A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field of `100Vm^(-1)`. If the mass of the drop is `1.6xx10^(3)g` the number of electrons carried by the drop is `(g=10ms^(-2))`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the mass of the oil drop from grams to kilograms
The mass of the oil drop is given as \(1.6 \times 10^{-3} \, \text{g}\). To convert grams to kilograms, we use the conversion factor \(1 \, \text{g} = 10^{-3} \, \text{kg}\).

\[ \text{Mass in kg} = 1.6 \times 10^{-3} \, \text{g} \times 10^{-3} \, \text{kg/g} = 1.6 \times 10^{-6} \, \text{kg} \]
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