MHT CET · Physics · Electrostatics
An electron of mass ' \(m\) ' and charge ' \(q\) ' is accelerated from rest in a uniform electric field of strength ' \(E\) '. The velocity acquired by the electron when it travels a distance ' \(L\) ' is
- A \(\sqrt{\frac{2 q E}{m L}}\)
- B \(\sqrt{\frac{2 E m}{q L}}\)
- C \(\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}\)
- D \(\sqrt{\frac{\mathrm{qE}}{\mathrm{mL}}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}\)
Step-by-step Solution
Detailed explanation
Force on the electron \(\mathrm{F}=\mathrm{qE}\)
Work done by the force \(\mathrm{W}=\mathrm{qEL}\)
Work done is equal to gain in kinetic energy
\(
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{qEL} \\
& \therefore \mathrm{v}=\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}
\end{aligned}
\)
Work done by the force \(\mathrm{W}=\mathrm{qEL}\)
Work done is equal to gain in kinetic energy
\(
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{qEL} \\
& \therefore \mathrm{v}=\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}
\end{aligned}
\)
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