An oil drop having a mass of 4.8 × 10-10 g and charge of 30 × 10-18 C stands still between two charged horizontal plates separated by a distance of 1 cm. If now polarity of the plates is changed, instantaneous acceleration of the drop is ( g =10 m / s 2)

Question

An oil drop having a mass of 4.8 × 10-10 g and charge of 30 × 10-18 C stands still between two charged horizontal plates separated by a distance of 1 cm. If now polarity of the plates is changed, instantaneous acceleration of the drop is ( g =10 m / s 2) (A) 10 m / s 2 (B) 15 m / s 2 (C) 25 m / s 2 (D) 20 m / s 2

Tardigrade Admin · Accepted Answer

Case I: when oil drop is in equilibrium

∴ mg = qE case II: when the polarity are reversed

mg + qE = ma mg + mg = ma 2 mg = ma 2 g = a 2 × 10 = a a = 20m/s2

Q. An oil drop having a mass of $4.8 \times 1 0^{- 10} g$ and charge of $30 \times 1 0^{- 18} C$ stands still between two charged horizontal plates separated by a distance of $1 c m$ . If now polarity of the plates is changed, instantaneous acceleration of the drop is $(g = 10 m / s^{2})$

$10 m / s^{2}$

$15 m / s^{2}$

$25 m / s^{2}$

$20 m / s^{2}$

Case I : when oil drop is in equilibrium

$∴ m g = qE$
case II : when the polarity are reversed

$m g + qE = ma$
$m g + m g = ma$
$2 m g = ma$
$2 g = a$
$2 \times 10 = a$
$a = 20 m / s^{2}$

Q. An oil drop having a mass of 4.8×10−10g and charge of 30×10−18C stands still between two charged horizontal plates separated by a distance of 1cm. If now polarity of the plates is changed, instantaneous acceleration of the drop is (g=10m/s2)

10m/s2

15m/s2

25m/s2

20m/s2