A charged oil drop is suspended in uniform field of $3 imes 10^{4}{\ Vm}^{- 1}$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $= 9.9 imes 10^{- 15}\ Kg\ and\ g = 10\ {ms}^{- 2})$

Question

A charged oil drop is suspended in uniform field of $3 	imes 10^{4}{\ Vm}^{- 1}$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $= 9.9 	imes 10^{- 15}\ Kg\ and\ g = 10\ {ms}^{- 2})$

Dr. Abubacker · Accepted Answer

(a) In steady state,
[image:11_image44.emf]
electric force on drop =weight of drop
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ qE = mg$
$\Rightarrow$ $q\  = \frac{mg}{E}$
=$\frac{9.9 	imes 10^{- 15} 	imes 10}{3 	imes 10^{4}}$
=$3.3 	imes 10^{- 18}$ C

Q. A charged oil drop is suspended in uniform field of $3 \times 10^{4}{\ Vm}^{- 1}$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $= 9.9 \times 10^{- 15}\ Kg\ and\ g = 10\ {ms}^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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