TS EAMCET · Physics · Capacitance
A oil drop having a mass \(4.8 \times 10^{-10} \mathrm{~g}\) and charge \(2.4 \times 10^{-18} \mathrm{C}\) stands still between two charged horizontal plates separated by a distance of \(1 \mathrm{~cm}\). If now the polarity of the plates is changed, instantaneous acceleration of the drop is : \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(5 \mathrm{~ms}^{-2}\)
- B \(10 \mathrm{~ms}^{-2}\)
- C \(15 \mathrm{~ms}^{-2}\)
- D \(20 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(B) \(10 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
\(a=\frac{F}{m}=\frac{q E}{m}=\frac{m g}{m}\) \(g=10 \mathrm{~m} / \mathrm{s}^2\)
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