A charged oil drop weighing `1.6 xx 10^(-15) N` is found to remain suspended in a uniform electric field of intensity `2 xx 10^3NC^(-1)`. Find the charge on the drop. 

Explanation

To solve the problem of finding the charge on a suspended oil drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Oil Drop**:
- The oil drop is suspended in an electric field, which means the electric force acting on it is equal to the gravitational force (weight) acting on it.
- The weight of the oil drop is given as \( W = 1.6 \times 10^{-15} \, \text{N} \).
- The intensity of the electric field is given as \( E = 2 \times 10^{3} \, \text{N/C} \).
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