What is the magnitude of the electric field?

AI Thread Summary
To determine the magnitude of the electric field affecting an electron released from rest, the acceleration can be calculated using kinematics, specifically the equation y = 1/2at^2. The electron travels 4.50 m in 3.00 microseconds, allowing for the calculation of acceleration. The force responsible for the acceleration is given by F = E * q, where q is the charge of the electron (1.62 * 10^-19 C) and the mass of the electron is approximately 9.11 * 10^-31 kg. By substituting known values into the equations, the electric field can be solved algebraically. This method effectively combines kinematics and electromagnetism to find the electric field's magnitude.
hitek131
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I need help with this question if anyone can give me an idea of what to do. Any help would be greatly appreciated.

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00 micro seconds after it is released.
What is the magnitude of the electric field?
 
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Do you know what force is in charge of accelerating the electron?
 
yes it is F=E * q

I know that q = 1.62 * 10^-19, but I still have an unknown variable.
 
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You can solve the acceleration using elementary kinematics. Then you'd need to relate that force to the acceleration you solved for.
 
yes i know, you use y = 1/2at^2, but you still get unknown variables
 
You know everything besides E. The electron's mass is standard tabulated stuff.
 
Is the mass 9.10938188(72) * 10 ^-31?
 
yeah that's it. it's all algebra from here on.
 
Thank you so much!
 
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