An electron is released from rest in a uniform electric field of magnitude \(9.1\times10^{2}\ \mathrm{N\,C^{-1}}\). Ignore gravity. The acceleration of the electron is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(m_e=9.1\times10^{-31}\ \mathrm{kg}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.6\times10^{14}\ \mathrm{m\,s^{-2}}\), opposite to the field
The field exerts a force of magnitude \( eE \) on the electron. By Newton’s second law, its acceleration is \( a=eE/m_e \).
\( \begin{aligned} a &= \frac{eE}{m_e} \\ &= \frac{1.6\times10^{-19}\times9.1\times10^{2}}{9.1\times10^{-31}} \\ &= 1.6\times10^{14}\ \mathrm{m\,s^{-2}} \end{aligned} \)
The electron has a negative charge. So the force on it, and hence its acceleration, points opposite to the field.
NCERT (2026-27):
Example 1.7 – “… the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is”
\( a_e=eE/m_e \)