A oil drop having a mass 4.8 × 10-10 g and charge 2.4 × 10-18 C stands still between two charged horizontal plates separated by a distance of 1 cm. If now the polarity of the plates is changed, instantaneous acceleration of the drop is: (g=10 ms -2)

Question

A oil drop having a mass 4.8 × 10-10 g and charge 2.4 × 10-18 C stands still between two charged horizontal plates separated by a distance of 1 cm. If now the polarity of the plates is changed, instantaneous acceleration of the drop is: (g=10 ms -2) (A) 5 ms -2 (B) 10 ms -2 (C) 15 ms -2 (D) 20 ms -2

Tardigrade Admin · Accepted Answer

a=(F/m)=(q E/m)=(m g/m) g=10 m / s 2

Q. A oil drop having a mass $4.8 \times 1 0^{- 10} g$ and charge $2.4 \times 1 0^{- 18} C$ stands still between two charged horizontal plates separated by a distance of $1 c m$ . If now the polarity of the plates is changed, instantaneous acceleration of the drop is : $(g = 10 m s^{- 2})$

$5 m s^{- 2}$

$10 m s^{- 2}$

$15 m s^{- 2}$

$20 m s^{- 2}$

$a = \frac{F}{m} = \frac{qE}{m} = \frac{m g}{m}$
$g = 10 m / s^{2}$

Q. A oil drop having a mass 4.8×10−10g and charge 2.4×10−18C stands still between two charged horizontal plates separated by a distance of 1cm. If now the polarity of the plates is changed, instantaneous acceleration of the drop is : (g=10ms−2)

5ms−2

10ms−2

15ms−2

20ms−2