Q103 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
A small particle carrying a negative charge of $1.6 \times 10^{-19} \mathrm{C}$ is suspended in equilibrium between two horizontal metal plates 8 cm apart having a potential difference of 980 V across them. The mass of the particle is $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right]$
- Correct answerA$2 \times 10^{-16} \mathrm{~kg}$
- B$2.2 \times 10^{-16} \mathrm{~kg}$
- C$20 \times 10^{-16} \mathrm{~kg}$
- D$4 \times 10^{-16} \mathrm{~kg}$
Solution
Magnitude of the force due to the electric field
$\mathrm{F}=\mathrm{qE}=\frac{\mathrm{qV}}{\mathrm{~d}}$
This should be equal to the weight of the particle $\begin{aligned} & \therefore \quad \mathrm{mg}=\frac{\mathrm{qV}}{\mathrm{~d}} \\ & \begin{aligned} \therefore \quad \mathrm{m} & =\frac{\mathrm{qV}}{\mathrm{gd}}=\frac{1.6 \times 10^{-19} \times 980}{9.8 \times 10^{-2} \times 8} \\ & =2 \times 10^{-16} \mathrm{~kg} \end{aligned} \end{aligned}$
This should be equal to the weight of the particle $\begin{aligned} & \therefore \quad \mathrm{mg}=\frac{\mathrm{qV}}{\mathrm{~d}} \\ & \begin{aligned} \therefore \quad \mathrm{m} & =\frac{\mathrm{qV}}{\mathrm{gd}}=\frac{1.6 \times 10^{-19} \times 980}{9.8 \times 10^{-2} \times 8} \\ & =2 \times 10^{-16} \mathrm{~kg} \end{aligned} \end{aligned}$
Hence, the correct answer is option: 'A'.