A small particle carrying a negative charge of $1.6 \times 10^{- 19} C$ is suspended in equilibrium, between the horizontal metal plates 5 cm apart, having a potential difference of 3000V across them. Then the mass of the particle is $m \times 10^{- 17} kg$ . The $m / 12$ is... [Take $g = 10 {ms}^{- 2}$ ]

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Detailed Solution

$E = \frac{- dv}{dr} = \frac{- 3000}{5 \times 10^{- 2}} = - 6 \times 10^{4} {Vm}^{- 1}$
The charged particles is in equilibrium.
$\begin{array}{l} ∴ F = mg = qE \Rightarrow m = \frac{qE}{g} = \frac{(- 1.6 \times 10^{- 19}) (- 6 \times 10^{4})}{10} \\ ∴ m = 9.6 \times 10^{- 16} = 96 \times 10^{- 17} kg \end{array}$