Problem: A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.What will be the final speed of an electron released from rest at the negative plate?Express your answer to two significant figures and include the appropriate unitsV=_________

FREE Expert Solution

Electric force,

F=qE

Newton's second law,

F=ma

UAM equations,

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

The magnitude of the electric force experienced by a proton or an electron is equal since they have the same charge. 

The ratio of electrons acceleration to proton's acceleration,

meae=mpapaeap=mpme

The separation of the plates of the capacitors, 

  • Δx = ?
  • vf = 45000 m/s
  • v0 = 0 m/s
  • a = ap
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Problem Details

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.

What will be the final speed of an electron released from rest at the negative plate?

Express your answer to two significant figures and include the appropriate units

V=_________