MHT CET · Physics · Electrostatics
An electron of mass ' \(m\) ' and charge ' \(q\) ' is accelerated from rest in a uniform electric field of intensity ' \(E\) '. The velocity acquired by it as it travels a distance ' \(l\) ' is ' \(v\) '. The ratio \(\frac{\mathrm{q}}{\mathrm{m}}\) in terms of \(\mathrm{E}, l\) and v is
- A \(\frac{\mathrm{v}^2}{2 \mathrm{E} l}\)
- B \(\frac{\mathrm{v}^2 l}{2 \mathrm{E}}\)
- C \(\frac{2 \mathrm{E}}{\mathrm{v}^2 l}\)
- D \(\frac{\mathrm{v}^2 l}{\mathrm{E}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{v}^2}{2 \mathrm{E} l}\)
Step-by-step Solution
Detailed explanation
We know
\(\begin{aligned}
& F=m a \text { and } F=q E \\
& \Rightarrow q E=m a \\
& \therefore \frac{\mathrm{qE}}{\mathrm{~m}}=\mathrm{a} \quad ...(i)
\end{aligned}\)
According to equation of motion, \(v^2-u^2=2 a L\)
\(\begin{aligned}
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& a=\frac{v^2}{2 L} \quad ...(ii)
\end{aligned}\)
From (i) and (ii),
\(\begin{aligned}
& \frac{\mathrm{v}^2}{2 \mathrm{~L}}=\frac{\mathrm{qE}}{\mathrm{~m}} \\
& \frac{\mathrm{q}}{\mathrm{~m}}=\frac{\mathrm{v}^2}{2 \mathrm{EL}}=\frac{\mathrm{v}^2}{2 \mathrm{E} \ell} \quad(\because \mathrm{~L}=\ell)
\end{aligned}\)
\(\begin{aligned}
& F=m a \text { and } F=q E \\
& \Rightarrow q E=m a \\
& \therefore \frac{\mathrm{qE}}{\mathrm{~m}}=\mathrm{a} \quad ...(i)
\end{aligned}\)
According to equation of motion, \(v^2-u^2=2 a L\)
\(\begin{aligned}
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& a=\frac{v^2}{2 L} \quad ...(ii)
\end{aligned}\)
From (i) and (ii),
\(\begin{aligned}
& \frac{\mathrm{v}^2}{2 \mathrm{~L}}=\frac{\mathrm{qE}}{\mathrm{~m}} \\
& \frac{\mathrm{q}}{\mathrm{~m}}=\frac{\mathrm{v}^2}{2 \mathrm{EL}}=\frac{\mathrm{v}^2}{2 \mathrm{E} \ell} \quad(\because \mathrm{~L}=\ell)
\end{aligned}\)
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