Q233 / 240 · KCET 2026 · Physics
An electron falls through a distance $1.5$ cm in $2.0 \times 10^4$ N/C from rest. The time taken to cover this distance in second is ____
$(e = 1.6 \times 10^{-19}\text{ C}, m_e = 9.11 \times 10^{-31}\text{ kg})$
$(e = 1.6 \times 10^{-19}\text{ C}, m_e = 9.11 \times 10^{-31}\text{ kg})$
- Correct answerA$2.9 \times 10^{-9}$
- B$2.9 \times 10^{9}$
- C$4 \times 10^{-6}$
- D$4 \times 10^{6}$
Solution
Force on the electron is $F = eE$
Acceleration of the electron is $a = \dfrac{eE}{m_e}$
Using the equation of motion $s = ut + \dfrac{1}{2}at^2$
Since the electron starts from rest, $u = 0$
$s = \dfrac{1}{2} \left( \dfrac{eE}{m_e} \right) t^2$
$t = \sqrt{\dfrac{2sm_e}{eE}}$
Substituting the given values:
$t = \sqrt{\dfrac{2 \times 1.5 \times 10^{-2} \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}}$
$t = \sqrt{\dfrac{27.33 \times 10^{-33}}{3.2 \times 10^{-15}}}$
$t = \sqrt{8.54 \times 10^{-18}}$
$t \approx 2.9 \times 10^{-9} \text{ s}$
Answer: $2.9 \times 10^{-9}$
Acceleration of the electron is $a = \dfrac{eE}{m_e}$
Using the equation of motion $s = ut + \dfrac{1}{2}at^2$
Since the electron starts from rest, $u = 0$
$s = \dfrac{1}{2} \left( \dfrac{eE}{m_e} \right) t^2$
$t = \sqrt{\dfrac{2sm_e}{eE}}$
Substituting the given values:
$t = \sqrt{\dfrac{2 \times 1.5 \times 10^{-2} \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}}$
$t = \sqrt{\dfrac{27.33 \times 10^{-33}}{3.2 \times 10^{-15}}}$
$t = \sqrt{8.54 \times 10^{-18}}$
$t \approx 2.9 \times 10^{-9} \text{ s}$
Answer: $2.9 \times 10^{-9}$
Hence, the correct answer is option: 'A'.