A proton has an initial speed of 3 10^5 m/s. It is brought to rest by an electri
ID: 1424118 • Letter: A
Question
A proton has an initial speed of 3 10^5 m/s. It is brought to rest by an electric field. What is the magnitude of the voltage difference needed to bring this charge to rest? At the point when the proton comes to rest (momentarily): its potential energy is, and it is a t a voltage as compared to initially. The electric field must be pointed the initial velocity of the proton, so that it is able to stop it. Describe the motion of the proton after it comes to rest (momentarily). (Why this is problem like a "ball being thrown upwards"?)Explanation / Answer
a) q*v = 0.5*m*v^2
=> v = 0.5*1.67*(10^-27)*9*(10^10)/(1.6*10^-19) = 469.6875 V
b) potential energy is larger and is at a larger voltage
c) in opposite direction as
d) Since it is in electric fied so after comming to rest it will starts acclerating in direction opposite to the initial velocity. accleration = q.E/m , q = charge on proton, E = electric field and m = mass of proton.
This is same as a boll being thrown upward because when we throw ball upward. It exprience gravitational force downwards. This force stops the ball thrown upward after some time. Then ball starts acclerating downwards due to gravitational force by earth. This is same case in quesion. instead of gravitational force there is electric force.