Electric Fields in Capacitors Video Lessons

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Problem: The electric field strength is 2.30×104 N/C inside a parallel-plate capacitor with a 0.600 mm spacing. An electron is released from rest at the negative plate.What is the electron's speed when it reaches the positive plate?

FREE Expert Solution

From Newton's second law:

F=mea

The force on electron moving under the influence of an electric field:

F=qE

We'll also need to use the kinematic equation:

vf2=v02+2ax

We get a by equating the first and second equations.

mea=qEa=qEme=(1.6×10-19)(2.30×104)9.11×10-31

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Problem Details

The electric field strength is 2.30×104 N/C inside a parallel-plate capacitor with a 0.600 mm spacing. An electron is released from rest at the negative plate.

What is the electron's speed when it reaches the positive plate?

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