KCET · Physics · Electrostatics
In the uniform electric field of \(\mathrm{E}=1 \times 10^{4} \mathrm{NC}^{-1}\), an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of \(2 \times 10^{-2} \mathrm{~m}\) is nearly \(\left(\frac{\mathrm{e}}{\mathrm{m}} \text { of electron } \approx 1.8 \times 10^{-11} \mathrm{C} \mathrm{} \mathrm{kg}^{-1}\right)\)
- A \(8.5 \times 10^{6} \mathrm{~ms}^{-1}\)
- B \(1.6 \times 10^{6} \mathrm{~ms}^{-1}\)
- C \(0.85 \times 10^{6} \mathrm{~ms}^{-1}\)
- D \(0.425 \times 10^{6} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(8.5 \times 10^{6} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
By work-energy theorem,
Work done by electric field = Change in KE of electron
or \(\mathrm{qE} x=\frac{1}{2} \mathrm{mv}^{2}\)
or \(\quad v=\sqrt{\frac{2 \mathrm{q} E x}{m}}\)
\(=\sqrt{2 \times 1.8 \times 10^{11} \times 1 \times 10^{4} \times 2 \times 10^{-2}} \)
\(=8.5 \times 10^{6} \mathrm{~ms}^{-1}\)
Work done by electric field = Change in KE of electron
or \(\mathrm{qE} x=\frac{1}{2} \mathrm{mv}^{2}\)
or \(\quad v=\sqrt{\frac{2 \mathrm{q} E x}{m}}\)
\(=\sqrt{2 \times 1.8 \times 10^{11} \times 1 \times 10^{4} \times 2 \times 10^{-2}} \)
\(=8.5 \times 10^{6} \mathrm{~ms}^{-1}\)
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