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A charged oil drop is suspended in a uniform field of 3 x 10 4 v/m so that it neither falls nor rises. The charge on the drop will (Take the mass of the charge = 9.9 x 10 -15 kg and g = 10 m/s 2 ) (a) 1.6 x 10 -18 C (b) 3.2 x 10- 18 C (c) 3.3 x 10 -18 C (d) 4.8 x 10 -18 C

A charged oil drop is suspended in a uniform field of 3 x 104  v/m so that it neither falls nor rises. The charge on the drop will (Take the mass of the charge  = 9.9 x 10-15 kg  and g = 10 m/s2  )
 
(a) 1.6 x 10-18 C
(b) 3.2 x 10-18 C
(c) 3.3 x 10-18 C
(d) 4.8 x 10-18 C

Grade:6

2 Answers

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago

(c) At equilibrium, electric force on drop balances weight of drop.

qE = mg ? q = mg /E

q = 9.9 x 10-15 x 10 / 3 x 104 = 3.3 x 10-18 C
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

At equilibrium, electric force on drop balances weight of drop.
qE = mg ? q = mg /E
q = 9.9 x 10^-15 x 10 / 3 x 10^4
= 3.3 x 10^-18 C

Thanks and Regards

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