Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:

Explanation:
From the question we are told that:
Electric field 
Distance 
At negative plate
Generally the equation for Velocity is mathematically given by

Therefore




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Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
=
(
-
) ........................................................ i
And,
v =
v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
=
(
-
)
or,
= (
)
=(
) 
or, y = 3.56 cm.
Answer:
In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion
In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.
In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies
Considering the unknown resistence as R and using the Ohm's First Law, we have:
The equivalent resistence is given by the resistor series with the lamp resistence.

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