Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density are released through a tiny hole at the center of the top disc. [2020]

Question

Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density $900 {kg m}^{- 3}$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of 200 V across the discs. As a result, an oil drop of radius $8 \times 10^{- 7} m$ stops moving vertically and floats between the discs. The number of electrons present in this oil drop is _____. (Neglect the buoyancy force, take acceleration due to gravity $= 10 {ms}^{- 2}$ and charge on an electron $(e) = 1.6 \times 10^{- 19} C$ ) [2020]

Smart Achievers · Accepted Answer

(6)

Let $n$ number of electrons present in the oil drop.

Electric field, $E = \frac{V}{d} = \frac{200}{0.01} = 2 \times 10^{4} V/m$

When terminal velocity is achieved,

$q E = m g$

As $q = n e$ and $m = \frac{4}{3} π R^{3} ρ,$

therefore, $n (1.6 \times 10^{- 19}) (2 \times 10^{4}) = \frac{4 π}{3} {(8 \times 10^{- 7})}^{3} (900) (10)$

$\Rightarrow n = \frac{4}{3} π \times \frac{{(8 \times 10^{- 7})}^{3} (900) (10)}{(2 \times 10^{4}) (1.6 \times 10^{- 19})}$

$∴$ $n = 6$

Solution

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Solution

Ans. (6) Let n number of electrons present in the oil drop. Electric field, E=Vd=2000.01=2×104 V/m When terminal velocity is achieved, qE=mg As q=ne and m=43πR3ρ, therefore, n(1.6×10-19)(2×104)=4π3(8×10-7)3(900)(10) ⇒n=43π×(8×10-7)3(900)(10)(2×104)(1.6×10-19) ∴ n=6

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