An electron having charge e and mass $\mathrm{m}$ starts from the lower plate of two metallic plates…

An electron having charge e and mass $\mathrm{m}$ starts from the lower plate of two metallic plates separated by a distance $\mathrm{d}$. If the potential difference between the plates is $\mathrm{V}$, the time taken by the electron to reach the upper plate is given by
  1. $\sqrt{\frac{2 \mathrm{md}^{2}}{\mathrm{eV}}}$
  2. $\sqrt{\frac{\mathrm{md}^{2}}{\mathrm{eV}}}$
  3. $\sqrt{\frac{\mathrm{md}^{2}}{2 \mathrm{eV}}}$
  4. $\frac{2 \mathrm{md}^{2}}{\mathrm{eV}}$

Solution

$\mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}}, \mathrm{F}=\mathrm{eE}=\mathrm{eV} / \mathrm{d}$
$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{eV}}{\mathrm{md}}$
$\mathrm{d}=\frac{1}{2} \mathrm{at}^{2}$ or $\mathrm{t}=\sqrt{\frac{2 \mathrm{~d}}{\mathrm{a}}}$
or $\mathrm{t}=\sqrt{\frac{2 \mathrm{~d} \mathrm{md}}{\mathrm{eV}}}=\sqrt{\frac{2 \mathrm{md}^{2}}{\mathrm{eV}}}$

Asked in: JEE Mains - Electrostatics - Test 4

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