Question

A uniform 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.

Answer 1

consider the motion of the proton between the plates

d = distance traveled by the proton = distance between the two plates = 1.70 cm = 0.0170 m

v₀ = initial velocity of the proton = 0 m/s

t = time taken to reach the opposite plate = 1.48 x 10⁻⁶ sec

a = acceleration of the proton

using the kinematics equation

d = v₀ t + (0.5) a t²

inserting the values

0.017 = 0 ( 1.48 x 10⁻⁶) + (0.5) a ( 1.48 x 10⁻⁶)²

a = 1.6 x 10¹⁰ m/s²

m = mass of proton = 1.67 x 10⁻²⁷ kg

q = charge on the proton = 1.6 x 10⁻¹⁹ C

E = electric field between the plates

force on the proton due to electric field is given as

F = ma

electric force is given as

F = qE

combining the two equations we get

qE = ma

(1.6 x 10⁻¹⁹) E = (1.67 x 10⁻²⁷) (1.6 x 10¹⁰)

E = 167 N/C

v = final speed of electron as it strikes the negative plate

Using the kinematics equation

v = v₀ + at

v = 0 + (1.6 x 10¹⁰) (1.48 x 10⁻⁶)

v = 23680 m/s