The magnitude of an electric field which can just suspend a deuteron of mass $3.2 \times 10^{-27}…
The magnitude of an electric field which can just suspend a deuteron of mass $3.2 \times 10^{-27} \mathrm{~kg}$ freely in air is
- $19.6 \times 10^{-8} \mathrm{NC}^{-1}$
- $196 \mathrm{NC}^{-1}$
- $1.96 \times 10^{-10} \mathrm{NC}^{-1}$
- $0.196 \mathrm{NC}^{-1}$
Solution
$\mathrm{m}=3.2 \times 10^{-27} \mathrm{~kg}, \mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
At equilibrium, $m g=qE$
$\Rightarrow \mathrm{E}=\frac{\mathrm{mg}}{\mathrm{q}}=\frac{3.2 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}}=19.6 \times 10^{-8} \mathrm{NC}^{-1}$
Asked in: AP EAMCET 2024 (22 May Shift 2)
Practice more Electrostatics questions on Aicharya