An oil drop having charge 2 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 / m 3, the radius of the drop will be

Question

An oil drop having charge 2 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 / m 3, the radius of the drop will be (A) 2.0 × 10-6 m (B) 1.7 × 10-6 m (C) 1.4 × 10-6 m (D) 1.1 × 10-6 m

Tardigrade Admin · Accepted Answer

In equilibrium, Q E=m g ⇒ Q ⋅ (V/d)=m g=((4/3) π r3 ρ) g ⇒ 2 × 1.6 × 10-19 × (12000/2 × 10-2) =(4/3) π r3 × 900 × 10 ⇒ r=1.7 × 10-6 m

Q. An oil drop having charge $2 e$ is kept stationary between two parallel horizontal plates $2.0 c m$ apart when a potential difference of $12000$ volts is applied between them. If the density of oil is $900 k g / m^{3}$ , the radius of the drop will be

$2.0 \times 1 0^{- 6} m$

$1.7 \times 1 0^{- 6} m$

$1.4 \times 1 0^{- 6} m$

$1.1 \times 1 0^{- 6} m$

In equilibrium, $QE = m g$
$\Rightarrow Q \cdot \frac{V}{d} = m g = (\frac{4}{3} π r^{3} ρ) g$
$\Rightarrow 2 \times 1.6 \times 1 0^{- 19} \times \frac{12000}{2 \times 1 0 ^{- 2}}$
$= \frac{4}{3} π r^{3} \times 900 \times 10$
$\Rightarrow r = 1.7 \times 1 0^{- 6} m$

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

2.0×10−6m

1.7×10−6m

1.4×10−6m

1.1×10−6m