Electric Field & Capacitor Problem HELP

AI Thread Summary
An electron enters a parallel plate capacitor with an initial speed of 4.00 x 10^6 m/s and exits after curving upward. The capacitor dimensions are 2.00 cm long and 0.150 cm apart, with a uniform electric field. To find the electric field's magnitude, the force on the electron can be calculated using F=ma=qE, where q is the charge of the electron (1.6 x 10^-19 coulombs). The time taken for the electron to travel horizontally can be determined using t=d/v, allowing for the calculation of acceleration and subsequently the electric field. The discussion emphasizes the importance of using kinematics and the known properties of the electron to solve the problem.
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Electric Field & Capacitor Problem...HELP!

An electron enters the lower left side of a parallel plate capacitor & exits at the upper right side making an upward curve. The initial speed of the electron is 4.00 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

So I understand that F=ma=qE. I also understand that I can use kinematics to find the acceleration and I can also use the mass of an electron to figure out the force. However, how do I find q? I'm so confused!:confused: Please help! Thanks!
 
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q is the charge on an electron. Look it up in your text.
 
Meir Achuz said:
q is the charge on an electron. Look it up in your text.

q=e=1.6x10^{-19} coulombs...
 
in the vertical direction, the electron has traveled 0.15 cm (from lower left corner to upper right corner). the initial speed in the vertical direction is 0
=> distance = a*t^2/2 (a= acceleration and t- time)
the time is found by: t= d/v (in horizontal direction) = 2 cm/ speed
so now you can find a, and then E.
Good luck.
 
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