i attempted this with these equations and numbers but its a mess. i would love to be enlightened....
charge of a proton- 1.6*10^-19
mass of a proton- 1.7*10^-27
formulas...
1. Volts=E*r
2. epe=q*V
3. F=q*E
4. velocity= square root( 2ke/mass proton)
Two uniformly and oppositely charged parallel
plates are arranged shown above, with the
positively charged plate a distance d = 1:00 cm
above the negatively charged plate. The uniform
electric _eld, pointing directly downwards from
the positive plate to the negative plate, has a
magnitude of j~E j = 2:00*10^4 N/C. Following the
usual convention, in this problem we assign the
electric potential to have the value V = 0 Volts
at the surface of the negatively charged plate.
(a) Suppose a proton is initially at rest on the positive plate. What is the electrical potential
(in Volts) at the location of the proton? What is the electrical potential energy (in J)
of the proton at this location?
(b) The proton is then released, and accelerates across the vacuum between the plates until
reaching the negatively charged plate. What is the electric potential (in Volts) at the
proton's new location? What is the electrical potential energy (in J) of the proton at
this location?
(c) How fast will the proton be traveling when it reaches the negative plate?
(d) Suppose a second proton starts from the negative plate with an upward velocity whose
magnitude is equal to 2/3 (two-thirds) the speed found in part (c). How close to the
positive plate will this second proton come before it stops and begins moving back
toward the negative plate?
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