A uniform electric field exists in the region between two oppositely charged parallel plates 1.70 cm 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 of 1.54*10-6 s.

a. Find the magnitude of the electric field.

b. Find the speed of the proton at the moment it strikes the negatively charged plate.

Given Information:

distance between parallel plates = r = 1.70 cm = 0.0170

time = t = 1.54*10⁻⁶ s

Required Information:

a) Electric field = E = ?

b) Speed = v = ?

Answer:

a) Electric field = 149.44 N/C

b) speed = 22076.67 m/s

Explanation:

Part (a)

The relation between electric field, time and distance is given by

E = 2mr/qt ²

E = 2*1.67x10⁻²⁷*0.0170*/1.602x10⁻¹⁹ * (1.54*10⁻⁶) ²

E = 149.44 N/C

Part (b)

From the kinematics,

v = v₀ + at

From Newton second law

F = ma

also we have,

F = qE

qE = ma

a = qE/m

v = v₀ + (qE/m) t

Where v₀ is zero

v = (1.602x10⁻¹⁹*149.44/1.67x10⁻²⁷) * 1.54*10⁻⁶

v = 22076.67 m/s