An electron is released from rest in a uniform electric field accelerates vertically upward. Acceleration of e is notice to have a magnitude of $ 1.6 \times 10^{12} m/s^2$.The direction and magnitude of electric field is :

$\text{Given e} = - 1.6 \times 10^{-19} C$

$\text{mass of e} = 9.10 \times 10^{-31}kg$ 

$\text{Magnitude of gravitational are g} = 10 m/s^2$

9.10 N/C in downward direction

The correct answer is Option (4) → 9.10 N/C in downward direction

Charge on electron, $e=-1.6×10^{-19}C$

Mass of electron, $m=9.10×10^{-31}Kg$

Acceleration, $a=1.6×10^{12}m/s^2$

Using Newton Second's law,

$F=m.a$

$=9.10×10^{-31}×1.6×10^{12}$

$=1.456×10^{-18}N$

and,

$F=q.E$  [E = Electric field]

$⇒E=\frac{F}{q}=\frac{1.456×10^{-18}}{1.6×10^{-19}C}$

$=9.1N/C$

Also, Since electrons feel force opposite to E, if the electron is accelerating upward, the electric field must be downward.