Question

A uniform electric field exists in the region between two oppositely charged parallel plates 1.70cmapart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.45

Answer 1

Well according to the laws of electrostatics, electric field exerts a force on every charged body. The
magnitude of the force can be calculated by the formula:-
F = q * E, where
F = Force,
q = Charge, and
E = Magnitude of electric field, and according to laws of Newtonian Mechanics, force (F) can be written as:-
F = m * a, where
m = Mass of body, and
a = Accelaration of body.
So we can deduce that
q * E = m* a
=> So accelaration (a) is given by:-
a = (q * E)/m
Now question says the proton moves from one plate to other i.e. covers a distance equal to the separation between the plates (s) in a given time (t).
Since the electric field is constant, the accelaration will also be so. Hence, we can calculate the value of accelaration by applying the second equation of motion:-
s = ut + {a(t^2)}/2, where
u = Initial velocity, which is zero in this case, since the proton starts moving from rest. So solve the equation:-
a = (2 * s) / (t^2), and once you obtain the value of "a", substitute this value in the previous equation i.e.
a = q * E/m
=> E = m * a/q
Solve it to get the value of Electric Field and always remember the direction of Electric Field is from positive to negative charge