MHT CET · Physics · Electrostatics
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 \(\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right]\)
- A \(2 \times 10^{-16} \mathrm{~kg}\)
- B \(2.2 \times 10^{-16} \mathrm{~kg}\)
- C \(20 \times 10^{-16} \mathrm{~kg}\)
- D \(4 \times 10^{-16} \mathrm{~kg}\)
Answer & Solution
Correct Answer
(A) \(2 \times 10^{-16} \mathrm{~kg}\)
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
& \therefore \quad \mathrm{mg}=\frac{\mathrm{qV}}{\mathrm{~d}} \\
& \begin{aligned}
\therefore \quad \mathrm{m} & =\frac{\mathrm{qV}}{\mathrm{gd}}=\frac{1.6 \times 10^{-19} \times 980}{9.8 \times 10^{-2} \times 8} \\
& =2 \times 10^{-16} \mathrm{~kg}
\end{aligned}
\end{aligned}\)
\(\mathrm{F}=\mathrm{qE}=\frac{\mathrm{qV}}{\mathrm{~d}}\)
This should be equal to the weight of the particle
\(\begin{aligned}
& \therefore \quad \mathrm{mg}=\frac{\mathrm{qV}}{\mathrm{~d}} \\
& \begin{aligned}
\therefore \quad \mathrm{m} & =\frac{\mathrm{qV}}{\mathrm{gd}}=\frac{1.6 \times 10^{-19} \times 980}{9.8 \times 10^{-2} \times 8} \\
& =2 \times 10^{-16} \mathrm{~kg}
\end{aligned}
\end{aligned}\)
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