What is the magnitude of the electric field?

Question

A uniform electric field exists in the region between two oppositely charged parallel plates 1.70cm apart. 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 1.57×10−6s .Use 1.60×10−19C for the magnitude of the charge on an electron and 1.67×10−27 KG for the mass of a proton.

Answer 1

Electric field E=F/q
also since distance under acceleration can be expressed as d= 0.5 a t^2

then we have a = 2d/t^2
we know that
F=m a
So finally
E=F/q
E= ma/q
since a =2 d/(t^2) [from above]
E= 2 m d/(q t^2 )
E= 2 x 1.67E−27 x 17.0 E-3 /( 1.60 E−19 x (1.57E−6s)^2)=

E= 144 N/C