron as spotted by @Steve4Physics , you have done another mistake in calculating the time to move vertically from the middle up (or down) to the plate due to the electric field.

You should use the formula $$y=\frac{1}{2}at^2$$ and not $$y=at$$ which is the one you used (y=0.01 since that charge starts from middle).
 
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