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 :
$\text{Given e} = - 1.6 \times 10^{-19} C$
$\text{mass of e} = 9.10 \times 10^{-31}kg$
$\text{Magnitude of gravitational are g} = 10 m/s^2$
Answer & explanation
Correct answer: option 4
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.