A charged oil drop is suspended in a uniform field of 3xx10^(4) V/m so that it neither falls nor rises. The charge on drop will be (Take mass of charge =9.9xx10^(-10) 10^(-15)kg and g=10m//s^(2)).

A charged oil drop is suspended in a uniform field of 3xx10^(4) V/m so

1.	A charged oil drop is suspended in a uniform field of 3xx10^(4) V/m so that it neither falls nor rises. The charge on drop will be (Take mass of charge =9.9xx10^(-10) 10^(-15)kg and g=10m//s^(2)).
Answer» `2.3xx10^(-18)C` `3.3xx10^(-18)C` `4.8xx10^(-18)C` `1.6xx10^(-18)C` Solution :`qE=mg` `:.Q=(mg)/(E)=(9.9xx10^(-15)xx10)/(3xx10^(4))` `:.q=3.3xx10^(-18)C`

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A charged oil drop is suspended in a uniform field of 3xx10^(4) V/m so that it neither falls nor rises. The charge on drop will be (Take mass of charge =9.9xx10^(-10) 10^(-15)kg and g=10m//s^(2)).

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