Answer :
The acceleration of the proton by using Coulomb's law is 2.806 m/s²
The acceleration of a proton can be defined by using coulombs force and centripetal acceleration. The value of coulomb force is
Fc = k.q1.q2 / d²
where Fc is coulomb force, k is Coulomb's constant, q is charge and d is distance.
Centrifugal force will also same as coulomb force, which can be determined as
F = m.a
where F is centrifugal force, m is mass and a is acceleration.
From the question above, we know that:
q1 = q2 = e = 1.602176634×10‾¹⁹ C
k = 8.988×10⁹ N⋅m²⋅C‾²
m = 1.673 x 10‾²⁴ kg
d = 7 x 10‾³ m
By combining these equations, we get
F = Fc
m.a = k.q1.q2 / d²
a = k.e² / m.d²
Substitute the parameter
a = 8.988×10⁹ . (1.6×10‾¹⁹)² / ( 1.673 x 10‾²⁴ . ( 7 x 10‾³ )² )
a = 2.806 m/s²
Hence, the acceleration of the proton by using Coulomb's law is 2.806 m/s²
Find more on Coulomb's law at: https://brainly.com/question/66110
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