An electron is released from rest in a uniform electric field and accelerates vertically upward. Acceleration of e is noticed to have a magnitude of \( 1.6 \times 10^{12}\,\mathrm{m/s^2} \). The direction and magnitude of electric field is :
Given \( e = -1.6 \times 10^{-19}\,\mathrm{C} \)
mass of e \( = 9.10 \times 10^{-31}\,\mathrm{kg} \)
Magnitude of gravitational acceleration \( g = 10\,\mathrm{m/s^2} \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9.10 N/C in downward direction
The electric force \( |e|E \) on the electron must give it the upward acceleration. By Newton's second law, with upward taken as positive,
\( \displaystyle |e|E - m g = m a \;\Rightarrow\; E = \frac{m(a + g)}{|e|} \)
Here \( g = 10\,\mathrm{m\,s^{-2}} \) is negligible beside \( a = 1.6\times10^{12}\,\mathrm{m\,s^{-2}} \), so \( a + g \approx a \).
\( \displaystyle E = \frac{9.10\times10^{-31} \times 1.6\times10^{12}}{1.6\times10^{-19}} = 9.10\,\mathrm{N\,C^{-1}} \)
The force on the electron is upward. Since the electron is negatively charged, the force on it is opposite to the field, so the field points downward.