If `g_E` and `g_M` are the acceleration due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio
electronic charge on the moon/electronic charge on the earth to be

Options

A. `1`
B. `0`
C. `g_(E)//g_(M)`
D. `g_(M)//g_(E)`

Answer

To solve the problem, we need to determine the ratio of the electronic charge on the Moon to the electronic charge on the Earth based on the context of Millikan's oil drop experiment.

### Step-by-Step Solution:

1. **Understanding the Experiment**: Millikan's oil drop experiment is designed to measure the charge of the electron. The experiment involves balancing the gravitational force acting on tiny oil droplets with an electric field.

2. **Recognizing the Role of Gravity**: The acceleration due to gravity affects the weight of the oil droplets, which in turn influences the electric field strength required to balance the droplet. However, the charge on the droplets themselves is an intrinsic property of the electrons.

3. **Charge Conservation**: According to the principle of charge conservation, the electronic charge is a fundamental constant of nature. This means that the charge of an electron does not change based on the location (whether on Earth or on the Moon).

4. **Setting Up the Ratio**: Let \( e_M \) be the electronic charge on the Moon and \( e_E \) be the electronic charge on the Earth. Since the charge is constant regardless of location, we can express this as:
\[ \frac{e_M}{e_E} = 1 \]

5. **Conclusion**: The ratio of the electronic charge on the Moon to the electronic charge on the Earth is equal to 1. This means that the electronic charge remains the same irrespective of the gravitational conditions of the two celestial bodies.

### Final Answer:
\[ \frac{e_M}{e_E} = 1 \]