Auniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.70 cm distant from the first, in a time interval of 1.48×10−6 s .

find the magnitude of the electric field.

find the speed of the proton when it strikes the negatively charged plate.

1) Magnitude of the electric field

The distance traveled by the proton is equal to the distance between the two plates:

d = 1.70 cm = 0.017 m

And the proton takes to cover this distance. From these data, we can find the acceleration experienced by the proton

But we know that the electric force F exerted on the proton is equal to the proton mass (m) times the acceleration (a):

and since the electric force is equal to the electric field strength, E, times the proton charge, q: F=qE, we can find E:

2) Speed of the proton as it strikes the negative plate

Since the proton starts from rest, its speed at time t is given by:

substituting the values of acceleration and time that we found in the previous part of the problem, we can calculate the speed of the proton:

before we can answer, we have to see the choices that are printed there with the question.   that's the only way we can point you to the correct one.

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right