Question

An electron is accelerated by a constant electric field of magnitude 410 N/C. Find the acceleration of the electron. The electron has a charge of e = 1.6 X 10-19 C.
a. 1.82 X 108 m/s2
b. 6.56 X 1013 m/s2
c. 0 d. 2.56 X 1021 m/s2
e. 3.90 X 10-22 m/s2

Answer 1

Answer:

option B

Explanation:

given,

magnitude of electric field, E = 410 N/C

charge of electron , q = 1.6 x 10⁻¹⁹ C

we know,

F = q E

and also

F = m a

equating both the force equation

m a = q E

m is the mass of electron = 9.11 x 10⁻³¹ Kg

so,

$a = \dfrac{q E}{m}$

$a = \dfrac{1.6\times 10^{-19}\times 410}{9.11\times 10^{-31}}$

     a = 7.2 x 10¹³ m/s²

The nearest answer is option B.