TS EAMCET · Physics · Electrostatics
If a particle of mass 10 mg and charge \(2 \mu \mathrm{C}\) at rest is subjected to a uniform electric field of potential difference 160 V, then the velocity acquired by the particle is
- A \(9 \mathrm{~ms}^{-1}\)
- B \(4 \mathrm{~ms}^{-1}\)
- C \(6 \mathrm{~ms}^{-1}\)
- D \(8 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(8 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{2} m v^2 = q \Delta V \) \( v = \sqrt{\frac{2 q \Delta V}{m}} = \sqrt{\frac{2 \times (2 \times 10^{-6} \mathrm{~C}) \times (160 \mathrm{~V})}{10 \times 10^{-6} \mathrm{~kg}}} = \sqrt{\frac{640 \times 10^{-6}}{10 \times 10^{-6}}} = \sqrt{64} = 8 \mathrm{~ms}^{-1} \)
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