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Q.

A charged oil drop of mass 2.5 x 10^-4 gm is in space between the two plates each of area 200 cm² of parallel plate capacitor. When the upper plate has a charge 5 x10^-7 C and the lower plate has an equal negative charge, the drop remains stationary. The charge of oil drop is

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a

$9 \times 10^{- 1} C$

b

$9 \times 10^{- 6} C$

c

$9 \times 10^{- 13} C$

d

$1 .8 \times 10^{- 14} C$

answer is C.

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Detailed Solution

For the charged oil drop to be stationary q E = mg
$∴ q (\frac{σ}{ε_{0}}) = mg or q (\frac{Q}{{Aε}_{0}}) = mg ∴ q [\frac{5 \times 10^{- 7}}{200 \times 10^{- 4} \times 8 .85 \times 10^{- 12}}] = (2 \cdot 5 \times 10^{- 4} \times 10^{- 3}) \times 9 \cdot 8 Solving for q, we get q \approx 9 \times 10^{- 13} C$

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