KCET · Physics · Electrostatics
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 ____
\((e = 1.6 \times 10^{-19}\text{ C}, m_e = 9.11 \times 10^{-31}\text{ kg})\)
- A \(2.9 \times 10^{-9}\)
- B \(2.9 \times 10^{9}\)
- C \(4 \times 10^{-6}\)
- D \(4 \times 10^{6}\)
Answer & Solution
Correct Answer
(A) \(2.9 \times 10^{-9}\)
Step-by-step Solution
Detailed explanation
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: \(2.9 \times 10^{-9}\)
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