Answer:
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
[tex]a =\frac{qE}{m}[/tex]
[tex]3.5\times 10^{10} =\frac{(1.6\times 10^{-19})E}{1.67\times 10^{-27}}[/tex]
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
