A small oil drop of mass 10^{-6} ,kg is hangi | vedclass

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(B) Given that the oil drop is at rest,the gravitational force acting downwards must be balanced by the electric force acting upwards.$qE = mg$Since t

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$2 	imes 10^{-11} \,C$

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(B) Given that the oil drop is at rest,the gravitational force acting downwards must be balanced by the electric force acting upwards.$qE = mg$Since the electric field $E$ between two plates is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates:$q \left(\frac{V}{d}ight) = mg$Rearranging to solve for the charge $q$:$q = \frac{mgd}{V}$Given values:$m = 10^{-6} \,kg$$g = 10 \,ms^{-2}$$d = 1 \,mm = 10^{-3} \,m$$V = 500 \,V$Substituting these values into the equation:$q = \frac{10^{-6} 	imes 10 	imes 10^{-3}}{500}$$q = \frac{10^{-8}}{500} = \frac{10^{-8}}{5 	imes 10^2} = 0.2 	imes 10^{-10} \,C = 2 	imes 10^{-11} \,C$

$A$ small oil drop of mass $10^{-6} \,kg$ is hanging at rest between two plates separated by $1 \,mm$ having a potential difference of $500 \,V$. The charge on the drop is $(g = 10 \,ms^{-2})$

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