AP EAMCET · PHYSICS · Electrostatics
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
- A \(19.6 \times 10^{-8} \mathrm{NC}^{-1}\)
- B \(196 \mathrm{NC}^{-1}\)
- C \(1.96 \times 10^{-10} \mathrm{NC}^{-1}\)
- D \(0.196 \mathrm{NC}^{-1}\)
Answer & Solution
Correct Answer
(A) \(19.6 \times 10^{-8} \mathrm{NC}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m}=3.2 \times 10^{-27} \mathrm{~kg}, \mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) At equilibrium, \(m g=qE\)…
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