Question

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in millikan Oil Drop experiment a charged drop of mass 1.8 ×10^-14 kg is stationary between its plates. the distance between its plate is 0.90 cm and potential difference is 2 kV the number of electron on the Drop is????

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Answer 1

Formula used:

$\boxed{\sf{QE = mg}}$

By using this formula, we get:

$Q = \frac{mg}{E}$

$n = \frac{mgd}{Ve}$

$n = \frac{1.8 \times 10 ^{ - 14} \times 10 \times 0.9 \times {10}^{ - 2} }{2 \times {10}^{3} \times 1.6 \times {10}^{ - 19} }$

$n = \frac{16.2 \times {10}^{ - 16} }{3.2 \times {10}^{ - 16} }$

$n = \frac{16.2}{3.2}$

$n = 5.0625$

Therefore:

Final answer: 5 approx.

Answer 2

Answer in the attachment