Relationships Between Force, Field, Energy, Potential Video Lessons

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Problem: The electric field strength is 5.00×104N/C inside a parallel-plate capacitor with a 1.90mm spacing. A proton is released from rest at the positive plate. What is the proton's speed when it reaches the negative plate?

FREE Expert Solution

Electric force:

F=qE

Newton's second law:

ΣF=ma

Uniform accelerated motion (UAM) equations:

 vf = v0 +atx= (vf+v02)tx= v0t+12at2 vf2= v02 +2ax

Charge of a proton = +e = +1.6 × 10-19 C

ΣF = eE = ma

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Problem Details

The electric field strength is 5.00×104N/C inside a parallel-plate capacitor with a 1.90mm spacing. A proton is released from rest at the positive plate. What is the proton's speed when it reaches the negative plate?

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