Auniform 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

a)

365.3 N/C

b)

3.85 x 10⁴ m/s

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

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

we need answer choices

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr