# An electron is released from rest in a uniform electric of magnitude 2.00×104 N/C. Calculate the acceleration of the electron (Ignore gravitation)?

Feb 17, 2016

$a = 3.51 \times {10}^{15} \frac{m}{s} ^ 2$
This would be magnitude of the electron’s acceleration. Since the electron has a negative charge the direction of the force on the electron and of course the acceleration is opposite the direction of the electric ﬁeld.

#### Explanation:

calculate Force first
Force $F = {q}_{e} E$, where ${q}_{e} = 1.602 \times {10}^{-} 19 C$ charge of electron
$E = 2 \times {10}^{4} \frac{N}{C}$ electric field
now divide force by mass of electron to have the acceleration.
$a = \frac{F}{m}$, where $m = 9.11 \times {10}^{-} 31$ mass of electron
$a = {q}_{e} \frac{E}{m} = \frac{1.602 \times {10}^{-} 19 C \times 2.0 \times {10}^{4} \frac{N}{C}}{9.11 \times {10}^{-} 31 k g}$
$a = 3.51 \times {10}^{15} \frac{m}{s} ^ 2$
This would be magnitude of the electron’s acceleration. Since the electron has a negative charge the direction of the force on the electron and of course the acceleration is opposite the direction of the electric ﬁeld.