A stationary particle of charge 2.0 C and mass 0.01 kg is subjected to a uniform electric field of magnitude 103 V/m at time t = 0. Its velocity at t = 5.0 s will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 106 m/s
The field exerts a constant force qE on the particle, so it moves with uniform acceleration from rest (its weight, 0.1 N, is negligible beside the electric force of 2000 N). By Newton’s second law:
\( \displaystyle a = \frac{qE}{m} = \frac{2.0\,\mathrm{C} \times 10^{3}\,\mathrm{V\,m^{-1}}}{0.01\,\mathrm{kg}} \)
\( a = 2\times10^{5}\,\mathrm{m\,s^{-2}} \)
\( v = at = 2\times10^{5}\,\mathrm{m\,s^{-2}} \times 5.0\,\mathrm{s} = 10^{6}\,\mathrm{m\,s^{-1}} \)