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Question: An oil drop having charge \(2 e\) is kept stationary between two parallel horizontal plates 2.0 cm a...

An oil drop having charge 2e2 e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m3, the radius of the drop will be

A

2.0×10−6 m2.0 \times 10 ^ { - 6 } \mathrm {~m}

B

1.7×10−6 m1.7 \times 10 ^ { - 6 } \mathrm {~m}

C

1.4×10−6 m1.4 \times 10 ^ { - 6 } \mathrm {~m}

D

1.1×10−6 m1.1 \times 10 ^ { - 6 } \mathrm {~m}

Answer

1.7×10−6 m1.7 \times 10 ^ { - 6 } \mathrm {~m}

Explanation

Solution

In equilibrium QE = mg

⇒ Q⋅Vd=mg=(43πr3ρ)gQ \cdot \frac { V } { d } = m g = \left( \frac { 4 } { 3 } \pi r ^ { 3 } \rho \right) g

⇒ 2×1.6×10−19×120002×10−2=43πr3×900×102 \times 1.6 \times 10 ^ { - 19 } \times \frac { 12000 } { 2 \times 10 ^ { - 2 } } = \frac { 4 } { 3 } \pi r ^ { 3 } \times 900 \times 10

⇒ r = 1.7 × 10–6 m