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, 2.10 cm distant from the first, in a time Interval of 1.10 x 10-6 s. (a) Find the magnitude of the electric field N/C (b) Find the speed of the proton when it strikes the negatively charged plate m/s

Answer 1

a)

365.3 N/C

b)

3.85 x 10⁴ m/s

Step-by-step explanation:

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

d = distance between the plates = 2.10 cm = 0.021 m

t = time interval = 1.10 x 10⁻⁶ sec

E = magnitude of electric field

v₀ = initial velocity = 0 m/s

a = acceleration of the proton

Using the equation

d = v₀ t + (0.5) a t²

0.021 = (0) (1.10 x 10⁻⁶) + (0.5) a (1.10 x 10⁻⁶)²

a = 3.5 x 10¹⁰ m/s²

acceleration of the proton inside electric field is given as

`a =(qE)/(m)`

`3.5* 10^(10) =((1.6* 10^(-19))E)/(1.67* 10^(-27))`

E = 365.3 N/C

b)

v = final speed of the proton

using the equation

v = v₀ + a t

v = 0 + (3.5 x 10¹⁰) (1.10 x 10⁻⁶)

v = 3.85 x 10⁴ m/s