MHT CET · Physics · Current Electricity
An electron of mass ' \(\mathrm{m}\) ' and charge ' \(\mathrm{q}\) ' is accelerated from rest in a uniform electric field of strength ' \(E\) '. The velocity acquired by the electron, when it travels a distance ' \(\mathrm{L}\) ', is
- A \(\sqrt{\frac{2 q E}{m L}}\)
- B \(\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}\)
- C \(\sqrt{\frac{2 \mathrm{Em}}{\mathrm{qL}}}\)
- D \(\sqrt{\frac{\mathrm{qE}}{\mathrm{mL}}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}\)
Step-by-step Solution
Detailed explanation
We know
\(\mathrm{F}=\mathrm{ma}\) and \(\mathrm{F}=\mathrm{qE}\)
\(\Rightarrow q E=m a\)
\(\therefore \quad \frac{\mathrm{qE}}{\mathrm{m}}=\mathrm{a}\)
According to equation of motion,
\(\begin{aligned}
& v^2-u^2=2 a L \\
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& v=\sqrt{2 a L} \\
\therefore \quad & v=\sqrt{\frac{2 q E L}{m}}
\end{aligned}\)
\(\mathrm{F}=\mathrm{ma}\) and \(\mathrm{F}=\mathrm{qE}\)
\(\Rightarrow q E=m a\)
\(\therefore \quad \frac{\mathrm{qE}}{\mathrm{m}}=\mathrm{a}\)
According to equation of motion,
\(\begin{aligned}
& v^2-u^2=2 a L \\
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& v=\sqrt{2 a L} \\
\therefore \quad & v=\sqrt{\frac{2 q E L}{m}}
\end{aligned}\)
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