What magnitude of electric field will balance the weight of a deuteron of mass \( 3.2\times10^{-27}\,\mathrm{kg} \) and charge \( 1.6\times10^{-19}\,\mathrm{C} \)? (Take \( g=10\,\mathrm{m\,s^{-2}} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\times10^{-7}\,\mathrm{N\,C^{-1}} \)
The deuteron stays balanced when the electric force \( qE \), acting upward, equals its weight \( mg \).
\( qE=mg \)
\( \begin{aligned} E &= \frac{mg}{q} \\ &= \frac{3.2\times10^{-27}\,\mathrm{kg}\times10\,\mathrm{m\,s^{-2}}}{1.6\times10^{-19}\,\mathrm{C}} \\ &= \frac{3.2\times10^{-26}\,\mathrm{N}}{1.6\times10^{-19}\,\mathrm{C}} \\ &= 2\times10^{-7}\,\mathrm{N\,C^{-1}} \end{aligned} \)