Question

Calculate the acceleration of a helium nucleus (charge +2e, where e-electronic charge), when it enters an electric field of strength 790 N/C. Write your answer in 10 m/s2, i.e. divide your answer by 109, and write what you get. (hint: use force from F = E*q, and then Newton's law F = ma to get acceleration)

Answer 1

Answer:

3.8 x 10^10 m/s^2

Explanation:

Charge, q = 2 e =  2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

Electric field strength, E = 790 N/C

mass of helium nucleus, m = 6.645 x 10^-27 Kg

the force due to electric filed on a charge particle is given by

F = q x E

Where, q be the charge on the charged particle and E be the strength of electric field.

By substituting the values

F = 3.2 x 10^-19 x 790

F = 2528 x 10^-19 N

According to the Newton's second law

F = m x a

Where, me be the mass and a be the acceleration

By substituting the values

$2528\times 10^{-19}=6.645\times 10^{-27}\times a$

a = 3.8 x 10^10 m/s^2