An oil drop of 12 excess electrons is held stationaty under a constant electric field of `2.55xx10^(4) NC^(-1)` in Millikan's oil drop experi,ment. The density of the oil is `1.26 g cm^(-3)`. Estimate the radius of the drop. `(g = 9.81 ms^(-2) , e = 1.60xx10^(19) C)`

Explanation

To solve the problem of estimating the radius of an oil drop in Millikan's oil drop experiment, we can follow these steps:

### Step 1: Calculate the Charge on the Oil Drop
The oil drop has 12 excess electrons. The charge of one electron is given as \( e = 1.6 \times 10^{-19} \) C. Therefore, the total charge \( Q \) on the oil drop can be calculated as:

\[ Q = n \cdot e = 12 \cdot (1.6 \times 10^{-19}) = 1.92 \times 10^{-18} \text{ C} \]
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