an oil drop having two electrons in excess remains stationary suspended in an electric filed of intensity 10^5 N/C .Find the weight of oil drop.
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For drop to be stationary
Mg=Eq
So m=Eq/g
=10^-5*2*1.6*10^—19/10
=3.2*10^-25kg
Mg=Eq
So m=Eq/g
=10^-5*2*1.6*10^—19/10
=3.2*10^-25kg
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